DEGREE SEQUENCES, FORCIBLY CHORDALGRAPHS, AND COMBINATORIAL PROOF SYSTEMS
DISSERTATION
Presented in Partial Fulfillment of the Requirements for
the Degree Doctor of Philosophy in the Graduate
School of the Ohio State University
By
Christian Altomare, B.S.
Graduate Program in Mathematics
The Ohio State University2009
Dissertation Committee:
Dr. G. Neil Robertson, Advisor
Dr. John Maharry
Dr. Akos Seress
ABSTRACT
We study the structure theory of two combinatorial objects closely related to graphs.
First, we consider degree sequences, and we prove several results originally mo-
tivated by attempts to prove what was, until recently, S.B. Rao’s Conjecture, and
what is now a theorem of Paul Seymour and Maria Chudnovsky, namely, that graphic
degree sequences are well quasi ordered. We give a new, surprisingly non-graph the-
oretic proof of the bounded case of this theorem. Next, we obtain an exact structure
theorem of degree sequences excluding a square and a pentagon. Using this result,
we then prove a structure theorem for degree sequences excluding a square and, more
generally, excluding an arbitrary cycle. It should be noted that taking complements,
this yields a structure theorem for excluding a matching.
The structure theorems above, however, are stated in terms of forcibly chordal
graphs, hence we next begin their characterization. While an exact characterization
seems difficult, certain partial results are obtained. Specifically, we first characterize
the degree sequences of forcibly chordal trees. Next, we use this result to extend the
characterization to forcibly chordal forests. Finally, we characterize forcibly chordal
graphs having a certain path structure.
Next, we define a class of combinatorial objects that generalizes digraphs and
partial orders, which is motivated by proof systems arising in mathematical logic. We
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give what we believe will be the basic theory of these objects, including definitions,
theorems, and proofs. We define the minors of a proof system, and we give two
forbidden minors theorems, one of them characterizing partial orders as proof systems
by forbidden minors.
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To Moomar.
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ACKNOWLEDGMENTS
First and foremost, I wish to thank Neil Robertson, my advisor. It is every student’s
wish to have an advisor with such depth of understanding, breadth of knowledge, and
raw intuition for his field of expertise. I have gained from him not only knowledge,
but an understanding of how research mathematics is carried out. His ability to find
the right generalization to prove, the right special case to consider, the right approach
to try, and the right question to ask at all, has continually amazed me.
Second, I would like to thank S.B. Rao for a beautiful conjecture.
Third, I would like to thank Christopher McClain for his generous and patient
help related to typesetting and document preparation, which are not my strong suits.
Fourth, I wish to thank Akos Seress and John Maharry for their time and effort
participating in my thesis committee.
Fifth, I would like to thank everyone in the Ohio State University Mathematics
Department who has helped me in my time since I started taking mathematics courses
here as a high school student. In particular, in the order in which I met them, I am
thankful to John Maharry, Alexander Dynin, Judie Monson, Yung-Chen Lu, Vitaly
Bergelson, Randall Dougherty, Tim Carlson, Cindy Bernlohr, Boris Pittel, and once
again my advisor for the amount of time, effort, and patience they were willing
to spend toward my career and development. I thank the countless others in the
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department who have helped me as well. Without their help, this would not be
possible.
Sixth, I thank my parents, Richard and Karen Altomare.
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VITA
April 7, 1980 . . . . . . . . . . . . . . . . . . . . . . . . . . Born - Columbus, OH
1998-2001 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Undergraduate,
The Ohio State University
2001 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.S. in Mathematics,
The Ohio State University
2001-Present . . . . . . . . . . . . . . . . . . . . . . . . . . Graduate Teaching Associate,
The Ohio State University
FIELDS OF STUDY
Major Field: Mathematics
Specialization: Graph Theory
vii
TABLE OF CONTENTS
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
CHAPTER PAGE
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Introduction to Degree Sequences . . . . . . . . . . . . . 11.2 Degree Sequence Basics, Notation, and Conventions . . . 51.3 Introduction to Combinatorial Proof Systems . . . . . . . 9
2 The Bounded Case of Rao’s Conjecture . . . . . . . . . . . . . . . . . . 12
3 Excluding Matchings and Cycles . . . . . . . . . . . . . . . . . . . . . . 24
4 Forcibly Chordal Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5 Forcibly Chordal Forests . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6 Forcibly Chordal Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 59
7 Combinatorial Proof Systems . . . . . . . . . . . . . . . . . . . . . . . 86
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 867.2 Proof Closure . . . . . . . . . . . . . . . . . . . . . . . . 877.3 The Merge . . . . . . . . . . . . . . . . . . . . . . . . . . 887.4 Preceding Set Proof Systems . . . . . . . . . . . . . . . . 937.4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 947.4.2 Motivation for Definition of Proof . . . . . . . . . . . . . . . . 977.4.3 Proof Definition and Basics . . . . . . . . . . . . . . . . . . . 97
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7.4.4 Autonomous Sets . . . . . . . . . . . . . . . . . . . . . . . . . 997.4.5 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.4.6 The Information in the Set of Proofs . . . . . . . . . . . . . . 1067.5 Autonomous Systems . . . . . . . . . . . . . . . . . . . . 1077.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.5.2 Definition of Proof Revisited . . . . . . . . . . . . . . . . . . . 1087.6 The Canonical Orders . . . . . . . . . . . . . . . . . . . . 1097.6.1 Canonical Order Definition and Basics . . . . . . . . . . . . . 1097.6.2 Descendability . . . . . . . . . . . . . . . . . . . . . . . . . . . 1117.6.3 Canonical Order Basic Theorems And Examples . . . . . . . . 1177.7 Partial Orders As Ausyses . . . . . . . . . . . . . . . . . 1227.8 Well Founded Autonomous Systems . . . . . . . . . . . . 1307.9 Blocking . . . . . . . . . . . . . . . . . . . . . . . . . . . 1357.9.1 The Blocking Order . . . . . . . . . . . . . . . . . . . . . . . . 1437.9.2 Blocking In Posets . . . . . . . . . . . . . . . . . . . . . . . . 1457.10 Ausys Lexicographic Sum . . . . . . . . . . . . . . . . . . 1477.11 Subausys, Dot, Homomorphisms, and Minors . . . . . . . 1537.12 Relations to Matching and Connectivity . . . . . . . . . . 166
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
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CHAPTER 1
INTRODUCTION
This work studies two classes of objects. The first class we study is the class of degree
sequences of finite graphs. The second class we study is a class of combinatorial proof
systems we call autonomous systems.
1.1 Introduction to Degree Sequences
We assume familiarity with basic graph theory. Definitions and conventions are as in
[3] unless otherwise stated. Our graphs are finite, simple, and undirected throughout
unless otherwise stated.
Definition 1.1.1. Let G be a graph with vertices v1, . . . , vn, listed such that d(v1) ≥
· · · ≥ d(vn). Then the degree sequence of G, denoted by D(G), is the sequence
(d(v1), . . . , d(vn)).
We make no use of the fact that, according to our definition, the degree sequence is a
decreasing sequence. It is rather simply the easiest way to make the degree sequence
of a graph unique, so we can refer to the degree sequence D(G) of G, as opposed to
a degree sequence of G.
We note that while a degree sequence does not technically have any vertices, it can be
1
very suggestive to think of the vertices of a degree sequence, which we sometimes do.
The degree sequence (2, 2, 2, 1, 1), for instance, would be said to have three vertices
of degree 2 and two of degree 1.
Definition 1.1.2. Let D be a degree sequence and let G be a graph. We say that G
realizes D, or that G is a realization of D, if D(G) = D. We denote by R(D) the set
of realizations of D.
Myriad theorems in combinatorics, and in particular graph theory, study the graphs
not “containing” a fixed graph, for various notions of containment. It is fruitful to
define a notion of containment for degree sequences as well so that similar questions
may be asked and theorems proved.
Definition 1.1.3. Let D1 and D2 be degree sequences. We write D1 ≤ D2 if there is
a graph G1 in R(D1) and a graph G2 in R(D2) such that G1 is an induced subgraph
of G2.
The reader may check that ≤ is a reflexive, transitive relation. One motivation
for making this definition is that the induced subgraph relation for graphs can be
extremely difficult to work with, even for questions that are tractable if the induced
subgraph relation is replaced with another containment relation. The relation ≤ for
degree sequences is similar to, but more tractable in many cases than, the induced
subgraph relation for graphs.
2
A discussion of claw free graphs and degree sequences best illustrates this point. A
claw is the unique graph up to isomorphism with degree sequence (3, 1, 1, 1). Suppose
we wish to find the structure of claw free graphs. What claw free means of course
depends on the containment relation used. If we work with the minor relation, we
are asking which graphs have no claw as a minor. It is trivial that a graph is claw
free in this sense iff it has no vertices of degree three or more. The claw free graphs
are trivially then exactly the disjoint unions of paths and cycles.
If instead of working with the minor relation, we rather work with the induced sub-
graph relation, the structure of claw free graphs is then a deep and difficult theorem
of Chudnovsky and Seymour, proved in a series of five papers totalling over 200 pages.
Now, if instead of working with graphs excluding a claw as an induced subgraph, we
instead ask which degree sequences exclude the degree sequence of a claw, then the
structure theorem given by Robertson and Song can be proved in under six pages.
Thus, in passing from induced subgraphs to the ≤ relation on degree sequences, we
have a theorem that is motivated by induced subgraphs, yet still more amenable to
analysis.
With this motivation, degree sequence analogues of questions asked for graphs are of-
ten asked for degree sequences. The celebrated Minor Theorem of Robertson and Sey-
mour says that finite graphs are well quasi ordered under the minor relation. A well
quasi order is a reflexive, transitive relation T on a set X such that if x1, x2, . . . , xn, . . .
is an infinite sequence in X then there exist i and j with i < j such that xiTxj. Anal-
ogous to the Minor Theorem, S.B. Rao’s famous conjecture, first stated in 1971 [10]
and proved in 2008 by M. Chudnovsky and P. Seymour and to appear in [1], says that
3
degree sequences of graphs are well quasi ordered under ≤. We refer to this theorem
as Rao’s Conjecture throughout.
In chapter 2, we give a proof that for each positive integer k, Rao’s Conjecture
holds for degree sequences of maximum degree at most k. Our proof was obtained
independently of Chudnovsky and Seymour’s proof of Rao’s Conjecture, and our proof
makes no use of the structure theory for degree sequences of those authors. In fact,
our proof has surprisingly little graph theory at all, which leads us to believe we may
be able to obtain results in a far more abstract, general setting in future works.
Just as Rao’s Conjecture is natural in light of the Minor Theorem, it is also natural, in
light of the many graph theorems excluding minors, topological minors, and so on, to
attempt to find the structure of degree sequences excluding a given degree sequence.
In chapter 3, we characterize degree sequences excluding (the degree sequence of)
certain matchings and cycles.
These exclusion results we obtain are stated in terms of pentagons, hexagons, the
complete bipartite graph K3,3, the split graphs first defined in [5], a binary operation
we call the half join first defined in [16] to characterize degree sequences with at most
one realization up to isomorphism, and in terms of forcibly chordal graphs. A graph
is chordal if no induced cycle has four or more vertices. A graph is forcibly chordal if
every graph with the same degree sequence is chordal.
While our exclusion theorems are exact, they are only valuable structure theorems to
the extent we understand the structure of the pentagons, hexagons, K3,3, split graphs,
and forcibly chordal graphs they are stated in terms of. Pentagons, hexagons, and
K3,3 may be considered well understood. The structure of split graphs has been
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found by Chudnovsky and Seymour in their proof of Rao’s Conjecture to appear in
[1]. We may thus take the structure of split graphs as known. That leaves the forcibly
chordal graphs. While the forcibly P and potentially P degree sequences have been
determined for many properties P (see [11] and [12] for excellent surveys), the forcibly
chordal graphs have not, to our knowledge, yet been characterized.
Our partial characterization of forcibly chordal graphs occupies us for the next three
chapters. In Chapter 4, we characterize the forcibly chordal trees. In Chapter 5, we
use these results to extend the characterization to forcibly chordal forests. In Chapter
6, we characterize connected, forcibly chordal graphs having a path structure, in a
sense to be defined in that chapter. We believe these results can be extended in
upcoming work to fully characterize forcibly chordal graphs.
1.2 Degree Sequence Basics, Notation, and Conventions
In order to make our presentation self contained and more efficient, we give the basic
notation, theorems, definitions, and conventions here for easy reference. First, we
must eliminate any possibility of ambiguity in the containment relation we will use
throughout chapters 2 through 6.
Definition 1.2.1. We say a graph G excludes a graph H if G contains no induced
subgraph isomorphic to H. We say a degree sequence D2 excludes a degree sequence D1
if D1 6≤ D2. We say that a degree sequence D excludes a graph G if D excludes D(G).
Note that while the subgraph and minor relations are far more commonly used in
graph theory than the induced subgraph relation, in light of the above definitions,
5
we will work exclusively with the induced subgraph relation. In light of this fact, we
make certain conventions to simplify wording throughout. If we say G contains H,
we mean as an induced subgraph, if we say G contains a hole, we mean an induced
hole, and so on. As such, when it causes no confusion, we will often “forget” to say
induced.
Another consequence of the fact that we work strictly with the induced subgraph
relation is that we can often simplify presentation by identifying the set X and the
induced subgraph G[X] of the graph G, which we often do when no confusion arises.
If we say a subset X of a graph G has a certain graph property, we mean that G[X]
does. Moreover, if G is a graph and X is a subset of V (G), we permit ourselves to
say X is a subset of G. Since we do not distinguish between X and G[X] in this
work, the reader should note that in particular, when we write X ⊆ G, we always
mean that X is an induced subgraph of G.
We use the notation G1
∐G2 to denote the disjoint union of graphs G1 and G2. Sim-
ilarly,∐n
i=1Gi denotes the disjoint union of graphs G1, . . . , Gn. If k is a nonnegative
integer, we use the notation k ·G or kG to denote the disjoint union of k isomorphic
copies of G.
Given subsets X and Y of a graph G, we say that X is complete to Y if each x in X
is adjacent to each y in Y . We say that X is complete if all pairs of distinct vertices
in X are adjacent. We say x in G is a universal vertex, or simply that x is universal,
if x is complete to G− x.
If G is a graph, Gc denotes the complement. If the degree sequence D is realized by
a graph G, we may speak of the complementary degree sequence Dc as the degree
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sequence of Gc. Though D may in general have more than one graph realizing it, it
is simple to check this definition of Dc does not depend on the choice of the realizing
graph, and Dc is thus well defined.
The set X is anti-complete in G if X is complete in Gc. The set X is anti-complete
to Y in G if X is complete to Y in Gc. In general, a graph or set is said to be
anti-P if property P holds on taking complements. An anti-hexagon, for instance, is
the complement of a hexagon, an anti-forcibly chordal graph is the complement of a
forcibly chordal graph, and so on.
Chapters 2 through 6 make extensive use of switchings, which we now define.
Definition 1.2.2. Let G be a graph. A switching is a tuple (a, b, c, d) of distinct
vertices in G such that a and b are adjacent, b and c are nonadjacent, c and d are
adjacent, and d and a are nonadjacent. The edges of the switching are ab and cd.
The nonedges of the switching are bc and da. If (a, b, c, d) is a switching in G then the
graph G− ab+ bc− cd+ da is said to arise from G by a switching in G. If there is a
sequence of graphs G1, . . . , Gn such that Gi+1 arises from Gi by a switching in Gi for
each i with 1 ≤ i < n then Gn is said to arise from G1 by a sequence of switchings.
Another way to state that (a, b, c, d) is a switching in G is that ab and cd are edges
of G while bc and da are nonedges of G. It is very important to note this definition
says nothing about whether or not ac is an edge or nonedge of G, and similarly for
bd. Moreover, we stress that if we say xy is not an edge of a switching, xy may or
may not be an edge of G. Similarly, if we say e is not a nonedge of the switching,
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while it is tempting to see this statement as a double negation equivalent to e being
an edge of the switching, this is not the case. The edge e may or may not be an edge
of the switching.
The reason for this behavior is simple. A switching has exactly two edges and two
nonedges. This leaves two pairs of vertices in {a, b, c, d} that are either edges of G
yet not edges of the switching, or nonedges or G yet not nonedges of the switching.
While care is needed on these points, no confusion arises if such care is taken, and we
speak of switchings rather informally by listing the two edges and the two nonedges.
We are rarely so formal as to present a switching as a tuple as in the definition.
We call two graphs equivalent if they have the same degree sequence. The reader may
note that if H arises from G by a switching in G then D(H) = D(G). Moreover, by
induction on the number of switchings, one sees that if H arises from G by a sequence
of switchings then D(H) = D(G). The following converse is a theorem first proved
in [6]. It is used at key points in chapter 2 and extensively throughout chapters 3
through 6 as our primary tool.
Theorem 1.2.3. Graphs G and H are equivalent iff H arises from G by a sequence
of switchings.
We now fix notation and conventions regarding the most important types of graphs
we use.
Definition 1.2.4. A graph G is called a split graph if V (G) can be partitioned into
8
(possibly empty) cells A and B such that G[A] is complete and G[B] is anti-complete.
The partition (A,B) is called a split partition.
We note the above definition allows for possibly empty split graphs. In general, we
allow empty graphs, but in cases where no problems arise, we casually disregard
empty graphs without comment if doing otherwise would unnecessarily complicate a
statement with trivialities.
We let Ck denote a cycle on k vertices, Pk denote a path on k vertices (not k edges),
Mk denote the matching kP2, and Kk the complete graph on k vertices. We often
say triangle for C3, square for C4, and so on. A hole in a graph is an induced cycle
on at least four vertices. A graph is called chordal if it has no holes.
1.3 Introduction to Combinatorial Proof Systems
In the second part of this work, we define and study certain combinatorial proof
systems that we call autonomous systems. No background in logic is required to
understand this part, though a basic understanding of partial orders, linear orders,
well foundedness and well orders, and transfinite induction and recursion is needed
at some points. The necessary facts may be found in [14] and [9]. The fact that we
need assume no previous exposure to logic from the reader arises from our abstract
approach, which of necessity starts from scratch, diverges early from that typically
studied by logicians, and soon far more closely resembles structural graph and partial
order theory than classical proof theory.
Proof theory is one of the main branches of mathematical logic. While proof theory as
9
understood by mathematical logicians does indeed study proofs, it is just as fair to say
that proof theorists study syntax and semantics, for the statements of typical results
in proof theory would be impossible to formulate, let alone prove, without syntactic
and semantic notions. While proof theory has many deep and difficult results, they
are deep and difficult results for proof systems with a great deal of structure beyond
the proofs themselves.
In 2001, the author had the goal of studying proofs in as general and abstract a setting
as possible. A proof is considered a (not necessarily finite or even well founded) partial
order such that for all x, the set of elements less than x is enough information to infer
x. We take, “is enough information to infer”, as a primitive notion. More precisely,
the proof system is a set together with a set of pairs (S, x), with S a subset of and x
a point in the domain, which we take to mean that S implies x.
We explicitly note that in this context, we have no syntax or semantics. We have
only implication and proofs. While it may seem a priori that this is too general to
prove anything, we in fact obtain nontrivial results. It is fair to say we obtain no
logical theorems. Our theorems are purely combinatorial. This was, in fact, a great
surprise to the author, who intended to prove logical results and found himself instead
working in structural combinatorics.
Roughly, just as there are rooted and unrooted trees, there are also rooted and un-
rooted proof systems. While rooted trees have singleton roots, rooted proof systems
allow arbitrary root sets, which are in fact the axioms of the proof system.
Unrooted proof systems generalize directed graphs. Roughly speaking, if a directed
edge from x to y is taken to mean x implies y, we can instead allow directed edges
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from an arbitrary set X to a point y to mean that set X of “formulas” implies y. An
unrooted proof system could, therefore, be thought of as a directed hypergraph.
Rooted proof systems, on the other hand, generalize well founded partial orders.
This work focuses on a generalization of rooted proof systems, which we will call
autonomous systems, and which we will define without reference to unrooted proof
systems. The topic of unrooted proof systems will be addressed in future work of the
author.
We give the basic definitions, theorems, and constructions related to these autonomous
systems that have proved useful in their study. We give three distinct axiomatiza-
tions of autonomous systems, give numerous characterizations of partial orders as
autonomous systems, and define what we call the canonical orders that encode con-
text dependent needing in a proof system and turn out to be an important structural
tool in proving even statements making no mention of these orders. We define the
notions of weak and strong aut descendability, two finiteness conditions on which
many autonomous system theorems and proofs depend. We define homomorphisms
and two containment relations that allow us to define the minors of a proof system.
We then use the canonical orders to prove two forbidden minors theorems that hold
under the assumption of strong aut descendability. (In particular, they hold for finite
and even finitary autonomous systems.) We also extend the definition of partial or-
der lexicographic sum to autonomous systems and prove the basic properties of the
lexicographic sum
11
CHAPTER 2
THE BOUNDED CASE OF RAO’S CONJECTURE
In this chapter, we answer a question posed by N. Robertson, who asked if graphic
degree sequences of bounded degree can be realized as disjoint unions of graphs with
bounded size components. Our answer in the affirmative implies the bounded case of
S.B. Rao’s Conjecture, which we state now.
Theorem 2.0.1. Graphic degree sequences of bounded degree are well quasi ordered.
There is surprisingly little graph theory in our proof. In fact, the graph theory
only comes in the initial lemmas constructing graphs with certain prescribed degree
sequences. Though there is an existence proof of all these initial lemmas using the
Erdos-Gallai inequalities proved in [4], our goal is to give a detailed construction from
first principles. We therefore avoid using any outside results in this proof.
We now turn to the proof.
Definition 2.0.2. A graph G is called k-regular if every vertex has degree k. A graph
is called regular if it is k-regular for some k.
Lemma 2.0.3. Let k be an even integer. Then there is an integer Lk such that for
all L ≥ Lk there is a k-regular graph G on L vertices.
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Proof. k is even, so let k = 2l, and let Lk = k + 1 = 2l + 1. For each L ≥ Lk, we
define a graph G on the integers 0, 1, . . . , L− 1 by letting
E(G) = {xy : 1 ≤ |x− y| mod L ≤ l|}
Obviously, G has L vertices, and since L is at least 2l+1, it follows that for all x, the
2l vertices x− l, x− l + 1, . . . , x− 1, x + 1, . . . , x + l are parwise distinct. Therefore
each x in G has degree 2l = k. Therefore G is a k-regular graph on L vertices as
needed, thus completing the proof.
The graphs in the above proof are called circulants.
Lemma 2.0.4. Let k be an odd integer. Then there is an integer Lk such that for all
even L ≥ Lk there is a k-regular graph G on L vertices.
Proof. Let Lk = 2k. It is enough to construct, for each even L ≥ Lk, a k-regular
bipartite graph G on L vertices. So take an even L ≥ Lk and let L = 2l. Note then
that l ≥ k. Take disjoint sets A = {v1, . . . , vl} and B = {w1, . . . , wl}. Define G as
the graph with vertex set A ∪B and edge set
E(G) = {viwj : 0 ≤ wj − vi mod l ≤ k − 1}
G is then a k-regular graph on L vertices, and the proof is complete.
Lemma 2.0.5. Given a positive integer k and a nonnegative integer j, there is a
graph G with exactly 2j vertices of degree k − 1 and all other vertices of degree k.
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Proof. Letm = max{k, j}. Take disjoint setsA = {v1, . . . , vm} andB = {w1, . . . , wm}.
Define G′ as the graph with vertex set A ∪B and edge set
E(G) = {viwj : 0 ≤ wj − vi mod m ≤ k − 1}
Let G = G′ − v1w1 − v2w2 − · · · − vjwj. Then dG(vi) = dG(wi) = k − 1 for 1 ≤ i ≤ j
and dG(vi) = dG(wi) = k for i > j. The claim follows.
Lemma 2.0.6. Given a positive integer k and nonnegative integer j such that 2j 6= k,
there is a graph G with exactly one vertex of degree 2j and all other vertices of degree k.
Proof. By the previous lemma, there is a graph G′ with exactly 2j vertices of de-
gree k − 1 and all other vertices of degree k. (Note the G′ of this lemma is the G of
the previous lemma.) Let v be a point not in G′. Let G be the graph on V (G′)∪{v}
such that xy ∈ E(G) iff one of the following conditions holds:
i xy ∈ E(G′).
ii x = v and dG′(y) = k − 1.
It follows by definition of G that v is adjacent in G to exactly the vertices of degree
k − 1 in G′. By choice of G′, there are exactly 2j of these. So dG(v) = 2j. It is
enough to show dG(x) = k for all other vertices in G, so take x 6= v. Then dG′(x)
is k or k − 1. If dG′(x) = k then it follows by definition of G that for all y in V (G),
the edge xy is in G iff it is in G′. Therefore dG(x) = dG′(x) = k. If dG′(x) = k − 1
then it follows from the definition of G that NG(x) = NG′(x) ∪ {v}. Therefore
dG(x) = dG′(x) + 1 = k − 1 + 1 = k.
14
Lemma 2.0.7. Given nonnegative integers j and k, there is a graph G with exactly
2j + 1 vertices of degree 2k and all others of degree 2k + 1.
Proof. Let m = max{2k + 1, k + j}. Let A = {v1 . . . , vm} and B = {w1, . . . , wm} be
disjoint sets. Define a graph G′′ on A ∪ B by letting A and B be anti-complete and
letting viwj ∈ E(G′′) iff 0 ≤ (wj − vi) mod m ≤ 2k. Note G′′ is a 2k + 1-regular
graph.
Let G′ = G′′ − v1w1 − v2w2 − · · · − vk+jwk+j. Then for 1 ≤ i ≤ k + j, we have
dG′(vi) = dG′′(vi)−1 = (2k+1)−1 = 2k. Similarly, dG′(wi) = 2k. For i > k+j, we see
from the definition of G′ that NG′(vi) = NG′′(vi) therefore dG′(vi) = dG′′(vi) = 2k+1.
Similarly, for i > k + j, we see that dG′(wi) = 2k + 1.
Let z be a point not in G′. Let G be the graph on V (G′) ∪ {z} such that E(G) =
E(G′) ∪ {zvi|1 ≤ i ≤ k} ∪ {zwi|1 ≤ i ≤ k}. We show G has the desired properties.
First, note that since z is not in G′, it follows directly from the definition of G that
dG(z) = 2k. Now consider vi with 1 ≤ i ≤ k. Then NG(vi) = NG′(vi) ∪ {z} therefore
dG(vi) = dG′(vi) + 1 = 2k + 1. Similarly, dG(wi) = 2k + 1 for 1 ≤ i ≤ k.
By similar reasoning, the reader may check that dG(vi) = dG(wi) = 2k+1 if k+j+1 ≤
i ≤ m and that dG(vi) = dG(wi) = 2k if k + 1 ≤ i ≤ k + j. Therefore G has exactly
2j + 1 vertices of degree 2k and the rest of degree 2k + 1 as claimed.
Lemma 2.0.8. Given distinct, nonnegative integers integers j and k, there is a
graph G with exactly one vertex of degree 2j+1 and all other vertices of degree 2k+1.
15
Proof. By the previous lemma, there is a graph G′ with exactly 2j + 1 vertices of
degree 2k and all other vertices of degree 2k+1. (The G′ of this lemma is the G of the
previous lemma.) Take y not in G′. Define G as the graph with vertex set V (G′)∪{y}
and edge set E(G′) ∪ {yx|x ∈ V (G′) and dG′(x) = 2k}. We show G has the desired
property by showing dG(y) = 2j + 1 and all other vertices have degree 2k + 1 in G.
First, by choice of G′, we know there are exactly 2j + 1 elements x in G′ such that
dG′(x) = 2k. Since NG(y) consists, by definition of G, of exactly these elements, we
see that dG(y) = |NG(y)| = 2j+1. We show all other vertices of G have degree 2k+1
in G.
So take x ∈ V (G)− y. Then dG′(x) is 2k or 2k+ 1. If dG′(x) = 2k then by definition
of G, we see that NG(x) = NG′(x) ∪ {y}. Therefore dG(x) = dG′(x) + 1 = 2k + 1. If
dG′(x) = 2k+1 then it follows by the definition of G that NG(x) = NG′(x). Therefore
dG(x) = dG′(x) = 2k+ 1. Therefore all vertices other than y have degree 2k+ 1 in G,
as was to be shown.
We note that in the following lemma, i and j may or may not be distinct. The
possibility that i = j must be allowed for use in a later proof.
Lemma 2.0.9. Let i and j be nonnegative integers. Let k be a positive, even integer.
Then there is a graph G with vertices v 6= w such that dG(v) = 2i+ 1, dG(w) = 2j+ 1
and dG(x) = k for all other vertices x in G.
Proof. We know there is a graph G′ with a vertex y of degree 2i+2j+2 and dG′(x) = k
for all other x. Let G be obtained from G′ by splitting the vertex y into nonadjacent
16
vertices v and w such that v is adjacent in G to 2i + 1 of the G′ neighbors of y and
w is adjacent in G to the remaining 2j + 1 G′ neighbors of y.
Definition 2.0.10. Let U be a class of graphs. U is called productive if the following
conditions hold:
(i) For every odd, nonnegative integer k, there is an integer LU,k such that for all
even L ≥ LU,k, there is a k-regular graph G in U of cardinality L.
(ii) For every even, nonnegative integer k, there is an integer LU,k such that for all
L ≥ LU,k, there is a k-regular graph G in U of cardinality L.
(iii) Given positive integers j, k, with 2j 6= k, there is a graph in U with exactly one
vertex of degree 2j and all other vertices of degree k.
(iv) Given distinct, nonnegative integers j and k, there is a graph with exactly one
vertex of degree 2j + 1 and all other vertices of degree 2k + 1.
(v) Given nonnegative integers i and j, and a positive, even k, there is a graph G in
U with vertices v 6= w such that dG(v) = 2i+ 1, dG(w) = 2j + 1 and dG(x) = k
for all other vertices x in G.
Corollary 2.0.11. The class of finite graphs is productive.
Proof. This is a restatement of the lemmas thus far proved.
17
Definition 2.0.12. We call a class U of graphs finitely representable if there is a
finite subset F of U such that for every graph G in U , there is a graph G′ such that
D(G′) = D(G) and G′ is the disjoint union of graphs in F .
Lemma 2.0.13. The finite union of finitely representable classes is also finitely rep-
resentable.
Proof. Let U1, . . . , Un be finitely representable and let
U =n⋃i=1
Ui
Since each Ui is finitely representable, there is for each i a finite subset Fi of Ui such
that every graph in Ui has the same degree sequence as a disjoint union of graphs
in Fi. Let
F =n⋃i=1
Fi
Then given a graph G in U , there is i such that G is in Ui. Therefore G has the same
degree sequence as the disjoint union of some graphs in Fi. Since F contains Fi, we
see G has the same degree sequence as the disjoint union of some graphs in F . Since G
is an arbitrary graph in U and F is finite, we see that U is finitely representable, as
was to be shown.
Lemma 2.0.14. If U is a finite set of graphs then U is finitely representable.
18
Proof. Let F = U .
We make use of the following basic fact from number theory.
Lemma 2.0.15. Let S be a nonempty set of positive integers and let g be its greatest
common divisor. Then there is a finite subset FS of S and a positive integer n such
that for all n′ ≥ n, we can write n′g as a1s1 + · · ·+ apsp for some positive integer p,
some nonnegative integers a1, . . . , ap, and some elements s1, . . . , sp of S.
Theorem 2.0.16. Let U be a class of graphs and let Uk be the class of k-regular
graphs in U . Then Uk is finitely representable.
Proof. If Uk is empty then it is vacuously true that Uk is finitely representable, so
suppose Uk is nonempty.
Let S be the set of cardinalities of graphs in Uk. Then S is nonempty. Let g be
its greatest common divisor. By the previous lemma, there exists n such that for
all n′ ≥ n, we can write n′g as a1s1 + · · · + apsp for some p positive integer p, some
positive integers a1, . . . , ap, and some elements s1, . . . , sp of S. Since each si is in S,
it follows by the definition of S that there are graphs G1, . . . , Gp in Uk such that
|Gi| = si for each i. Let F = {G1, . . . , Gp} ∪ {H ∈ Uk : |H| < ng}. Note that since
there are only finitely many graphs on less than ng vertices, F is a finite set.
By definition of finite representability, it is enough to show that given G in Uk, there
is a graph G′ with the same degree sequence as G such that G′ is the disjoint union
of graphs in F . So take a graph G in Uk.
19
If |G| < ng then G is in F by definition of F , so we see that G itself is a graph
with the same degree as G that is the disjoint union of elements of F . So suppose
|G| ≥ ng. Then we may write |G| = a1s1 + · · · + apsp as in the previous lemma.
Consider the graph
G′ =
p∐i=1
aiGi
Then |G′| =∑p
i=1 ai|Gi| =∑p
i=1 aisi = |G|. Also note that G and G′ are both k-
regular. Since G and G′ are k-regular graphs of the same cardinality, they have the
same degree sequence. Clearly, we have expressed G′ as the disjoint union of graphs
in F . This completes the proof of the lemma.
Definition 2.0.17. A degree class sequence C is an infinite sequence c0, c1, c2, c3, . . .
with values in {1, 2, 3, . . . ,∞} such that ci is eventually 1.
Definition 2.0.18. Let U be a class of graphs and C = (ci)∞i=1 a degree class sequence.
Then UC denotes the class of graphs G in U such that for all i, the graph G has less
than ci vertices of degree i.
Definition 2.0.19. Let X = {xi}i≥0 be a sequence. We define the support S(X)
of X as {i : xi 6= 1}. We define the infinity support S∞(X) of X as {i : xi =∞}.
Lemma 2.0.20. Let U be a productive class. Let C be a degree class sequence such
that S(C) is finite. Then UC is finitely representable.
20
Proof. The proof is by induction on |S∞(C)|. If |S∞(C)| = 0 then UC is finite, so
we know by Lemma 2.0.14 that UC is finitely representable. Suppose the result is
true for all degree class sequences with infinity support of cardinality at most N . We
must prove that UC is finitely representable for each degree class sequence C such
that |S∞(C)| = N + 1.
For every proper subset X of S∞(C) and every positive integer M , let XM be the
degree class sequence such that XM(i) = C(i) for all i except that XM(i) = M
for all i in S∞(C) − X. Since the infinity support of XM is a proper subset of
the infinity support S∞(C), we know by the induction hypothesis that UXMis finitely
representable for each such XM . Since the finite union of finitely representable classes
is finitely representable, we see that for each M , the class⋃X(S∞(C)
UXM
is finitely representable. To show UC is finitely representable, it is therefore enough
to show that
WM := UC −⋃
X(S∞(C)
UXM
is finitely representable for some M .
Note that WM is the class of graphs G in UC such that if C(i) = ∞ then G has at
least M vertices of degree i. We have only to show this class is finitely representable
for some large enough M . This is immediate from the definition of productivity and
Theorem 2.0.16. If M is large enough, we simply take out vertices in G whose degree
d is in S(C) − S∞(C) by using the almost regular graphs whose vertices all have
the same degree except possibly one or two. More precisely, we subtract the degree
21
sequence of these almost regular graphs from the degree sequence D of G. Call the
remaining degree sequence D′, which we do not yet know is graphic.
The degree sequence D′ has an even number of vertices of odd degree. We may pair
them up. (More formally, we partition the set of vertices of odd degree into double-
tons.) For each such pair {2i+1, 2j+1} in turn, by condition (iv) of Definition 2.0.10,
we may choose a graph Gi,j with all vertices of degree 2i + 1 except one of degree
2j + 1. Let Di,j be the degree sequence of Gi,j. Let D′′ be the degree sequence re-
sulting from subtracting each Di,j with i paired to j from the degree sequence D′. It
is again important to note that, at this point, we have not yet shown that D′ or D′′
is realizable.
However, by choosing M large enough, D′ and D′′ are indeed realizable. Our degree
sequence remaining has an even number of vertices of each odd degree and at least M
vertices of each degree in S∞(C). By Theorem 2.0.16, D′′ may therefore be realized
as the disjoint union of finitely many regular graphs. Letting G′′ realize G′, it is
clear from the definitions of D′ and D′′ that we may unite G′′ with almost regular
graphs to obtain a graph graph H with the same degree sequence as G. Since S(C)
is finite, only finitely many such almost regular graphs are used. WM is therefore
finitely representable as needed.
Theorem 2.0.21. For any fixed bound k, degree sequences bounded by k are finitely
realizable.
22
Proof. The class of degree sequences with all degrees at most k is simply UC where
U is the class of finite graphs and C is the degree class sequence satisfying C(i) =∞
if i ≤ k and C(i) = 1 for i > k. Since U is productive, we may apply the previous
result.
We now prove Theorem 2.0.1.
Proof. We know that degree sequences of degree at most k can be realized as disjoint
unions from a finite set F of graphs. Let G1, G2, . . . be a sequence of graphs each
of which is a disjoint union of graphs in F = {F1, . . . , Ft}. Then for each i, we may
write
Gi = ci,1F1
∐· · ·∐
ci,tFt,
for some nonnegative integers ci,t. We may choose a strictly increasing sequence
(in)n≥0 such that cin,j is an increasing sequence of n for each j. Then Gin is an in-
creasing sequence of graphs under the induced subgraph relation. Since the sequence
G1, G2, . . . was an arbitrary sequence of disjoint unions in F , this shows finite dis-
joint unions in F are well quasi ordered under induced subgraph. In particular, their
degree sequences are well quasi ordered under ≤.
23
CHAPTER 3
EXCLUDING MATCHINGS AND CYCLES
In this chapter, we derive structure theorems for some classes of degree sequences
excluding the matching M2 and/or cycles. More precisely, we first recall the char-
acterization of split graphs by forbidden induced subgraphs. Next, we use this to
characterize the degree sequences that exclude the matching M2 and a square. We
next use this result to characterize degree sequences excluding only a square and,
more generally, degree sequences excluding an arbitrary cycle. For each theorem
one proves characterizing the degree sequences having a property X by excluding
graphs in the set S, one may also prove the complementary theorem that the degree
sequences whose complementary degree sequence has property X are exactly those
that exclude graphs whose complement is in S. Taking the complementary theorem
to the result on excluding a square, we characterize degree sequences excluding the
matching M2.
However, each of these theorems is stated in terms of another class: split graphs,
forcibly chordal graphs, and, generalizing forcibly chordal graphs, the class of graphs
which forcibly have all chordless cycles of length at most k. This leads naturally
to the problem of characterizing forcibly chordal graphs, which we address in the
following chapters.
24
We make use of the following propositions, the first of which is a folklore theorem
that may be taken as an exercise.
Proposition 3.0.22. The following are equivalent for a graph G:
(i) G excludes M2, C4, and C5.
(ii) G excludes M2 and all holes.
(iii) G is a split graph.
Corollary 3.0.23. Split graphs are chordal.
Proof. By Proposition 3.0.22, split graphs contain no holes. Therefore, they are
chordal.
Proposition 3.0.24. The following are equivalent for a degree sequence D:
(i) D excludes the degree sequences (1, 1, 1, 1), (2, 2, 2, 2), and (2, 2, 2, 2, 2).
(ii) D excludes the degree sequence (1, 1, 1, 1) and the degree sequences of all cycles
on at least 4 vertices.
(iii) D is the degree sequence of a split graph.
25
Proof. D satisfies condition (i) of this theorem iff it is realized by a graph that satisfies
condition (i) of Proposition 3.0.22. Similarly for conditions (ii) and (iii). Since the
three conditions of Proposition 3.0.22 are equivalent, it thus follows that the three
conditions of this theorem are equivalent.
The following proposition follows from the well known characterization of split graphs
as those graphs for which at least one of the Erdos-Gallai inequalities is equality.
Proposition 3.0.25. Let D be the degree sequence of a split graph. Then every
realization of D is a split graph.
In other words, the above proposition states that every split graph is forcibly split.
In particular, we know the following.
Corollary 3.0.26. Every split graph is forcibly chordal.
Proof. If a graph is split then, by Proposition 3.0.22 it has no holes and is thus
chordal. Every split graph is forcibly split, therefore every realization of the degree
sequence of a split graph is split, and hence chordal. Since every realization of the
degree sequence of a split graph is chordal, it follows that every split graph is forcibly
chordal.
Our next lemmas will make use of the notion of half join, which we define below.
Informally, the half join is obtained by joining an arbitrary graph H completely to
26
the complete part of a split graph S and anti-completely to the anti-complete part
of S.
Definition 3.0.27. Let S be a split graph with partition into a complete part A and
anti-complete part B. Let H be an arbitrary graph. Then the half join (S,A,B,H) of
S and H with respect to the split partition (A,B) is defined as the graph with vertex
set V (S) ∪ V (H) and edge set
E(S) ∪ E(H) ∪ {xy : x ∈ H and y ∈ A}
The above definition of half join arises naturally and often when working with split
graphs, and is used, for instance, in [16] to state a decomposition theorem for split
graphs, though we make no use of this theorem. Tyshkevich does not use the word
half join, or any other word, simply using notation to denote the operation, but we
find it convenient to have a word denoting it, so we choose half join.
We also note that A and B are mentioned in addition to S because a split graph may
have more than one split partition, but in practice, when talking about half joins, we
are usually far less formal, and simply say “the half join of S and a pentagon” and
similar. We will permit this abuse of language when it causes no confusion.
We will use the following lemma several times in proving the structure theorems of
this chapter.
Lemma 3.0.28. Let S be a split graph with split partition (A,B) and let H be an
arbitrary graph. Let G be a (not necessarily connected) graph on at least three vertices
27
with no induced triangles, no isolated vertices, and which is not a star. If G is an
induced subgraph of the half join (S,A,B,H) then G is an induced subgraph of S
or H.
Proof. Since (A,B) is a split partition, A is by definition complete. Therefore A∩G
is complete and thus any three vertices of A ∩G comprise an induced triangle in G.
Since G is triangle free by assumption, it follows that A ∩ G is empty, has exactly
one vertex, or has exactly two vertices. We consider these three cases.
First, assume A∩G is empty. B is anti-complete to H and B itself is anti-complete,
therefore every vertex of B ∩ G is an isolated vertex of G. Since G has no isolated
vertices by assumption, it follows that B ∩ G is empty. Therefore G is an induced
subgraph of H as the lemma claims.
Second, assume A∩G contains exactly one vertex x. Consider any other two vertices y
and z in G. We show that y and z are non-adjacent. First, suppose one of y or z
is in B. Without loss of generality, we may assume y is in B. Note that z is in B
or H since x is the only element of A∩G. Since B is anti-complete and anti-complete
to H, and since z is in either B or H, it follows that y and z are not adjacent. Now
suppose neither y nor z is in B. Then both y and z are in H. Since H is complete
to A, it follows that y and z are both adjacent to x. Since G has no induced triangles
by assumption, it therefore follows that y and z are not adjacent.
This shows that for all choices of y and z in G distinct from x, the vertices y and z
are non-adjacent. Now take any element y of G distinct from x. If y is in H then y is
adjacent to x since H is complete to A. Otherwise, y is in B. Since G has no isolated
vertices, we see that y must be adjacent to some vertex of G. Since vertices of B are
28
at most adjacent to vertices of A, we see that y is adjacent to some element of A∩G.
Since x is the only such vertex by assumption, we see that y is adjacent to x.
We have thus shown that x is complete to G − x. Since we have also shown G − x
is anti-complete, we see that G is a star, contrary to assumption. This contradiction
shows that A ∩G can not have exactly one vertex.
Finally, assume A ∩ G contains exactly two vertices. Call them x and y. Since A
is complete, x and y are adjacent. Suppose H ∩ G contains a vertex z. Since A is
complete to H, it follows that z is adjacent to x and y and hence x, y, z comprise a
triangle, contrary to choice of G as triangle free. Therefore H ∩ G is empty, which
means G is an induced subgraph of S as claimed.
In all three cases, G is an induced subgraph of H or S, thus completing the proof.
In the following proposition, we characterize graphs excluding M2 and C4. This
proposition is notable in two ways. First, we prove the result for graphs, which is
stronger than simply proving the analogous result for degree sequences, and it is
somewhat surprising a nice characterization exists for graphs at all. Later in the
chapter, for instance, when we exclude M2 alone, it will be quite necessary to use
degree sequences rather than graphs.
Second, we have pointed out each exclusion theorem has a complementary theorem,
but the following proposition is self complementary. The results later in the chapter
lose the property of self complementarity as well.
Theorem 3.0.29. The following are equivalent for a graph G:
29
(i) G excludes M2 and C4.
(ii) G is a split graph or the half join of a split graph and a pentagon.
Proof. To see that (ii) implies (i), note that by Proposition 3.0.22, we know that if
G is a split graph then G has no induced M2 and no induced holes, and in particular
no induced square. Now suppose G is the half join of a split graph and a pentagon.
Note that since M2 and C4 have at least three vertices, have no induced triangles, no
isolated vertices, and are not stars, it follows from Lemma 3.0.28 that if M2 or C4 is
an induced subgraph of the half join of a split graph and a pentagon then M2 or C4
must be an induced subgraph of a split graph or an induced subgraph of a pentagon.
It is easy to see a pentagon contains no induced M2 or C4, and we have already noted
a split graph contains no induced M2 or C4, therefore the half join of a split graph
and a pentagon has no induced M2 or C4.
For the other direction, suppose G has no induced M2 and no induced C4. If G also
has no induced pentagon then by Proposition 3.0.22, we know that G is a split graph
as desired. So suppose G has an induced pentagon C. We show that G[V (G)−C] is
a split graph and that G is the half join of G[V (G)− C] and C.
Toward this end, we first show that every vertex x of V (G)−C is either complete or
anti-complete to C. So let x be in V (G)−C. Let C = {a, b, c, d, e} with the vertices
in that cyclic order. Suppose x is adjacent to at least one vertex in C. Without loss of
generality, x is adjacent to a. If x has degree 1 in G[C ∪ x] then {x, a} and {c, d} are
independent edges and G thus has an induced M2, contrary to hypothesis. Suppose
30
x has degree 2 in G[C ∪ x]. Then x is either adjacent to a vertex of C adjacent to a
or a vertex of C at distance 2 from a in C. Suppose x is adjacent to a vertex adjacent
to a. Without loss of generality, x is adjacent to b. Then {x, a} and {c, d} are again
independent edges, contrary to hypothesis that G has no induced M2. So suppose
x is adjacent to a vertex at distance 2 from a in C. Without loss of generality, x is
adjacent to c. Then x, a, b, c is an induced 4 cycle in G, contrary to hypothesis. Both
possible graphs in which x has degree 2 in G[C ∪ x] result in a contradiction, thus x
has degree greater than 2 in this graph.
Now suppose x has degree 3 or 4 in G[C∪x]. Consider the complement K of G[C∪x].
The complement of a pentagon is a pentagon, soK consists a pentagon together with a
vertex adjacent to either 1 or 2 vertices of that pentagon. By the previous paragraph,
such a graph contains an induced M2 or C4, so K has an induced M2 or C4. If K has
an induced C4 then by taking complements, G[C∪x] has an an induced M2. Similarly,
if K has an induced M2 then by taking complements, G[C ∪ x] has an induced C4.
Both these possibilities are contrary to assumption that G has no induced M2 or C4.
Thus x can not have degree 3 or 4 in G[C ∪ x] either.
The only remaining possibility is that x has degree 5 in G[C ∪ x], or in other words,
x is complete to C. We have thus shown that if x is not anti-complete to C then x
is complete to C. Since x was arbitrary, we have shown that every vertex outside C
is either complete or anti-complete to C.
Let A be the set of vertices outside C and complete to C, and let B be the set of
vertices outside C and anti-complete to C. We know that A ∪B ∪ C = G. We show
that A is complete and B is anti-complete.
31
First, to show A is complete, let x and y be vertices of A. Suppose they are not
adjacent. Then letting a and c be non-adjacent vertices of C, we see that x, a, y, c is
an induced square, contrary to choice of G. Thus x and y are adjacent. Since x and
y are arbitrary vertices of A, it follows that A is complete.
We have thus shown that in a graph excluding M2 and C4 but containing a pentagon,
the set of vertices complete to the pentagon is itself complete. By taking complements,
and noting the complement of a pentagon is a pentagon, we see that the set of vertices
anti-complete to the pentagon is itself anti-complete. Thus B is anti-complete.
We now know that G = A ∪ B ∪ C, that A is complete, B is anti-complete, A
is complete to C, and B is anti-complete to C. It thus follows directly from the
definition of half join that G is the half join of a split graph and a pentagon, as was
to be shown.
Lemma 3.0.30. Consider the half join (S,A,B,H) of a graph H and split graph S
with complete part A and anti-complete part B. Let x, y, z, w be a switching operation
in the half join. Then {x, y, z, w} ⊆ H or {x, y, z, w} ⊆ S. In other words, the
switching either lies entirely in H or entirely in the split graph.
Proof. Let x, y, z, w be a switching operation in the half join (S,A,B,H). Simple
checking shows G = (S,A,B,H)[x, y, z, w] is a graph on four vertices with no isolated
vertices, no induced triangle, and which is not a star. By lemma Lemma 3.0.28, it
follows that G is contained in S or H, which proves our claim.
32
Lemma 3.0.31. If a degree sequence D is realized by the half join of a graph H and
a split graph S then every realization of D is the half join of a graph with the same
degree sequence of H and a split graph with the same degree sequence as S.
Proof. By assumption, D is realized by a half join of a graph H and a split graph S.
Call this graph K. Let K ′ be some other realization of D. Then K ′ can be obtained
from K by a sequence of switching operations, so take K1, . . . , Kn such that K1 = K,
Kn = K ′, and Ki+1 is obtained from Ki by some single switching operation.
We show by induction on i that Ki satisfies the conclusion of the corollary for each
i. For i = 1, K1 = K satisfies the conclusions of the corollary by choice of K. Now
suppose the conclusion of the corollary holds for i and consider Ki+1. By the inductive
hypothesis, Ki is the half join of a split graph Si with the same degree sequence as
S and a graph Hi with the same degree sequence as H. Ki+1 arises from Ki by a
switching operation. By the Lemma 3.0.30, the vertices of this switching operation
must lie entirely in Si or entirely in Hi. If the switching operation lies in Si then let
Si+1 be the graph obtained from Si by performing the switching and let Hi+1 = Hi.
If the switching operation lies in Hi then let Hi+1 be the graph obtained from Hi by
performing the switching and let Si+1 = Si. Then Hi+1 has the same degree sequence
as Hi, Si+1 is a split graph with the same degree sequence as Si, and Ki+1 is the
half join of Hi+1 and Si+1, completing the induction. Thus Ki satisfies the conclusion
of the corollary for each i. In particular, K ′ = Kn satisfies the conclusion of the
corollary. Since K ′ was an arbitrary realization of D, it follows that every realization
of D satisfies the conclusion of the corollary, as was to be shown.
33
Theorem 3.0.32. The following are equivalent for a degree sequence D:
(i) D excludes (1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences.
(ii) D is the degree sequence of a split graph or D is the degree sequence of the half
join of a split graph and a pentagon.
Proof. Assume (i) and let G realize D. Then G excludes M2 and C4 as induced
subgraphs. Therefore G is a split graph or the half join of a pentagon and a split
graph by Theorem 3.0.29. Therefore D is realized by a graph as required in condition
(ii).
For the other direction, assume condition (ii) holds. Then D is realized by a split
graph or the half join of a split graph and a pentagon. We consider each possibility.
If D is realized by a split graph then every realization is a split graph since split
graphs are forcibly split. Split graphs have no induced C4 or M2, therefore every
realization of D excludes M2 and C4 as induced subgraphs. Therefore D excludes
(1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences, as was to be shown.
Similarly, if D is realized by the half join of a split graph and a pentagon then by
Lemma 3.0.31, every realization is a half join of a split graph and a pentagon. By
Theorem 3.0.29, it follows that every realization excludes M2 and C4 as induced
subgraphs, and thus D excludes (1, 1, 1, 1) and (2, 2, 2, 2) as degree sequences. Thus
in each case, D excludes both degree sequences, as was to be shown.
34
Lemma 3.0.33. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.
Let G be a realization of D containing a cycle C isomorphic to Ck. Then each vertex
of G− C is complete or anti-complete to C.
Proof. Assume not. Then there is a vertex x outside of C adjacent to some vertex y
of C and non-adjacent to some other vertex z of C. Let v be a neighbor of z in C
distinct from y. Let K = G[C ∪ x]. Define K ′ as the graph obtained from K/{v, z}
by subdividing the edge {x, y}. Simple checking shows that K and K ′ have the same
degree sequence. But K ′ − x is isomorphic to Ck−1. Therefore K ′ contains Ck−1 as
an induced subgraph. Therefore K does not exclude D(Ck−1), and hence G does not
exclude D(Ck−1) either. This implies that D(Ck−1) ≤ D, contrary to hypothesis.
This contradiction shows that every vertex outside C is complete or anti-complete to
C as claimed.
Lemma 3.0.34. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.
Let G be a realization of D containing a cycle C isomorphic to Ck, and let A be the
set of vertices of G− C that are complete to C. Then A is complete.
Proof. Assume there are non-adjacent vertices x and y, both complete to C. Write C
in cyclic order as c1, c2, . . . , ck. Then we can use c1, c3, x, y as a switching to ob-
tain a graph G′ with the same degree sequence as G. The reader can check that
G′[c1, c3, c4, . . . , ck] is a cycle in that cyclic order. Therefore G′ contains an induced
35
Ck−1. We thus see that D does not exclude Ck−1, contrary to hypothesis. This con-
tradiction shows x and y must be adjacent. Since x and y are arbitrary elements
of A, it follows that A is complete as claimed.
Lemma 3.0.35. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.
Let G be a realization of D containing a cycle C isomorphic to Ck, and let B be the
set of vertices of G− C that are anti-complete to C. Then B is anti-complete.
Proof. Let x and y be distinct vertices in B. It is enough to show x and y are not
adjacent. Suppose they are adjacent. Then G[C ∪ {x, y}] is isomorphic to Ck∐P2,
which has the same degree sequence as Ck−1
∐P3. ThereforeD does not exclude Ck−1,
contrary to assumption. This contradiction completes the proof.
Lemma 3.0.36. Let k ≥ 5. Suppose D excludes Ck−1, but D does not exclude Ck.
Let G be a realization of D containing a cycle C isomorphic to Ck, and let A and B
be the sets of vertices of G−C that are complete and anti-complete to C, respectively.
Then G is the half join (G[A ∪B], A,B,C) of C and G[A ∪B].
Proof. We have only to show that every vertex of G−C is complete or anti-complete
to C, that vertices complete to C are pairwise adjacent, and that vertices anti-
complete to C are pairwise non-adjacent. This is exactly the content of Lemma 3.0.33,
Lemma 3.0.34, and Lemma 3.0.35.
36
Theorem 3.0.37. A degree sequence D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2) iff D is
forcibly chordal or D is the degree sequence of the half join of a split graph and a
hexagon.
Proof. First, we prove the only if. Assume D is forcibly chordal. Then by definition,
every realization of D has no holes, and thus D excludes the degree sequence of every
hole, in particular (2, 2, 2, 2) and (2, 2, 2, 2, 2).
The next part of the only if direction is to prove if D is the degree sequence of the
half join of a split graph and a hexagon then D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2).
It is enough to show every realization of D excludes the square and the pentagon as
induced subgraphs, so let G realize D. Since the degree sequence of a hexagon can
only be realized as a hexagon or two disjoint triangles, it follows from Lemma 3.0.31
that G is either the half join of a split graph and a hexagon or the half join of a
split graph and the disjoint union of two triangles. Suppose G has an induced square
or pentagon C. Then by Lemma 3.0.28, it follows that C is an induced subgraph
of a hexagon, an induced subgraph of the disjoint union of two triangles, or an
induced subgraph of a split graph, which is in all three cases a contradiction. This
contradiction completes the proof of the only if direction of the lemma.
For the other direction, letD be a degree sequence excluding (2, 2, 2, 2) and (2, 2, 2, 2, 2).
We must show D is forcibly chordal or D is the degree sequence of the half join of a
split graph and a hexagon.
First, we note no realization of D contains a hole with 7 or more vertices, for assume
some realization G contains such a cycle Cn with n ≥ 7 as an induced subgraph. This
37
graph has the same degree sequence as the disjoint union of Cn−4 and a square, and
this disjoint union obviously contains the square as an induced subgraph. Therefore
(2, 2, 2, 2) ≤ D, contrary to assumption.
Either no realization of D contains an induced hexagon or some realization does.
We consider these two cases. First, assume no realization of D contains an induced
hexagon. Since D excludes (2, 2, 2, 2) and (2, 2, 2, 2, 2), we also see that no realiza-
tion contains an induced square or pentagon. By the last paragraph, no realization
contains an induced cycle on 7 or more vertices. Together, this implies that no real-
ization contains a hole. In other words, every realization is a chordal graph, and D
is forcibly chordal as claimed.
Next, assume some realization G of D contains a hexagon C. Then D excludes D(C5)
but not D(C6). Let A and B be the sets of vertices not in C that are complete and
anti-complete, respectively, to C. Then by Lemma 3.0.36 with k = 6, we see that G
is the half join of the split graph G[A∪B] and the cycle C, thus completing the proof
of the lemma.
Theorem 3.0.38. A degree sequence D excludes (2, 2, 2, 2) iff one of the following
conditions holds:
(i) D is forcibly chordal.
(ii) D is the degree sequence of the half join of a split graph and a pentagon.
(iii) D is the degree sequence of the half join of a split graph and a hexagon.
38
Proof. We know by Theorem 3.0.29 and Theorem 3.0.37 that if D is one of the three
above mentioned types of degree sequences then D excludes (2, 2, 2, 2). We show the
converse.
D excludes (2, 2, 2, 2, 2) or D does not exclude (2, 2, 2, 2, 2). Suppose it does. Then
by Theorem 3.0.37, it follows that D is forcibly chordal or D is the degree sequence
of the half join of a split graph and a hexagon. The lemma is thus proved in this
case.
We therefore assume D does not exclude (2, 2, 2, 2, 2). Then there is a realization G
of D containing a pentagon C as an induced subgraph. Let A and B be the vertices
of G − C that are complete and anti-complete to C, respectively. Since D excludes
(2, 2, 2, 2) = D(C4) by hypothesis, we see from Lemma 3.0.36 with k = 5 that G is
the half join (G[A ∪B], A,B,C). This completes the proof.
By noting the complement of M2 is C4, we get a theorem characterizing degree se-
quences excluding M2 as a corollary.
Theorem 3.0.39. A degree sequence D excludes (1, 1, 1, 1) iff one of the following
conditions holds:
(i) D is forcibly anti-chordal.
(ii) D is the degree sequence of the half join of a split graph and a pentagon.
(iii) D is the degree sequence of the half join of a split graph and K3,3.
39
Proof. Just take complements, use Theorem 3.0.38, and note anti-chordal graphs are
the complements of chordal graphs by definition, the pentagon is self-complementary,
and the complement of a hexagon has the same degree sequence as K3,3.
We next wish to generalize the characterization of degree sequences excluding a square
found in Theorem 3.0.38 to longer cycles.
Theorem 3.0.40. Let n ≥ 4. A degree sequence D excludes the degree sequence of
Cn iff one of the following conditions holds:
(i) No realization of D has a chordless cycle on n or more vertices.
(ii) D is the degree sequence of the half join of a split graph and Cn+1.
(iii) D is the degree sequence of the half join of a split graph and Cn+2.
Proof. We prove this essentially by generalizing the proof of Theorem 3.0.38. By
similar reasoning to that in previous proofs, it is easy to see degree sequences in the
above three classes exclude the degree sequence of Cn. We prove the converse.
So, let D exclude the degree sequence of Cn. We must show D falls into one of the
above three classes as claimed.
First, note D excludes the degree sequence of Cn+k for all k ≥ 3. To see this, assume
not. The degree sequence of Cn+k is the same as the degree sequence of the disjoint
union of Cn and Ck. Ck exists since k ≥ 3 by assumption. So D does not exclude the
degree sequence of this disjoint union. Therefore D has a realization G containing
40
the disjoint union as an induced subgraph. In particular, G contains an induced Cn,
contrary to assumption that D excludes the degree sequence of Cn. This contradiction
proves our claim.
Next, we break into cases. The first case we consider is that D excludes the degree
sequence of Cn+1 and Cn+2. D excludes the degree sequence of Cn by hypothesis,
and by the previous paragraph, D excludes the degree sequence of Cn+k for k at least
three. Therefore D excludes the degree sequence of all cycles on at least n vertices.
Therefore, as claimed, no realization has a chordless cycle on n or more vertices.
The other case is that D does not exclude both the degree sequence of Cn+1 and Cn+2.
Then D has a realization G containing either Cn+1 or Cn+2 as an induced subgraph.
Let k = n + 1 if G contains an induced Cn+1 and let k = n + 2 otherwise. In either
case, G contains an induced Ck but no induced Ck−1. Let Ck = C. Then it follows
by Lemma 3.0.36 that G is the half join (G[A∪B], A,B,C) of G[A∪B] and C, thus
completing the proof.
41
CHAPTER 4
FORCIBLY CHORDAL TREES
In the last chapter, we proved, among other things, a structure theorem for the
degree sequences excluding the degree sequence of a square. However, this structure
theorem is not sharp, for it is stated in terms of the class of forcibly chordal graphs,
and it is in no way obvious a priori what this class of graphs is. While the forcibly P
degree sequences have been determined for many properties P (see [11] and [12] for
excellent surveys), the forcibly chordal graphs have not, to our knowledge, yet been
characterized. As such, the next natural step is to give a precise characterization of
the forcibly chordal graphs. As the first major step in this process, in this chapter,
we characterize the forcibly chordal trees.
The following lemma is used freely without comment.
Lemma 4.0.41. Let H be an induced subgraph of G and let H ′ have the same degree
sequence as H. Then there is a graph G′ with the same degree sequence as G such
that H ′ is an induced subgraph of G′.
Proof. Enumerate the vertices of G as v1, . . . , vn, and assume, without loss of gener-
ality, that there is some i with 1 ≤ i ≤ n such that H = G[v1, . . . , vi]. Since H and
42
H ′ have the same degree sequence, it follows that there is a sequence of switchings in
the vertices v1, . . . , vi that transforms H into H ′. This same sequence of switchings,
applied with G instead of H as the starting graph, produces a graph G′ that contains
H ′ as an induced subgraph by construction.
The following lemma is trivial, but worth explicitly stating as we freely use it without
comment.
Lemma 4.0.42. If G and G′ have the same degree sequence then G is forcibly chordal
iff G′ is.
Proof. Immediate from the definition of forcible chordality.
To fix terminology before stating the next lemma, we note that for us, Pn denotes an
n vertex path, not an n edge path. We will only use parts (i) and (ii) of the following
lemma in this chapter, and the reader may skip the other parts until they are later
needed, but we include all parts listed here for easy reference.
Lemma 4.0.43. Forcibly chordal graphs exclude the following graphs as induced sub-
graphs:
(i) P6
(ii) Trees such that at least three neighbors of some vertex are nonleaves.
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(iii) P5
∐P3
(iv) 4P3
(v) 2P4
(vi) P4
∐2P3
(vii) 2C3
Proof. We prove the contrapositives of these statements.
For (i), suppose G contains an induced P6. P6 has the same degree sequence as the
disjoint union H of a square and an edge, therefore by Theorem 4.0.41, there is a
graph G′ with the same degree sequence as G such that G′ contains H, and hence
a square, as an induced subgraph. Therefore G′ is not forcibly chordal, so G is not
forcibly chordal either, thus proving (i).
For (ii), suppose G contains such an induced tree T . Let r be a vertex of T with three
nonleaf neighbors. Consider T as a rooted tree with root r. Let x, y, and z be three
nonleaf neighbors of r. Let x′, y′, and z′ be successors of x, y, and z, respectively.
Then G[r, x, y, z, x′, y′, z′] = H has degree sequence (3, 2, 2, 2, 1, 1, 1). But this is also
the degree sequence of a graph H ′ isomorphic to a 6 point path with a pendant vertex
adjoined in the middle. If G contains T then by Theorem 4.0.41, there is a graph G′
containing H ′ such that G and G′ have the same degree sequence. Since H ′ contains
P6, it follows that G′ contains P6 as well. By part (i), it follows that G′ is not forcibly
chordal. Therefore G is not forcibly chordal, proving part (ii).
44
For (iii), suppose G contains H = P5
∐P3. H has the same degree sequence as
H ′ = P6
∐P2, therefore there is a graph G′ with the same degree sequence as G such
that G′ contains H ′, and hence G′ contains P6. By part (i), it follows that G′ is not
forcibly chordal. Therefore G is not forcibly chordal.
For (iv), suppose G contains H = 4P3. H has the same degree sequence as H ′ =
C4
∐4P2, therefore there is a graph G′ with the same degree sequence as G such
that G′ contains H ′, and hence G′ contains C4. Therefore G′ is not forcibly chordal.
Therefore G is not forcibly chordal.
For (v), suppose G contains H = 2P4. H has the same degree sequence as H ′ =
C4
∐2P2, therefore there is a graph G′ with the same degree sequence as G such
that G′ contains H ′, and hence G′ contains C4. Therefore G′ is not forcibly chordal.
Therefore G is not forcibly chordal.
For (vi), note P4
∐2P3 has the same degree sequence as P6
∐2P2.
For (vii), note 2C3 has the same degree sequence as C6.
In order to state the characterization of forcibly chordal trees, we first need to define
the term X-stars. Intuitively, an X-star is obtained from a graph X by letting a star
grow out of each vertex.
Definition 4.0.44. Let X and G be graphs. G is called an X-star if there is a subset
Y of V (G) such that G[Y ] is isomorphic to X and such that each vertex of G− Y is
adjacent to exactly one vertex of Y and no other vertices. The set Y is called the root
set, and the vertices of Y will be referred to as the root vertices. Edges with exactly
one end in Y will be called pendant edges. A P1-star is called a star.
45
The following lemmas will be used in characterizing the forcibly chordal trees.
Lemma 4.0.45. A switching in a P3-star yields a P3-star or the disjoint union of a
C3-star and an edge.
Proof. Take a P3-star G with root vertices u, v, and w such that v is adjacent to u
and w, and hence u and w are non-adjacent. Now consider an arbitrary switching in
G, and let G′ denote the graph obtained from G by this switching. We must show
G′ is a P3-star or the disjoint union of an edge and a C3-star.
Suppose first that neither edge of the switching is a pendant edge. Then one of the
edges is uv and the other edge is vw. But these edges are not disjoint, hence can
not determine a switching. Therefore either one or two edges of the switching are
pendant edges. We consider these two cases.
First, assume exactly one edge determining the switching is a pendant edge. Then
we may assume, without loss of generality, that the nonpendant edge determining
the switching is uv. The pendant edge of the switching must then be disjoint to
uv, therefore the root vertex of the pendant edge of the switching is w. Let wx be
the pendant edge of the switching. Then either wu or wv is one of the nonedges
determining the switching, and v and w are adjacent by hypothesis, therefore wu is
one of the nonedges of the switching, which implies vx is the other nonedge of the
switching.
We claim this switching yields a P3-star with R = {u, v, w} as the root set. To see
this, we must show first that G′[R] is isomorphic to P3, and second, that in G′, every
46
vertex outside R is adjacent to exactly one vertex of R and no other vertices. Now,
the edge vw remains unaffected by the switching, u and v become nonadjacent after
switching, and v and w become adjacent after switching. Hence R is isomorphic to
P3 after switching. Now take a vertex y outside R. Note that by definition of an
X-star, before the switching each such y is adjacent to exactly one vertex of R and
no other vertices. If y 6= x then the neighborhood of y is unchanged by the switching.
If y = x then after switching, y is adjacent to v and only v, and v is in R. Therefore
in either case, after switching, y is adjacent to exactly one vertex of R and no other
vertices, as required. This proves a switching with exactly one pendant edge yields a
P3-star.
Next, assume both edges of the switching are pendant edges. Since the edges of
the switching are disjoint, they can not have the same root vertex. Let us call the
root vertices of the pendant edges of the switching r and r′. rr′ is a nonedge of the
switching or it is not. We consider these two cases.
First, suppose rr′ is a nonedge of the switching. Then rr′ is in particular a nonedge
in G, therefore neither r nor r′ equals v, so without loss of generality, r = u, r′ = w,
the edges of the switching are rx and r′y for some x, y, and the nonedges of the
switching are rr′ and xy. We claim switching yields the disjoint union of a C3-star
and an edge with R = {u, v, w} as the root set.
Now, x and y lie outside the root set R of G, and hence have degree 1 in G. Switching
does not affect the degree of any vertex, therefore x and y have degree 1 in G′ as
well. Since x and y are adjacent to one another in G′, it follows that G′ is the disjoint
union of the edge xy and G′ − xy. We show G′ − xy is a C3-star.
47
Toward that end, note that uw is an edge of G′, and uv and vw are unaffected by the
switching, so R is indeed isomorphic to C3 in G′ − xy. Now take any vertex of z in
G′− xy−R. Then z lies outside R and hence is a pendant vertex of G. Therefore in
G, z is adjacent to exactly one vertex of R and no other vertices. Since no vertices
outside R∪{x, y} have their neighborhoods changed by the switching, it follows that
in G′ too, z is adjacent to exactly one vertex of R and no other vertices. Therefore
G′ − xy is a C3-star as claimed.
Second, suppose rr′ is not a nonedge of the switching. Then the edges of the switching
are rx and r′y and the nonedges of the switching are ry and r′x. In this case, switching
essentially just renames x and y, and G′ is isomorphic to G, hence G′ is a P3-star.
Thus in all cases, G′ is a P3-star or the disjoint union of an edge and a C3-star,
completing the proof.
Lemma 4.0.46. A switching in the disjoint union of a C3-star and an edge yields a
P3-star or the disjoint union of a C3-star and an edge.
Proof. Let G be the disjoint union of an edge xy and a C3-star with root set R =
{u, v, w}. Take a switching S in G and let G′ be the graph arising from G after
applying S. We show G′ is a P3-star or the disjoint union of an edge and a C3-star.
We break into cases by which edges and nonedges of G determine the switching S.
First, suppose xy is not in the switching S. Then both edges of S lie in the C3-star
G−xy. The root set R is isomorphic to C3, so any two edges of the root set intersect.
The two edges of S, on the other hand, are disjoint, therefore at most one edge of S
48
is contained in the root set R. If some edge of S does lie in R then without loss of
generality it is uv. The other edge of S must be disjoint to uv, so its root vertex is w.
But then wu or wv must be a nonedge of S, contrary to the fact that w is adjacent
to both u and v in G. Therefore neither edge of S lies in R.
Since neither edge of S lies in R then both edges of S are pendant edges, and since
they are disjoint, they must have different root vertices. Without loss of generality,
we may assume the edges of the switching are ua and vb for some a and b. Since u
and v are adjacent in G, uv is not a nonedge of S, therefore the nonedges of S are ub
and va. Therefore G′ essentially arises from G by renaming a and b, and hence G and
G′ are isomorphic. In particular, G′ is the disjoint union of an edge and a C3-star as
desired.
Next, suppose xy is an edge of the switching S. The other edge of the switching is
either a pendant edge or an edge contained in R. We consider these two cases.
First, assume the other edge of the switching is a pendant edge, without loss of
generality ua for some a. Without loss of generality, ux and ay are the nonedges of
the switching. This switching essentially just renames a and x, therefore G and G′
are isomorphic, and in particular, G′ is the disjoint union of an edge and a C3-star,
as desired.
Finally, assume the edge of S other than xy is a nonpendant edge, which hence lies
in R. Without loss of generality, this edge is uv. Then xy and uv are the edges of S
and without loss of generality, xu and yv are the nonedges. Trivial checking shows
that G′ is a P3-star with root set {u, v, w}.
49
Thus G′ is a P3-star or the disjoint union of an edge and a C3-star in all cases,
completing the proof.
Corollary 4.0.47. If G is a P3-star and G′ has the same degree sequence as G then
G′ is a P3-star or the disjoint union of a C3-star and an edge.
Proof. Since two graphs have the same degree sequence iff one can be obtained from
the other by a sequence of switchings, it is enough to show that every switching in a
P3-star yields a P3-star or the disjoint union of a C3-star and an edge, and that every
switching in the disjoint union of a C3-star and an edge yields either a P3-star or the
disjoint union of a C3-star and an edge. These facts are exactly the last two lemmas.
Corollary 4.0.48. P3-stars are forcibly chordal.
Proof. Let G be a P3-star. We must show every realization of the degree sequence
of G is chordal. Let G1, . . . , Gn be a sequence of graphs such that G = G1 and Gi+1
arises from Gi by a switching in Gi for 1 ≤ i < n. It is enough to show by induction
that each Gi is chordal. Noting that P3-stars and the disjoint union of an edge and
a C3-star are both chordal, it is enough to show by induction that each Gi falls into
one of these two classes. G1 = G is a P3-star or the disjoint union of an edge and
a C3-star by assumption. If Gi is a P3-star or the disjoint union of an edge and a
50
C3-star then, by Theorem 4.0.45 and Theorem 4.0.45, it follows that Gi+1 is as well,
thus completing the proof.
Finally, we can use the results just proved to characterize the forcibly chordal trees.
Theorem 4.0.49. The forcibly chordal trees are exactly the stars, P2-stars, and P3-
stars.
Proof. We must show all forcibly chordal trees are of one of these three types, and
conversely that graphs of these three types are in fact forcibly chordal.
First, we show every forcibly chordal tree T is in fact a star, P2-star, or P3-star.
Choose a root r for T . We note that by part (ii) of Theorem 4.0.43, r has at most
two nonleaf neighbors. Since r has zero, one, or two nonleaf neighbors, we consider
these three cases.
First, if r has no nonleaf neighbors, then every neighbor of r is a leaf, therefore T
consists only of r and its leaves, and T is a star.
Second, if r has exactly one nonleaf neighbor, call it v, let r = w, and then T consists
of two adjacent vertices v and w together with pendant vertices adjoined to each. By
definition, T is thus a P2-star with root set {v, w}.
Third, assume r has exactly two nonleaf neighbors. Call them u and v and let r = w.
Then T [u, v, w] is a three point path. T consists of u, v, w, and pendant vertices
adjoined to them. Therefore by definition, T is a P3-star as claimed. We have thus
shown all forcibly chordal trees are stars, P2-stars, or P3-stars as claimed.
51
For the converse, let G be a star, a P2-star, or a P3-star. We must show G is forcibly
chordal. Note that in all three cases, G is an induced subgaph of a P3-star H. H is
forcibly chordal by Theorem 4.0.49, and an induced subgraph of a forcibly chordal
graph is forcibly chordal, therefore G is forcibly chordal as claimed.
It is worth restating the previous theorem in two equivalent ways.
Corollary 4.0.50. The forcibly chordal trees are exactly the X-stars for which the
graph X is a tree on at most three elements.
Corollary 4.0.51. The forcibly chordal trees are exactly the trees with at most three
vertices of degree at least two.
Proof. Let T be a tree. If T has at most three vertices of degree at least two then the
same is true of every graph G with the same degree sequence as T . Such a graph G
does not have enough vertices of degree at least two to contain a hole. Therefore G
is chordal. Since G is an arbitrary graph with the same degree sequence as T , we see
that T is forcibly chordal.
Conversely, let T be a tree with at most three vertices of degree at least two. If T is
a single vertex or an edge then T is trivially forcibly chordal as needed, so suppose
not. Let X be the set of vertices v in T such that dT (v) ≥ 2, so that X has at most
three vertices by hypothesis. Take any vertex y not in X. Then y must have degree
at least one since T is a tree that is not itself an isolated vertex, and y must have
52
degree at most one since y is not in X, so dT (y) = 1. Let z be the unique vertex of
T adjacent to y. If z has degree 1 also then {y, z} is a component of T , contrary to
assumption that T is not an edge. Therefore z has degree at least two, which means
z is in X. Since y is an arbitrary vertex of V (T )−X, we see that T is an X-star.
We need only show T [X] is a tree. T [X] is acyclic since T is acyclic. To show T [X] is
connected, let v and w be disntinct vertices of X, and let let v1, . . . , vn be the unique
T -path from v to w, with v = v1 and w = vn. Since all interior points of this path
have degree at least two, they must all lie in X. Since v and w are arbitrarily chosen
distinct vertices of X, we see that T [X] is connected. Therefore T [X] is a tree, and
the proof is complete.
53
CHAPTER 5
FORCIBLY CHORDAL FORESTS
In the last chapter, we characterized the forcibly chordal trees. In this chapter, we use
this characterization to characterize the forcibly chordal forests. It should be noted
this is not as simple as saying, “the forests whose components are forcibly chordal
trees”. That condition is obviously necessary, but it is not sufficient. For instance,
the forest 2P4 is the disjoint union of two forcibly chordal trees, but has the same
degree sequence as 2P2
∐C4, and is therefore not a forcibly chordal forest.
Definition 5.0.52. Let m ≥ 0 and let G be a graph. Then mG denotes the disjoint
union of m copies of G.
Lemma 5.0.53. Let G = H∐mP2 where H is an X-star with root set R. Let G′
have the same degree sequence as G. Then there is m′ ≥ 0 and graphs H ′, X ′ such
that G′ = H ′∐m′P2 and H ′ is an X ′-star with the same root set R.
Proof. Since G and G′ have the same degree sequence, we may assume without loss
of generality that V (G) = V (G′). It is immediate from the definitions of matching
and root set that dG(x) = 1 for all vertices x in V (G)−R. Since G and G′ have the
54
same degree sequence, we see that dG′(x) = 1 for all x in V (G′)− R as well. Let M
be the set of all x in V (G′) − R whose unique neighbor is also in V (G′) − R. Then
G[M ] is a matching. Moreover, it then follows that the unique neighbor of every x
in V (G′) − R −M is in R. The lemma then follows with m′ = |M |/2, X ′ = G′[R],
and H ′ = G′[V (G)−M ].
Lemma 5.0.54. For 1 ≤ i ≤ n, let Gi be an Xi-star. Then∐n
i=1Gi is a∐n
i=1Xi
star.
Proof. Immediate from the definition of X-star.
Theorem 5.0.55. A forest is forcibly chordal iff it is of one of the following types:
(i) H∐iP2
∐jP1, where H is a P3-star and i, j ≥ 0.
(ii) H∐iP2
∐jP1, where H is a P2-star, and i, j ≥ 0.
(iii) H∐iP2
∐jP1, where H is the disjoint union of a P1-star and a P2-star, and
i, j ≥ 0.
(iv)(⋃m
n=1Hn
)∐iP2
∐jP1, where i, j ≥ 0, each Hn is a P1-star, and 0 ≤ m ≤ 3.
Proof. First, we show that if G is of one of these four types then G is a forcibly chordal
forest. Obviously, they are all forests. Note that since adding or removing isolated
55
vertices does not affect the forcible chordality of a graph, we may assume j = 0. Note
also that by the last lemma, a graph as in part (iii) is the disjoint union of a P1
∐P2-
star and a matching, and a graph as in (iv) is the disjoint union of an mP1-star and
a matching for some m with 0 ≤ m ≤ 3. Therefore in cases (i), (ii), (iii), and (iv),
the graph is the disjoint union of a matching and an X-star, where X is a graph on
at most 3 vertices. Moreover, by Lemma 5.0.53, it follows that any graph with the
same degree sequence as one of these four types is the disjoint union of a matching
and such an X-star as well. It is easy to see that if |X| ≤ 3 then the disjoint union
of an X-star and a matching is chordal. Since all realizations of the four classes of
graphs listed in the lemma arise as just such a disjoint union with |X| ≤ 3, it follows
that these forests are all in fact forcibly chordal.
For the converse, we show that if G is a forcibly chordal forest then G is of one of
the four types claimed in the theorem, so take a forcibly chordal forest G. Note each
component of G is a forcibly chordal tree. Therefore by Theorem 4.0.49, we know
each component of G is a star, a P2-star, or a P3-star. (It is worth noting at this
point that an X-star is not assumed to have pendant vertices, so by our definition,
an isolated vertex is a star.)
First, assume one of the components C is a P3-star that is not a P2-star. Then
C contains P5. If G has another component B such that B is not an edge or an
isolated vertex then B contains P3. Therefore G contains P5
∐P3. By part (iii) of
Theorem 4.0.43, we see that G is not forcibly chordal, contrary to assumption. This
contradiction shows B must be an edge or isolated vertex. Since B was an arbitrarily
chosen component of G other than C, it follows that G is the disjoint union of C, a
56
(possibly empty) set of edges, and a (possibly empty) set of isolated vertices, as in
part (i) of this lemma, thus the lemma is proved in the case that G has a component
that is a P3-star but not a P2-star.
So, we may now assume all components of G are either stars or P2-stars. Suppose G
has at least two P2-star components that are not stars. Then each such component
contains P4, therefore G contains 2P4. Therefore by part (v) of Theorem 4.0.43, it
follows that G is not forcibly chordal, contrary to assumption. This contradiction
shows G has only one P2-star component C that is not a star. Suppose G contains
two star components other than C that are not edges or isolated vertices. Each
such star component contains P3. Therefore G contains P4
∐2P3, so by part (vi)
of Theorem 4.0.43, it follows that G is not forcibly chordal, contrary to assumption.
This contradiction shows that G has at most one component other than C that is
not an edge or an isolated vertex, and if there is one, then it is a star. If there is no
such star component other than the edges and isolated vertices, we are in case (ii) of
this theorem. Otherwise, we are in case (iii). Hence in either case, we are done.
So we may now assume all components of G are stars. If G has at least 4 components
that are not edges or isolated vertices then G contains 4P3, therefore by part (iv)
of Theorem 4.0.43, it follows that G is not forcibly chordal, which is contrary to
assumption. This contradiction shows there are at most three components that are
not edges or isolated vertices. Therefore G is as claimed in part (iv) of this lemma,
and the proof is complete.
Just as we stated the characterization of forcibly chordal trees in two equivalent ways
57
after the main theorem of the last chapter, we similarly give two equivalent statements
of the above theorem now.
Corollary 5.0.56. The forcibly chordal forests are exactly the graphs of the form
H∐
iP2
∐jP1
where H is an X-star for which the graph X is a forest on at most three elements.
Corollary 5.0.57. The forcibly chordal forests are exactly the forests with at most
three vertices of degree at least two.
58
CHAPTER 6
FORCIBLY CHORDAL GRAPHS
In this chapter, we continue the partial characterization of forcibly chordal graphs.
First, we prove results concerning the structure of G[X ∪ Y ], where X and Y are
maximal cliques of the forcibly chordal graph G. We will show that G[X ∪ Y ] has
one of two structures, depending on whether or not there are edges from X − Y to
Y −X.
These structures, in addition to yielding interesting results in their own right, will be
used to motivate a definition of adjacency of maximal cliques in a forcibly chordal
graph. We believe that this definition of adjacency can, with further research, be
used to give a complete characterization of connected, forcibly chordal graphs. As
a partial result toward this end, we characterize forcibly chordal graphs that have a
path structure to be defined below.
First, we note some basic definitions and results regarding chordal graphs to be used
in the course of the chapter. The following definition and Proposition 6.0.59 may be
found in [3].
Definition 6.0.58. Let G and H be graphs such that G ∩H is a complete graph K,
and such that there are no edges between V (G)− V (H) and V (H)− V (G). Then the
graph G∪H is said to arise from G and H by pasting along the complete subgraph K.
59
When it is not necessary to make K explicit, we simply say G ∪ H arises from G
and H by pasting along a complete subgraph.
Proposition 6.0.59. A graph G is chordal iff it is a clique, or can be obtained from
two smaller chordal graphs by pasting along a complete subgraph.
A universal vertex in a graph is a vertex that is adjacent to all other vertices. Our
next two lemmas show that addition or deletion of universal vertices does not affect
the forcible chordality of a graph. This will allow us the convenience of ignoring the
intersection of two cliques in certain cases, by allowing us to first prove a statement
in the case of disjoint cliques, and then add universal vertices to show the statement
is still true, mutatis mutandis, in the case that the cliques are not necessarily disjoint.
Lemma 6.0.60. Let G be a graph and suppose v in G is complete to G− v. Then G
is forcibly chordal iff G− v is forcibly chordal.
Proof. First, note that induced subgraphs of forcibly chordal graphs are forcibly
chordal so if G is forcibly chordal then so is G− v. For the converse, suppose G− v
is forcibly chordal. We must show G is also.
So let H have the same degree sequence as G. We must show H is chordal. Since G
and H have the same degree sequence, we see that v has the same degree in H as
it does in G. Therefore v is complete to H − v. Note G − v and H − v have the
same degree sequence, and G− v is forcibly chordal by assumption, therefore H − v
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is chordal. So suppose H has a chordless cycle C on at least 4 vertices. Then v is
in C. Therefore v is nonadjacent to at least one other vertex of C, contrary to the
fact that v is complete to H − v. This contradiction proves H has no chordless cycle.
H is therefore chordal, which completes the proof.
Lemma 6.0.61. Let G be a graph, A ⊆ G, and suppose v is complete to G − v for
all v in A. Then G is forcibly chordal iff G− A is forcibly chordal.
Proof. Let |A| = n. Apply the last lemma n times.
Now, we begin determining the structure of G[X∪Y ] for maximal cliques X and Y in
a forcibly chordal graph G. The first step in doing so is studying the neighborhoods
in Y of vertices in X, and the neighborhoods in X of vertices in Y . In the next lemma,
it is enough to assume that G is chordal, and we need not assume the cliques X and Y
are maximal. Eventually, we will need to assume both forcible chordality of G, and
maximality of the cliques X and Y .
Lemma 6.0.62. Let X and Y be two disjoint cliques in a chordal graph G, and
choose v and w in X. Then NG(v)∩Y and NG(w)∩Y are comparable under inclusion.
Proof. Suppose not. Let NG(v)∩ Y = Yv and NG(w)∩ Y = Yw. Then Yv and Yw are
incomparable under inclusion, so take s in Yv−Yw and t in Yw−Yv. Simple checking
from the definitions shows that s, v, w, t comprise a four cycle, in that cyclic order.
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Lemma 6.0.63. Let X and Y be disjoint, nonempty cliques in a chordal graph G.
Then there is n ≥ 1 and a partition of X into nonempty sets X1, . . . , Xn such that,
given s in Xi and t in Xj, the following conditions hold:
(i) If i = j then Y ∩NG(s) = Y ∩NG(t).
(ii) If i < j then Y ∩NG(s) ( Y ∩NG(t).
Proof. Define a relation ≤ on X by letting s ≤ s′ iff
Y ∩NG(s) ⊆ Y ∩NG(s′)
If s ≤ s′ ≤ s′′ then
Y ∩NG(s) ⊆ Y ∩NG(s′) ⊆ Y ∩NG(s′′)
Therefore
Y ∩NG(s) ⊆ Y ∩NG(s′′)
Therefore s ≤ s′′. This proves ≤ is a transitive relation. It follows by definition of ≤
that ≤ is reflexive as well.
A transitive, reflexive relation is known as a pre-order. It is a well known fact, and a
simple exercise to show, that if we define s ∼ s′ iff s ≤ s′ and s′ ≤ s for a pre-order
≤ on X then ∼ is an equivalence relation on X, and that ≤ induces a partial order
≤2 on the set X/ ∼ of ∼ equivalence classes. Namely, S ≤2 S′ for equivalence classes
S and S ′ iff s ≤ s′ for some s ∈ S and for some s′ ∈ S ′.
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Now take two equivalence classes S and S ′, and choose representatives s and s′,
respectively. By Lemma 6.0.62, we know that Y ∩NG(s) ⊆ Y ∩NG(s′) or Y ∩NG(s′) ⊆
Y ∩NG(s). Therefore s ≤ s′ or s′ ≤ s. Therefore S ≤2 S′ or S ′ ≤2 S. This shows ≤2
is in fact a total order on the ∼ classes.
Since X is finite and nonempty, there are finitely many nonempty ∼ classes. Let X1
be the ≤2 least ∼ class. Having defined X1, . . . , Xi for some i less than n, let Xi+1
be the ≤2 least ∼ class not already chosen as some Xj. The result follows with this
choice of X1, . . . , Xn by definition of ≤ and ≤2.
Lemma 6.0.64. Let X and Y be disjoint, maximal cliques in a forcibly chordal
graph G. If X is partitioned into sets X1, . . . , Xn as in the previous lemma and Y is
similarly partitioned into sets Y1, . . . , Ym then m = n ≤ 2, and if n = 1 then there
are no edges from X to Y .
Proof. We know the stated partition into Xi’s exists by the previous lemma. For
each i, choose si in Xi. Then for i < j, the neighborhood of si in Y is a proper subset
of the neighborhood of sj in Y .
Suppose n ≥ 3. Then there is t1 in Y adjacent to sn−1 but not sn−2, and there is t2
in Y adjacent to sn but not sn−1 or sn−2. Since Y is a maximal clique, it follows that
no vertex of X is complete to Y . In particular, sn is not complete to Y . Therefore
there is t3 in Y such that t3 and sn are not adjacent. Since sn has the largest possible
neighborhood in Y under inclusion of all vertices of X, it follows that t3 is not adjacent
to sn−1 or sn−2 either.
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Consider H = G[sn−2, sn−1, sn, t1, t2, t3]. Simple checking shows that dH(sn−2) =
2, dH(sn−1) = 3, dH(sn) = 4, dH(t1) = 4, dH(t2) = 3, and dH(t3) = 2. Thus
D(H) = (4, 4, 3, 3, 2, 2), which by simple checking, is also the degree sequence of
H ′ = C6+v1v3+v1v4+v2v4, where C6 is taken to be the cycle on vertices v1, v2, . . . , v6
in that cyclic order. Trivial checking showsH ′[v1, v4, v5, v6] is a four cycle in that cyclic
order. Therefore H ′ is not chordal. Since H and H ′ have the same degree sequence,
it follows that H is not forcibly chordal. Since H is an induced subgraph of G, it
follows G is not forcibly chordal, contrary to assumption. This contradiction shows
that n ≤ 2 as the lemma claims.
We must now show that m = n, and that if n = 1 then there are no edges from X
to Y . We consider the cases n = 1 and n = 2.
First, suppose n = 1. Then X = X1, and any two vertices of X therefore have the
same neighborhood in Y . Suppose this neighborhood contains a vertex t in Y . Then t
is adjacent to every vertex of X, so X ∪ {t} is a clique. Since X and Y are assumed
disjoint, it follows that t is not in X, therefore X ∪{t} is a clique properly containing
the maximal clique X, a contradiction. This contradiction shows the neighborhood
in Y of each vertex of X is empty. It then follows that the neighborhood in X of
each vertex of Y is empty, and thus, as the lemma claims, there are no edges from X
to Y . Moreover, since all vertices of Y have the same neighborhood in X, namely
the empty set, we see that m = 1 as claimed.
Second, assume n = 2. We must show m = 2. The same argument we used to
show n ≤ 2 also shows that m ≤ 2. If m = 1 then by symmetry, the argument used
64
in the last paragraph shows that n = 1, contradiction. Therefore m = 2 as claimed,
and the proof is complete.
Lemma 6.0.65. Let X and Y be disjoint, maximal cliques of a forcibly chordal
graph G, let X1, . . . , Xn and Y1, . . . , Yn be as in the previous lemma, and 1 ≤ i, j ≤ n.
Then Xi is complete to Yj or anti-complete to Yj.
Proof. Suppose Xi is not anti-complete to Yj. It is enough to show Xi is complete
to Yj. Since Xi is not anti-complete to Yj, we can choose x in Xi and y in Yj
such that x is adjacent to y. Now take x′ in Xi and y′ in Yj. Since x′ and y′ are
arbitrary, it is enough to show x′ is adjacent to y′. The vertex x is adjacent to y by
assumption. By definition of Xi, any two vertices of Xi have the same neighborhood
in Y . Therefore x′ is adjacent to y as well. Since y is adjacent to x′ and since any two
vertices of Yj have the same neighborhood in X, it follows that y′ is adjacent to x′ as
well. This completes the proof.
Corollary 6.0.66. Let X and Y be disjoint, maximal cliques of a forcibly chordal
graph G with at least one edge between them. Then X can be partitioned into cells A
and B, and Y into cells C and D, such that the following conditions hold:
(i) A is anti-complete to C
(ii) A is anti-complete to D
65
(iii) B is anti-complete to C
(iv) B is complete to D
(v) A is complete to B
(vi) C is complete to D
Proof. We have two disjoint, maximal cliques X and Y as in the previous lemma, so
take n = m as in the statement of Lemma 6.0.63. Lemma 6.0.64 shows that n ≤ 2,
and that if n = 1 then there are no edges from X to Y . Since we have at least one
edge from X to Y by assumption, we see that n = 2. Let A = X1, B = X2, C = Y1,
and D = Y2, so that {A,B} is a partition of X and {C,D} is a partition of Y . We
show this choice of A, B, C, and D satisfies the desired properties.
First, note that A is trivially complete toB since they are both subsets of the cliqueX,
and C is trivially complete to D since they are both subsets of the clique Y . These
two facts were simply stated for easy reference. We now prove the remaining four.
We know by the previous lemma that A is complete or anti-complete to D. Suppose A
is complete to D. Since the neighborhood of B in Y contains the neighborhood of A
in Y , it follows that B is complete to D as well. Therefore D is complete to X.
Since D is nonempty, it follows that X ∪D is a clique strictly containing X, contrary
to choice of X as maximal. This contradiction shows that A is anti-complete to D.
By symmetry, we also know that C is anti-complete to B. Since C is anti-complete
to B, and since the neighborhood of A in Y is contained in the neighborhood of B
in Y , it follows that A is anti-complete to C.
66
Suppose B is anti-complete to D. Since we’ve already shown A is anti-complete
to C, A is anti-complete to D, and C is anti-complete to B, it follows that X is
anti-complete to Y , contrary to assumption. This contradiction shows B is not anti-
complete to D. Therefore B is complete to D by the previous lemma, thus completing
the proof.
Corollary 6.0.67. Let X and Y be disjoint, maximal cliques of a forcibly chordal
graph G such that there is at least one edge between them. Then there is a in X and c
in Y such that the following conditions hold.
(i) a is not adjacent to c.
(ii) a is anti-complete to Y − c
(iii) X − a is anti-complete to c
(iv) X − a is complete to Y − c
(v) a is complete to X − a
(vi) c is complete to Y − c
Proof. The hypotheses of this corollary are the same as those of the one preceding
it. First, let A, B, C, and D be as in the previous corollary. We show A and C have
exactly one vertex.
First, note that there is an edge from X to Y by hypothesis, and by the last lemma,
this edge must be from B to D. Therefore B and D are nonempty. Suppose A is
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empty. Then X = B. Since B is complete to D, it follows that X is complete to D.
Since D is nonempty, it follows that X ∪D is a clique strictly containing X, contrary
to maximality of X. This contradiction shows that A is nonempty. By symmetry, C
is nonempty as well.
Suppose A has at least two vertices. Call them a and a′. We have already noted
that B, C, and D are nonempty, so take b in B, c in C, and d in D. The reader may
check that a, a′, and b are pairwise adjacent, d is adjacent to b and c, and there are
no other edges between these 5 vertices. Therefore G[a, a′, b, c, d] has degree sequence
(3, 2, 2, 2, 1), which is also the degree sequence of a 4 cycle with a pendant vertex.
A 4 cycle with a pendant verex is of course not chordal, therefore G is not forcibly
chordal, contrary to hypothesis. This contradiction shows A does not have at least
two vertices. Since A is nonempty, it follows that A has exactly one vertex as claimed.
By symmetry, C has exactly 1 vertex as well.
Since A and C are singletons, it follows that A = {a} and C = {c} for some a and c.
Then B = X − a and D = Y − c. The reader may check that the six conclusions
of this corollary are exactly the six conclusions of the previous corollary for these
choices of A, B, C, and D. The result follows.
The following remark is key to what follows. The previous lemma has a converse as
well, in the sense that if vertices a and c and cliques X and Y in a graph G satisfy the
six conclusions of the lemma then G[X ∪ Y ] is a forcibly chordal graph. To see this,
note that those six statements imply G[X ∪ Y ] is a split graph, and is thus forcibly
68
chordal. Therefore the above lemma can be considered the full structure theorem for
two disjoint, maximal cliques in a forcibly chordal graph G.
The next proposition, a corollary of the previous, generalizes this structure to the case
that the cliques X and Y are not necessarily disjoint, and can similarly be considered
the full structure theorem in that somewhat more general setting.
Proposition 6.0.68. Let X and Y be maximal, not necessarily disjoint cliques of
a forcibly chordal graph G such that there is at least one edge between X − Y and
Y −X. Then there is a in X and c in Y such that the following conditions hold.
(i) a is not adjacent to c.
(ii) a is anti-complete to Y −X − c
(iii) X − Y − a is anti-complete to c
(iv) X − Y − a is complete to Y −X − c
(v) a is complete to X − Y − a
(vi) c is complete to Y −X − c
Proof. This follows immediately from the previous lemma by noting that X −Y and
Y −X are disjoint, maximal cliques of the graph G− (X ∩ Y ).
Note that in the previous lemma, we referred to edges from X − Y to Y −X. This
comes up often enough that the following definition will simplify the wording of a
number of statements to come.
69
Definition 6.0.69. Let X and Y be cliques in a graph G. We say that X and Y
have extra edges between them if there is at least one edge from X − Y to Y − X.
Otherwise, the cliques are said to have no extra edges between them.
So, the previous lemma gave the structure theorem for two maximal cliques in a
forcibly chordal graph in the case that there is at least one extra edge between them.
Now that we have a structure theorem for the union of two cliques in this case, we
wish to do the same in the case that there are no extra edges between them.
Proposition 6.0.70. Let X and Y be two distinct, maximal cliques of a graph G
such that V (G) = X∪Y , and such that there are no extra edges from X to Y . Then G
is forcibly chordal iff |X − Y | ≤ 2 or |Y −X| ≤ 2.
Proof. Note that every vertex v of X ∩ Y is complete to G − v. Since addition or
deletion of universal vertices does not affect the forcible chordality of a graph, we
may assume without loss of generality that X ∩ Y = ∅. So X and Y are disjoint,
maximal cliques of G. Therefore X − Y = X and Y − X = Y , so we may simply
refer to X and Y throughout instead of X −Y and Y −X, and refer simply to edges
from X to Y instead of extra edges from X to Y .
Suppose first that |X| = 2. Then G is the disjoint union of a clique Kn and an edge.
We must show G is forcibly chordal. We do this by showing there are in fact only two
graphs up to isomorphism with the same degree sequence as G, and noting that each
of these graphs is chordal. So let H be a graph with the same degree sequence as G.
70
Then H has exactly two vertices x and y of degree 1. Either x and y are adjacent or
they are not. We consider these two cases.
First, assume x and y are adjacent. Since D(H) = D(Kn
∐P2), we see the remain-
ing n vertices of H each have degree n−1. Since x and y have degree 1 in G and they
are adjacent to one another, we see they are adjacent to no other vertices. Therefore
for any vertex z in H − {x, y}, the neighborhood of z in H is contained in the n− 1
vertex set H − {x, y, z}. Since z has degree n − 1, it follows that z is complete to
H−{x, y, z}. Since z is an arbitrary vertex of H−{x, y}, it follows that H−{x, y} is
complete. Therefore H is isomorphic to Kn
∐P2, and this graph is trivially chordal.
Second, assume x and y are not adjacent. Each has degree 1. Suppose first that x
and y are adjacent to the same vertex z. We know dH(z) = n − 1. Therefore z is
adjacent to exactly n − 3 vertices of the n vertex set H − {x, y}. Therefore there is
some vertex w that is not adjacent to z. Then w is adjacent to at most n− 2 vertices
of H−{x, y}. Since x and y are adjacent to z and have degree 1, we know also that w
is not adjacent to x or y. Therefore dH(w) ≤ n − 2, contrary to the fact that each
vertex of H has degree 1 or n− 1. This contradiction shows that x and y can not be
adjacent to the same vertex z.
So let x be adjacent to z and y be adjacent to w. Then z and w are each adjacent
to exactly n− 2 other vertices of H − {x, y}, and every other vertex of H − {x, y} is
adjacent to n−1 vertices of H−{x, y}. Therefore H−{x, y} = Kn−zw. Therefore H
is uniquely determined up to isomorphism in the case that x and y are not adjacent.
It is trivial to check that this graph H is chordal.
71
This shows that each of the two possible realizations of the degree sequence of G is
chordal in the case that |X| = 2. Therefore G is forcibly chordal in this case.
Suppose now that the cardinality of X less than 2. Then G is contained in the disjoint
union of a clique and an edge, which is, by the previous paragraphs, forcibly chordal.
Since induced subgraphs of forcibly chordal graphs are forcibly chordal, it follows
that G is forcibly chordal in this case as well. Similarly if |Y | ≤ 2. This shows that G
is forcibly chordal if X or Y has at most two vertices, as claimed.
For the converse, we must show that G is not forcibly chordal if both X and Y have
at least three vertices. So suppose both X and Y have at least 3 vertices. Then
both X and Y contain C3. Therefore G contains 2C3 as an induced subgraph, since
there are no edges from X to Y . Therefore by part (vii) of Theorem 4.0.43, we see
that G is not forcibly chordal. This completes the proof.
By Proposition 6.0.59, every chordal graph can be obtained by starting with cliques
and repeatedly pasting along complete subgraphs. In particular, the same is true
for forcibly chordal graphs. It is therefore natural to study the structure of forcibly
chordal graphs by viewing a forcibly chordal graph as a tree of cliques.
Viewing a chordal graph as a tree of cliques is by no means new. It is well known [3]
that every chordal graph has a tree decomposition whose parts are cliques. While a
discussion or definition of tree decompositions is outside the scope of this writing, suf-
fice it to say such a tree decomposition of a chordal graph is certainly most naturally
thought of as a tree of cliques.
However, in the case of forcibly chordal graphs, we can do better, and pick out certain
72
trees as most natural. If X and Y are maximal cliques of a forcibly chordal graph G
such that X and Y have extra edges between them, we see that the graph G[X ∪ Y ]
actually has three maximal cliques. These cliques are X, Y , and Z = X ∪Y −{a, c},
in the notation of Proposition 6.0.68. Now, Z ∩X and Z ∩ Y both strictly contain
X ∩ Y . Thus intuitively, Z seems more connected to X and Y than X and Y do to
one another. We make this intuition precise in the following definition.
Definition 6.0.71. Let G be a graph. Define the graph MG of maximal cliques of G
as the graph such that
(i) V (MG) is the set of maximal cliques of G.
(ii) Maximal cliques in G are adjacent in MG iff they intersect nontrivially and have
no extra edges between them.
We wish to view each connected, forcibly chordal graph as a tree of cliques in such a
way that adjacent cliques have no extra edges between them. Such a tree exists, as
a (not necessarily induced) spanning subtree of MG, if and only if MG is connected.
Our next two lemmas show this is indeed the case.
Lemma 6.0.72. Let G be a connected, forcibly chordal graph and let X and Y be
distinct, maximal cliques with extra edges between them. Then X and Y are in the
same component of MG.
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Proof. We choose a and c as in Proposition 6.0.68. X∪Y −{a, c} is a clique. Enlarge
this to a maximal clique M in G. Suppose there is an extra edge from M to X. M
contains all vertices of X−a, so the extra edge must be from some vertex m of M−X
to a. But then X ∪{m} is a clique in G, contradicting choice of X as maximal. This
shows there are no extra edges from M to X. By symmetry, there are no extra edges
from M to Y .
But we know M intersects both X and Y . Therefore M is adjacent to both X and Y
in MG. Therefore X and Y are in the same component of MG, completing the proof.
Lemma 6.0.73. Let G be a connected, forcibly chordal graph. Then MG is connected.
Proof. Suppose not. Then MG contains at least two components. We know by the
previous lemma that any two maximal cliques in G with extra edges between them
lie in the same component of MG, so that by the contrapositive, any two maximal
cliques of G in distinct components of MG have no extra edges between them.
Suppose that X and Y are disjoint for every two maximal cliques X and Y of G
in distinct components of MG. Since any such pair also has no extra edges between
them, and since MG has at least two components, it follows that G is disconnected,
contrary to hypothesis. This contradiction shows that there must be some cliques X
and Y in distinct components of MG such that X ∩ Y is nonempty. Since X and Y
are not adjacent in MG, it follows that there are extra edges between them. Therefore
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they lie in the same component of MG by the previous lemma, contrary to hypothesis.
This contradiction proves the lemma.
Since we know that MG is connected for each forcibly chordal graph G, we may
simply take a spanning tree of MG in order to view G as a tree of cliques in such
a way that adjacent cliques have no extra edges between them. It is our belief that
this line of reasoning will lead to a complete characterization of connected, forcibly
chordal graphs. As a partial result, we characterize forcibly chordal graphs whose
cliques have a path structure. First, we rigorously define this path structure.
Definition 6.0.74. Let G be a graph, and let P be an (induced) path in MG. Then G
is said to have a P path structure if every pair of nonadjacent points X, Y in P
satisfies X ∩ Y = ∅ when considered as subsets of G.
We say that G has a path structure if G has a P path structure for some P . We
now characterize the forcibly chordal graphs with a path structure. Any path P is
isomorphic to Pn for some nonnegative integer n, so it is enough to characterize, for
each n, the forcibly chordal graphs with path structure Pn. Toward this end, we first
state a trivial lemma.
Lemma 6.0.75. A graph has a P1 structure iff it is a clique.
The above lemma, in its triviality, did not require the assumption of forcible chordal-
ity. For n ≥ 2, that assumption is needed.
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Theorem 6.0.76. The forcibly chordal graphs G with a P2 structure are exactly the
graphs G such that V (G) is the disjoint union of sets X1, X2, and X3 such that the
following conditions hold:
(i) Each Xi is nonempty.
(ii) X2 is complete to X1 and X3.
(iii) X1 is anti-complete to X3.
(iv) Either X1 or X3 has at most two vertices.
Proof. There are two directions to show. First, suppose there are three such sets X1,
X2, and X3 in a graph G. We must show G is a forcibly chordal graph with a P2
structure.
First, we show G is forcibly chordal. Note that X2 is complete to G−X2. Therefore,
by Lemma 6.0.61, we see that G is forcibly chordal iff G−X2 = G[X1 ∪X3] is. But
G[X1 ∪ X3] is the disjoint union of two cliques with no edges between them. By
Proposition 6.0.70, we know such a graph is forcibly chordal iff one of the two cliques
has at most two elements. But this is exactly condition (iv), therefore G is indeed
forcibly chordal.
Next, we show G has a P2 structure. Simple checking shows the maximal cliques of G
are exactly X1 ∪X2 and X2 ∪X3. Since X2 is nonempty, we see these two maximal
cliques have nontrivial intersection. Moreover, since X1 is anti-complete to X3, we
see the maximal cliques have no extra edges between them. So consider the path P
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whose vertex set is the doubleton {X1∪X2, X2∪X3}, and such that these two vertices
are adjacent. Then G has a P path structure by definition, as was to be shown.
This completes the proof of one direction of the lemma. For the converse, we sup-
pose G is a forcibly chordal graph with a P2 structure. We must show three sets
X1, X2, and X3 with the desired properties exist. Let X and Y be the two vertices
of P . Letting X1 = X − Y , X2 = X ∩ Y , and X3 = Y − X, we see that V (G) is
the disjoint union of X1, X2, and X3. Since X and Y are adjacent in P , we see that
X2 = X ∩ Y 6= ∅. Moreover, X and Y are distinct, maximal cliques of G. Therefore
X1 = X − Y and X3 = Y −X are nonempty, so condition (i) holds.
For condition (ii), note that since X is a clique, in particular X1 = X−Y is complete
to X2 = X ∩ Y . Similarly, X3 is complete to X2.
For condition (iii), note that since X and Y are adjacent in P , it follows that the
cliques X and Y have no extra edges between them. By definition, this means there
are no edges from X1 = X−Y to X3 = Y −X. Therefore X1 is anti-complete to X3,
as was to be shown.
For condition (iv), note that since G is forcibly chordal, so is G[X1 ∪ X3]. But
G[X1 ∪ X3] is the disjoint union of two cliques with no edges between them. By
Proposition 6.0.70, this can only be forcibly chordal if X1 or X3 has at most two
elements. This completes the proof of the lemma.
Next, we characterize the forcibly chordal graphs with a P3 structure. Intuitively,
we do this by thinking of P3 as two P2’s spliced together, and noting that each of
those P2’s must yield the structure described in the previous lemma.
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Theorem 6.0.77. The forcibly chordal graphs G with a P3 structure are exactly
the graphs G such that V (G) is the disjoint union of sets X1, . . . , X5 such that the
following conditions hold:
(i) X1, X2, X4, and X5 are nonempty.
(ii) Xi is complete to Xi+1 for 1 ≤ i ≤ 4 and X2 is complete to X4.
(iii) All other pairs of Xi and Xj are anti-complete.
(iv) X1 and X5 are singletons.
Proof. There are two directions to show. First, suppose there are five such sets
X1, . . . , X5 in a graph G. We must show G is a forcibly chordal graph with a P3
structure.
First, we show G is forcibly chordal. Note that X2 ∪ X3 ∪ X4 is a complete graph.
Also, the singletons X1 and X5 are anti-complete to each other. Therefore G is a
split graph, and in particular forcibly chordal.
Next, we show G has a P3 structure. Simple checking shows the maximal cliques of G
are exactly X = X1 ∪X2, Y = X2 ∪X3 ∪X4, and Z = X4 ∪X5. Let P be the path
with vertex set {X, Y, Z} such that X and Y are adjacent, Y and Z are adjacent,
but X and Z are not adjacent. Obviously, P is isomorphic to P3. We show G has
a P structure.
Since each Xi except possibly X3 is nonempty, we see that both X ∩ Y = X2 and
Y ∩ Z = X4 are nonempty. It is also immediate from the definitions of X, Y , and Z
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that there are no extra edges from X to Y , and no extra edges from Y to Z. Finally,
by definition of X and Z, we see that X ∩ Z = ∅. Together, these facts imply, by
definition of path structure, that G has path structure P , a P3 path structure, as was
to be shown.
This completes the proof of one direction of the lemma. For the converse, we sup-
pose G is a forcibly chordal graph with a P3 structure P . We must show five sets
X1, . . . , X5 with the desired properties exist. Let X, Y , and Z be the three ver-
tices of P , where P is a three point path such that G has path structure P . Let
X1 = X − Y , X2 = X ∩ Y , X3 = Y − X − Z, X4 = Y ∩ Z, and X5 = Z − Y . We
show that X1, . . . , X5 have the desired properties.
First, note that since X, Y , and Z are the three vertices of P , it follows that X,
Y , and Z are exactly the maximal cliques of G. In particular, V (G) = X ∪ Y ∪ Z.
Moreover, simple checking shows that
X ∪ Y ∪ Z =5⋃i=1
Xi
Therefore V (G) =⋃5i=1Xi. Simple checking also shows Xi ∩ Xj = ∅ for i 6= j, so
V (G) is the disjoint union of the Xi’s.
For condition (i), we must show each Xi is nonempty except possibly X3. Since X
and Y are distinct, maximal cliques of G, it follows that X1 = X − Y is nonempty.
Similarly, X5 = Z−Y is nonempty. X and Y are adjacent in P , therefore X2 = X∩Y
is nonempty. Similarly, X4 = Y ∩ Z is nonempty. This proves condition (i).
Next, we prove condition (ii). For each i with 1 ≤ i ≤ 4, the reader can check that
Xi ∪ Xi+1 is contained in one of the three cliques X, Y , or Z. Therefore Xi is
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complete to Xi+1. Similarly, X2 ∪ X4 is contained in the clique Y . Therefore X2 is
complete to X4. This proves condition (ii).
For condition (iii), we must show Xi is anti-complete to Xj for all other pairs i and j.
In other words, we must show X1 is anti-complete to X3, X4, and X5, and that X2
and X3 are anti-complete to X5. That X1 is anti-complete to X3 and X4 follows from
the fact that there are no extra edges from X to Y . That X2 and X3 are anti-complete
to X5 follows from the fact that there are no extra edges from Y to Z.
Suppose X1 is not anti-complete to X5. Then there is an edge from x1 to x5 for some
x1 ∈ X1 and x5 ∈ X5. Choose x2 in X2 and x4 in X4. By the pairs we have shown to
be complete and anti-complete, respectively, it follows that x1, x2, x4, x5 comprise a
cycle in that cyclic order, contrary to choice of G as forcibly chordal. Therefore X1
is anti-complete to X5. This completes the proof of (iii).
For (iv), we note that by symmetry, it is enough to show that X1 is a singleton.
To see that X1 is a singleton, suppose not. We know X1 is not empty, so if it is
not a singleton then it has at least two elements. Call them v and w. Choose xi
in Xi for i = 2, 4, 5, and consider H = G[u, v, x2, x4, x5]. Simple checking shows
{u, v, x2} is a triangle, x2 is adjacent to x4, x4 is adjacent to x5, and H has no other
edges. Therefore D(H) = (3, 2, 2, 2, 1). Simple checking shows this is also the degree
sequence of a square with a pendant edge, which of course is not chordal. Therefore H
is not forcibly chordal. Since H is an induced subgraph of G, we see that G is not
forcibly chordal, contrary to assumption. This contradiction shows that X1 is in fact
a singleton, thus proving (iv), and completing the proof of the lemma.
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Theorem 6.0.78. The forcibly chordal graphs G with a P4 structure are exactly the
graphs G such that V (G) is the disjoint union of sets Y1, . . . , Y5 such that the following
conditions hold:
(i) Each Yi is nonempty.
(ii) Yi is complete to Yi+1 for 1 ≤ i ≤ 4.
(iii) All other pairs of Yi and Yj are anti-complete.
(iv) Y1, Y2, Y4, and Y5 are singletons.
(v) Y3 is a clique.
Proof. There are two directions to show. First, suppose there are five such sets
Y1, . . . , Y5 in a graph G. We must show G is a forcibly chordal graph with a P4
structure.
First, we show G is forcibly chordal. For i = 1, 2, 4, 5, let the singleton Yi equal {yi}.
Then y1 and y2 are adjacent, y2 and y4 are not, y4 and y5 are adjacent, and y5 and y1
are not. Therefore we may apply the switching operation y1, y2, y4, y5 to obtain a new
graph G′. The reader can check that G′[y1, y5] is an edge and G′[Y2 ∪ Y3 ∪ Y4] is a
clique. Also, the reader may check that G′ has no edges from {y1, y2} to Y2 ∪Y3 ∪Y4.
Therefore G′ is the union of two disjoint cliques, one of which has only two vertices.
By Proposition 6.0.70, we know G′ is forcibly chordal. Since G and G′ have the same
degree sequence, it then follows that G is forcibly chordal.
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Next, we show G has a P4 structure. Simple checking shows the maximal cliques of G
are exactly X = Y1∪Y2, Y = Y2∪Y3, Z = Y3∪Y4, and W = Y4∪Y5. Let P be the path
with vertex set {X, Y, Z,W} such that X and Y are adjacent, Y and Z are adjacent,
Z and W are adjacent, and no other pairs of vertices are adjacent. Obviously, P is
isomorphic to P4. We show G has a P structure.
Since each Yi is nonempty, we see the sets X ∩ Y = Y2, Y ∩Z = Y3, and Z ∩W = Y4
are nonempty. It is also immediate from the definitions of X, Y , Z, and W that there
are no extra edges from X to Y , from Y to Z, or from Z to W . Finally, by definition
of X, Y, Z,W we see that
X ∩ Z = X ∩W = Y ∩W = ∅
Together, these facts imply, by definition of path structure, that G has path struc-
ture P , a P4 path structure, as was to be shown.
This completes the proof of one direction of the lemma. For the converse, we sup-
pose G is a forcibly chordal graph with a P4 structure P . We must show five sets
Y1, . . . , Y5 with the desired properties exist. Let X, Y , Z, and W be the four vertices
of P , where P is a four point path such that G has path structure P .
Note that G[X ∪ Y ∪ Z] has path structure P [X, Y, Z]. Since P [X, Y, Z] is a three
point path, we may use to define sets X1, . . . , X5 as in that lemma. Namely, we let
X1 = X − Y , X2 = X ∩ Y , X3 = Y −X −Z, X4 = Y ∩Z, and X5 = Z − Y , and we
know by the last lemma that these five sets satisfy that lemma’s conditions.
Letting Y1 = X1, Y2 = X2, Y3 = X4, Y4 = X5, and Y5 = W − Z, we show that
Y1, . . . , Y5 have the desired properties.
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First, we show each Yi is nonempty. Note that X1, X2, X4, X5 are nonempty by the
previous lemma, so Y1, Y2, Y3, Y4 are nonempty. Since W and Z are distinct, maximal
cliques, we know neither is a subset of the other. Therefore Y5 = W −Z is nonempty
as well. This completes the proof of condition (i).
Next, we show condition (ii), namely, that Yi is complete to Yi+1 for i = 1, 2, 3, 4. By
the previous lemma, we know that Y1 = X1 is complete to Y2 = X2, that Y2 = X2 is
complete to Y3 = X4, and that Y3 = X4 is complete to Y4 = X5.
It remains to show Y4 is complete to Y5. To do this, we first need to show that
Z − Y = Z ∩W . Let v be a vertex of Z ∩W . We know Y and W are disjoint by the
path structure of G and non-adjacency of Y and W in P . Therefore v is in Z − Y .
Since v was chosen arbitrarily, we see that Z ∩W ⊆ Z − Y . But Z − Y = X5 is a
singleton by the previous lemma. Moreover, Z ∩W is nonempty since Z and W are
adjacent in P . Since the nonempty set Z ∩W is a subset of the singleton Z − Y , it
follows that these two sets are equal as claimed.
We now note that Y4 = X5 = Z − Y = Z ∩W and Y5 = W − Z are both subsets of
the clique W . In particular, Y4 is complete to Y5 as condition (ii) claims.
Next, we must prove condition (iii), that all other pairs Yi and Yj are anti-complete.
Without loss of generality, i < j. If j 6= 5 then the fact that Yi and Yj are anti-
complete follows immediately from the last lemma and the definition of each Yi and Yj
as an Xk for some k. It remains to show that Y5 is anti-complete to Y1, Y2, Y3. Note
that Y1 = X1 = X − Y ⊂ X, Y2 = X2 = X ∩ Y ⊂ X, and Y3 = X4 = Y ∩ Z.
Therefore Yi is a subset of X ∪ Y for i = 1, 2, 3, and Y5 = W − Z is a subset of W .
X ∪ Y and W are anti-complete by the path structure of G and the fact that W is
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not adjacent to X or Y in P . In particular, Y5 is anti-complete to Y1, Y2, and Y3.
This completes the proof of condition (iii).
Now, we must show that Y1, Y2, Y4, and Y5 are singletons. Y1 = X1 and Y4 = X5 are
singletons by the previous lemma. The previous lemma shows, among other things,
that if G has a three point path structure such that the path consists of the vertices
A,B,C in that order, then A−B and C −B are singletons. We can apply this also
to G[Y ∪ Z ∪W ], which has path structure P [Y, Z,W ], to see that W − Z = Y5 is
a singleton, and that Y − Z is a singleton. Now, Y2 = X2 = Y ∩X. Moreover, if v
is a vertex of Y ∩ X then v is not in Z since X and Z are disjoint. Therefore v is
in Y − Z. Since v is arbitrary, we see that the nonempty set Y ∩X is subset of the
singleton Y − Z. Therefore Y2 = Y ∩ X = Y − Z is a singleton, proving condition
(iv).
To show (v), note that if Y3 is not a clique then Y3 must contain two nonadjacent
vertices v and w. The reader may then check that G[y2, v, w, y4] is a four cycle,
contrary to forcible chordality of G. Therefore Y3 is a clique as claimed.
We have only to show that V (G) is the disjoint union of Y1, . . . , Y5. We know by
the previous lemma that X ∪ Y ∪ Z is the disjoint union of X1, . . . X5. Since W is
disjoint to X and Y , it follows that W − Z = Y5 is disjoint to X ∪ Y ∪ Z. Therefore
V (G) = X ∪ Y ∪ Z ∪W is the disjoint union of X1, . . . , X5 and Y5. Since Y1 = X1,
Y2 = X2, Y3 = X4, and Y4 = X5, the proof will thus be complete if we show that X3
is empty.
Suppose not. Then choose xi in Xi for 2 ≤ i ≤ 5 and y5 in Y5. Having shown
which sets are complete and anti-complete to one another, respectively, it is simple
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to check which of these vertices are adjacent. Doing this, one can check that the
degree sequence of H = G[x2, x3, x4, x5, y5] is (3, 2, 2, 2, 1), which is also the degree
sequence of a square with a pendant edge. As noted previously, this graph is not
chordal. From this we see that H is not forcibly chordal, and therefore G is not
forcibly chordal.
This contradiction shows that in fact X3 is empty, which completes the proof.
Corollary 6.0.79. Let n ≥ 5. Then there are no forcibly chordal graphs with a Pn
structure.
Proof. It is enough to show there are no forcibly chordal graphs with a P5 structure.
We give a proof by contradiction. Suppose a forcibly chordal graph G has a P5
structure P . Call the vertices of the path X, Y, Z,W, V , in that order. We then get
a P [X, Y, Z,W ] structure for G[X ∪ Y ∪ Z ∪W ]. This path structure gives us sets
Y1, . . . , Y5 as in the previous lemma. Choose yi in Yi for 1 ≤ i ≤ 5. By the path
structure P [Y, Z,W, V ] for G[Y ∪ Z ∪W ∪ V ], we know that V −W is a singleton.
Without loss of generality, V − W = {v}. It is a matter of checking, using the
complete and anti-complete pairs of the previous lemma, to see that G[y1, . . . , y5, v]
is a six point path with the vertices in that order. A six point path is not chordal.
In particular, G is not forcibly chordal, a contradiction.
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CHAPTER 7
COMBINATORIAL PROOF SYSTEMS
7.1 Introduction
This chapter begins the study of proof systems. While this may seem more related to
logic than combinatorics, we study proof systems in an abstract, general setting. In
this general setting, proof systems provide a common, yet natural generalization of
graphs and partial orders, and their study is essentially combinatorial in nature. For
instance, our characterization of finite partial orders by forbidden subdots in Corol-
lary 7.11.17 is similar to the many forbidden minor theorems of graph theory, and we
will also see that proof systems yield a common framework in which to simultaneously
treat questions on graph matching and graph connectivity.
Our eventual formalization of “proof system” is what we will call an autonomous
system, or ausys, which is essentially what one obtains by axiomatizing partial orders
in terms of their downward closed sets then discarding the assumption that downward
closed sets are closed under arbitrary intersection. Mathematically, we could make
this definition now and immediately start proving ausys theorems. Pedagogically,
the reader would have little to no intuition for what is going on. To address this
issue, we start with motivation and the definition of what we call preceding set proof
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systems. Preceding set proof systems are easily motivated, and we then define the
mathematically more convenient ausyses in terms of them.
7.2 Proof Closure
We will be looking at abstract proof systems in which proofs are partial orders such
that the set of predecessors of any element in the partial order is considered enough
information to “infer” that element.
Given a set T of partial orders of subsets of a set P , we want to know what other
partial orders <′ of subsets of P are to be considered proofs if every partial order in
T is. We want a way to say for various T and ≤ that if every partial order in the set
T is considered a proof then we must consider ≤ to be a proof also.
First, we make a definition that facilitates the discussion and makes things easier to
understand.
Definition 7.2.1. Let (P,≤P ) be a partial order and x in P . Then {≤P x} denotes
the set {y ∈ P : y ≤P x}. Similarly for {<P x},{≥P x}, and {>P x}.
Now, suppose we consider every partial order in a set T of partial orders to be a
proof. Let x be in the domain of <, where < is in T . The set {< x} is considered
enough information to infer x since < is in T and we are considering everything in
T to be a proof. So if {<′ x} contains {< x} for some partial order ≤′, then {<′ x}
should be considered enough information to infer x too.
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By this logic, if for every x in the domain of <′ there is some order < in T such that
{<′ x} contains {< x}, then for every x in the domain of <′, the set {<′ x} should
be considered enough information to infer x. So <′ should be considered a proof.
This leads us to the following definition:
Definition 7.2.2. Let T be a set of partial orders of subsets of the set P . A partial
order <′ whose domain is a subset of P is in the proof closure of T if for every x in
the domain of <′ there is < in T such that the set {<′ x} contains the set {< x}. T
is proof closed if T equals the proof closure of T .
It is simple to check the name proof closure is justified in the sense that it is indeed
a closure operator on the set of partial orders of a given set.
To illustrate the definition of proof closure, note that for example, every downward
closed subset of a partial order in T is in the proof closure of T . Once the merge
is defined in Definition 7.3.1, we will see the merge of a sequence in T is also in T .
Given two partial orders ≤1 and ≤2 on a set P , we say that ≤2 lengthens ≤1, or ≤1
fattens ≤2, if x ≤1 y implies x ≤2 y for all x and y in P . With this convention, we
may note a lengthening of any partial order in T is in the proof closure of T .
7.3 The Merge
The merge is a partial order construction useful for constructing new proofs from old.
We will use this construction at several key points in enlarging the domain of a proof
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while maintaining certain properties, combining proofs {(Pi,≤i)}i∈I to obtain a proof
on the union of their domains, and so on.
Definition 7.3.1. Let T be a set of partial orders and let <T be a well order of T .
For x in⋃P∈T P let L(x) be the <T least partial order in T with x in its domain.
The merge <m of <T is defined to be the partial order whose domain D is⋃P∈T P
and such that for x, y in D, x <m y iff
(i) L(x) <T L(y) or
(ii) L(x) = L(y) and x <L(x) y.
We will sometimes say we are merging the partial orders in T , “forgetting” that <T
matters. In such cases, the specific well order used is either unimportant or clear
from context.
To use the merge to get new proofs from old, we will need to know the merge is in
the proof closure of the orders being merged.
Lemma 7.3.2. Let T be a set of partial orders and let <T well order T . Then the
merge <m of <T is in the proof closure of T .
Proof. Let D be the union of the domains of partial orders in T (that is, D is the
domain of the merge) and let x in D. You must show there is a partial order <′ in
T such that every <′ predecessor of x is a <m predecessor of x. Let <′ be L(x). If
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y <′ x then y is in L(x) so L(y) ≤ L(x). If L(y) < L(x) then y <m x by the first
part of the definition of <m. If L(y) = L(x) then y <m x by the second part of the
definition of <m and the fact that y <L(x) x.
It is worth noting the following proposition.
Lemma 7.3.3. Let T be a set of well founded partial orders and ≤T a well order of
T . Then the merge ≤m of ≤T is well founded.
Proof. Suppose not. Then there is an infinite strictly decreasing ≤m sequence
x1 >m x2 >m x3 >m . . .
of elements in the union of the domains of the posets in T . Call xi and xj equivalent
if L(xi) = L(xj). There are either finitely many or infinitely many eqivalence classes.
We consider each case.
If there are only finitely many, then some class contains infinitely many elements
y1 >m y2 >m . . .
and L(yi) = L(yj) for all i, j. By definition of merge this means that
y1 >L(y1) y2 >L(y1) y3 >L(y1) . . . ,
contrary to the fact that L(y1) is a partial order in T and thus well founded by
assumption.
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So there must be infinitely many equivalence classes. Picking from each class one
member of the sequence of xi’s we get a sequence
y1 >m y2 >m y3 . . .
such that for all i, j we have L(yi) 6= L(yj). Since L(yi) 6= L(yj) it follows from the
definition of the merge that
L(y1) >T L(y2) >T L(y3) >T . . . ,
contrary to the assumption that ≤T is well founded.
The following two lemmas are two of the primary properties used when merging proofs
to obtain a new proof.
Lemma 7.3.4. If (Pi,≤i) for i in I is a set of partial orders then there is a partial
order in the proof closure of {Pi : i ∈ I} with domain⋃i∈I Pi.
Proof. This is trivial from what has already been said. Take a well order ≤ of I. Now
take the merge of the sequence of partial orders. We have proved the merge is in the
proof closure of the Pi’s, and by definition of the merge the domain is the union of
the Pi’s.
Before stating the next lemma, we need the following definition.
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Definition 7.3.5. Let (A,≤A) be a partial order. An end extension of (A,≤A) is a
partial order (B,≤B) such that A is a ≤B downward closed subset of B and ≤B |A
is the same partial order as ≤A.
Lemma 7.3.6. Let (A,≤A) and (B,≤B) be partial orders with A ⊆ B. Then there
is a partial order (B,≤′) of B in the proof closure of (A,≤A) and (B,≤B) that is an
end extension of (A,≤A). If ≤A and ≤B are well founded then ≤′ may be chosen as
well founded as well.
Proof. Well order the set {≤A,≤B} by choosing ≤A to be less than ≤B. Now take
the merge of these two partial orders . We choose ≤′ to be the merge. We know the
merge is a partial order on the union of the domains of A and B, which in this case
is B itself since A ⊆ B. We also know the merge is in the proof closure of ≤A and
≤B.
We now show the merge is an end extension of ≤A. This is equivalent to showing two
things. First, we have to show A is a downward closed subset of B in the merge ≤′.
Second, we must show ≤A and ≤′ |A are the same partial order .
To show A is ≤′ downward closed subset of B, let y ∈ A and x ≤′ y. We have to
show x ∈ A. Well, x ≤′ y which means x precedes y in the merge, by definition of ≤′.
This implies L(x) ≤ L(y). We can not have L(x) < L(y) since y ∈ A and A is the
first set being merged. So L(x) = L(y). So x ∈ A. It follows that A is ≤′ downward
closed .
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To show ≤A is the same partial order as ≤′ |A is to show that for all x, y ∈ A, you
have x ≤′ y iff x ≤A y. Since both x, y are in A, we see that L(x) = L(y). Hence,
by definition of merge (and the fact that we chose ≤′ as the merge) it follows that
x ≤′ y iff x ≤L(x) y iff x ≤A y.
That ≤′ is well founded if ≤A and ≤B are well founded is simply Lemma 7.3.3.
7.4 Preceding Set Proof Systems
In every abstraction, some information is kept, and much is discarded. For instance,
though the complex numbers have an interesting algebraic structure, the topology
of the plane simply has nothing to say about this. Topology speaks of open sets
and anything that can be defined from them. So the first point to address is which
information we want to capture in defining proof systems abstractly, and which is
outside the scope of our subject.
We state now, just as topology takes “open set” as a basic notion and studies that
which can be defined in terms of them, our preceding set proof systems essentially
take “T ⇒ x” as a basic notion, and studies that which can be defined in terms of
these implications. Our proof systems have nothing to say about syntax, semantics,
truth or falsehood, models, and so on. While we may intuitively think of points of a
proof system as “formulas”, in general, the points of our proof systems have no more
structure than points of graphs or topological spaces. We now make this precise in
the following definition.
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Definition 7.4.1. A preceding set proof system is a set S, called the set of formulas,
together with a set I of ordered pairs (T, x) with T ⊆ S and x ∈ S, called the rules
of inference, such that the following property hold:
(i) Let T1, T2 be subsets of S and x ∈ S. If (T1, x) ∈ I and T1 ⊂ T2 then (T2, x) ∈ I.
Intuitively, every element in S is a formula. The ordered pair (T, x) says “from T
infer x”. We infer a formula from a set of formulas. The condition states that we are
dealing with monotonic proof systems, which in short means if you assume more you
can prove more. Though there are exceptions, the vast majority of naturally occurring
proof systems are monotonic, and we assume this monotonicity throughout.
When I is clear from the context, we will make the agreement that statements such
as, “T implies x”, “T is enough information to infer x”, etc, means the pair (T, x) is
in I. For convenience, we define T ⇒I x to mean the pair (T, x) is in I, and we may
write T ⇒ x when I is clear from context.
7.4.1 Examples
First, in a typical proof system you might have modus ponens as a rule of inference.
In other words, given formulas φ and φ ⇒ ψ, from {φ, φ ⇒ ψ} to infer ψ. In the
notation of our definition, {φ, φ⇒ ψ} would be our set T and ψ would be our element
x. Similarly for any other rule of inference found in a typical proof system arising in
mathematical logic.
Second, given a partial order (P,≤), we can let I be the set of pairs of the form
(T, x) such that T ⊆ P , x ∈ P , and T contains all y < x. It is essentially this
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definition, when made for ausyses (to be defined below) instead of preceding set proof
systems, that allows generalizing partial order notions to proof systems. Note this
definition works equally well with infinite partial orders and even non-well founded
partial orders, which is part of the motivation for allowing the very general setting
we will allow, as opposed to making restrictions of finiteness or well foundedness in
our general definitions.
Third, let S be the set of Borel sets of reals. We want to make S into a preceding
set proof system that captures how Borel sets are constructed from the open sets.
The Borel sets are defined by starting with the open sets and allowing (transfinite
iteration of) complementation and countable union. If we think of a rule (T, x) as
saying, “once you have constructed everything in T , you may now construct x”, then
our definition of preceding set proof system can capture the common mathematical
theme of certain starting objects and certain construction rules. The starting objects,
such as the open sets of the Borel hierarchy, are analogous to the axioms of proof
systems of mathematical logic. To say we may “start” with an open set O, we simply
put pairs of the form (∅, O) in I. To say once we have a set B, we may now construct
its complement Bc, we put pairs of the form ({B}, Bc) into I. Similarly, to allow
countable union as a construction rule, we put pairs of the form ({Bn}n≥0,⋃n≥0Bn)
into I. Note that by convention, when we say a rule (T, x) goes into I, we implicitly
mean that (T ′, x) goes into I for all T ′ containing T as well, to maintain monotonicity.
The previous example should be seen as far more general than the Borel sets. Such
definitions and theorems, with starting objects and construction rules, are extremely
common througout mathematics, and all yield natural examples of preceding set proof
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systems. Also, the previous example should be noted as another natural example of
an infinitary proof system, further motivating the level of generality we maintain
throughout.
Finally, we note that graphs can be made into preceding set proof systems in two
natural ways. Informally, given a graph (V,E), one associated preceding set proof
system with domain V ∪ {x} (for some element x not in V ) has all vertices in V as
axioms, and allows infering the nonaxiom x from each doubleton in V corresponding
to an edge. Another associated preceding set proof system chooses a vertex v in V
as an axiom, and allows inferring elements adjacent to any vertex already proved.
While these definitions may at first seem unnatural, in fact, the former is related
to graph matching and the latter to graph connectivity in such a way that graph
matching and graph connectivity questions can be be simultaneously asked by asking
the corresponding question regarding the proof systems. In particular, we give a
problem whose solution would specialize to Konig’s Duality Theorem for the former
type of proof systems and to (even the infinite version of) Menger’s Theorem for the
latter.
While we do not need syntax or semantics, in order to develop an abstract theory of
proofs we certainly need to define the notion of proof in a preceding set proof system.
For brevity, preceding set proof systems will simply be called proof systems for the
rest of this section. First, we give the motivation for our definition.
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7.4.2 Motivation for Definition of Proof
What is a proof of a theorem? It is a sequence of claims such that for any claim x, the
set of claims which were given before x is considered enough information to infer x. So
we want all of these to be considered proofs. Namely, given any finite sequence such
that for every x in the sequence, the set of predecessors of x is enough information to
infer x, this sequence should be considered a proof. Making, “is enough information
to infer”, precise by saying “(T, x) is in I”, this would yield a rigorous definition of
proof. We could say a proof is a finite sequence such that for all x in the sequence,
we have {< x} ⇒I x.
However, there is no reason to insist a proof is a finite sequence. A finite sequence
is just a finite total order. The reason we think of proofs as being totally ordered is
because that is the way they always appear on the pages of a book. For the general
theory, it is actually more convenient and natural to allow partial orders in general to
be considered proofs also, so long as the set of predecessors of any element is enough
information to infer that element. For the general theory, we do not require these
proofs to be finite or even well founded.
We are now ready to give the definition of proof in a proof system.
7.4.3 Proof Definition and Basics
Definition 7.4.2. Let (S, I) be a preceding set proof system. An I proof is a partial
order (P,≤) such that {< x} ⇒I x for all x ∈ P .
The astute reader may notice this does not in fact give the “right” definition of proof
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for naturally arising proof systems in mathematical logic, as the rules of inference
given there would, if non-well founded proofs are allowed, cause unprovable formulas
to become provable. While this is easily remedied, we do not yet have the necessary
definitions to state the remedy. For now, suffice it to say that the general theory
is much smoother with the above definition, and roughly, for specific proof systems
arising in mathematical logic, the solution is to define a first proof system as stated
above, define a second proof system that is intuitively a well founded version of the
first, and then apply the general theory to the well founded version. Though this
well founded version of the proof system still allows non-well founded proofs, the
non-well founded proofs in this proof system no longer yield proofs that “should not
be” proofs. We make this precise in Lemma 7.8.5.
After defining proof systems, the following question is natural. Given a set S, for
which sets T of partial orders of subsets of S can rules of inference I be found such
that T is exactly the set of I proofs? It is not hard to see that T has to be proof
closed, and that this is enough is the content of the next proposition.
Proposition 7.4.3. Let T be a set of partial orders of subsets of the set S. Then
there is a preceding set proof system (S, I) with T the set of I proofs iff T is proof
closed.
Proof. That the set of proofs of a preceding set proof system is proof closed is imme-
diate from the definition of proof and proof closure.
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For the converse, suppose T is proof closed. If {< x} ⊆ C ⊆ S then let C ⇒ x be
a rule in I. Moreover, let rules of inference arising in this way be the only ordered
pairs in I. It is enough to show (S, I) is a preceding set proof system that has T as
its set of proofs.
That (S, I) is a preceding set proof system is just trivial checking of the definition.
What we have to show is (S, I) has T as its set of proofs. We have to show every
partial order in T is a proof in (S, I) and every proof in (S, I) is a partial order in T .
First, let (P,≤) be a partial order in T . We have to show it is a proof in (S, I).
Equivalently, we must show {< x} ⇒I x for all x ∈ P . This is trivial from the
definition of I. We can choose any subset C of S containing {< x}, in particular
{< x} itself.
Now, let (P,≤) be a proof in (S, I). We have to show this partial order is in T . T is
proof closed by hypothesis, so it is enough to show ≤ is in the proof closure of T . By
definition of proof closure, this means we have to show for all x ∈ P , there is some
partial order (Q,≤′) in T such that the set of < predecessors of x contains the set of
<′ predecessors of x. Again, this is clear from the definition of I.
7.4.4 Autonomous Sets
In these proof systems, we now wish to define the autonomous sets. The autonomous
sets are, roughly speaking, sets that are self contained. Everything in an autonomous
set can be proved without using anything outside that set. Their role in the theory
of proof systems is analogous to the role of open sets in metric spaces. They do not
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carry all of the information of preceding sets, but all the information we want for
many cases, and the study of their structure is quite interesting.
The set of open sets of a metric space satisfies certain properties, which are then used
as the defining axioms of a topological space as a set together with a set of subsets,
called open sets, such that certain properties hold.
We will show the set of autonomous sets satisfies certain properties and then define
an autonomous system, or ausys, to be a set together with a set of subsets, called
autonomous sets, such that certain properties hold.
There is one important way in which the relation of metric spaces to topological spaces
differs from that of proof systems to autonomous systems. Not every topological space
is metrizable, but every ausys is induced by a preceding set proof system, as we will
show.
Now, we give a definition that is basic to all that follows.
Definition 7.4.4. Let (S, I) be a preceding set proof system. Let P ⊆ S. P is
autonomous if there is a proof (P,≤) with domain exactly P .
By the above definition, the autonomous sets of a proof system are obviously uniquely
determined by the set of proofs. The next lemma shows the set of proofs is in fact
uniquely determined by the set of autonomous sets as well, so that the set of proofs
and the set of autonomous sets contain the same information.
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Proposition 7.4.5. Let (S, I) be a preceding set proof system. Let (P,≤) be a par-
tial order of a subset of S. Then ≤ is a proof iff every ≤ downward closed set is
autonomous.
Proof. If ≤ is a proof, then for every ≤ downward closed set D, ≤ |D is a proof with
domain D, which shows D is autonomous.
Conversely, suppose every ≤ downward closed subset of P is autonomous. We must
show ≤ is a proof. That is, we must show for all y in the domain P of ≤, that the
rule {< y} ⇒ y is in I.
Well, {≤ y} is a ≤ downward closed subset of P and therefore autonomous. Since it
is autonomous it has a proof ≤′. Since ≤′ is a proof it follows that the rule {<′ y} ⇒ y
is in I. Now, {<′ y} ⊂ {< y}, so by upward closure of the rules of a preceding set
proof system, the rule {< y} ⇒ y is in I too, and we are done.
So we now know the proofs and the autonomous sets carry the same information.
From either you can determine the other uniquely. And we know exactly when a set
T of proofs is the set of proofs of some preceding set proof system, namely, when T is
proof closed. So it is natural to ask the corresponding question for the autonomous
sets. The answer is given by the following lemma. This lemma is extremely important,
for it gives us the axiomatization we will use in the definition of autonomous system.
The role of this lemma in studying autonomous systems is completely analogous to
the role played in topology by the lemma that says in a metric space, the whole space
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and null set are open, the open sets are closed under arbitrary union, and the open
sets are closed under finite intersection.
Proposition 7.4.6. Let S be a set and T a set of subsets. Then T is the set of
autonomous sets of a preceding set proof system on S iff the following two conditions
hold:
(i) T is closed under arbitrary union.
(ii) (The Order Property) If A ∈ T then there is a total order ≤′ with domain A
such that every ≤′ downward closed subset B of A is in T .
Proof. First, we prove if T is the set of autonomous sets of a preceding set proof
system on S then (i) and (ii) are satisfied. Equivalently, we prove that if T is the set
of I autonomous sets of some set I of rules of inference on S, then (i) and (ii) are
satisfied. So take any such I.
First, we show (i). So let Aj ∈ T for j in an index set J and let A =⋃j∈J Aj be their
union. We must show A is in T . By choice of I we know each Aj is I autonomous
which means each Aj has an I proof ≤j. Let ≤w be a well order of J and merge
all of the ≤j’s in that order to get ≤m. By the definition of the merge we know ≤m
has domain A =⋃Aj and we proved in Lemma 7.3.2 that the merge is in the proof
closure of the orders being merged. Hence ≤m is in the proof closure of all the ≤j’s.
But each ≤j is an I proof therefore ≤m is in the proof closure of a set of I proofs.
By Proposition 7.4.3, the set of I proofs is proof closed, and hence must contain ≤m.
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In other words, ≤m is an I proof with domain A, hence A is I autonomous, and (i)
follows.
For (ii), suppose A is in T . We must show there is a total order≤′ of A such that every
≤′ downward closed subset B of A is in T . Well, A is I autonomous and therefore has
an I proof ≤. This proof is a partial order, and it is well known every partial order
can be lengthened to a total order ≤′. The reader may check that the lengthening
of an I proof is in its proof closure and is therefore an I proof. Thus ≤′ is an I
proof. We proved in Proposition 7.4.5 that every downward closed subset of a proof
is autonomous. Hence every ≤′ downward closed subset B of A is I autonomous, and
therefore in T . Hence ≤′ is as needed, thus proving (ii).
Now for the converse. We assume T satisfies (i) and (ii), and we must show T is the
set of autonomous sets of a preceding set proof system on S. Equivalently, we must
show T is the set of I autonomous sets of a set I of rules of inference on S.
We need to define I from T . The idea is to put into I all the rules which have to be
there if each A in T is autonomous, but to put in no other rules. Then we show the
set of all rules which “have to” be there in fact has T as its set of autonomous sets,
thus completing the proof.
To see what rules have to be there if each A in T is autonomous, take A in T . If A
is I autonomous then A has an I proof ≤. Take f ∈ A. Since ≤ is an I proof we
know {< f} ⇒I f . Now, {< f} is of course a subset of A− f , so by monotonicity of
the proof system, C ⇒I f for any subset C of S containing A − f . We use all rules
C ⇒I f obtained in this way as our I.
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More formally, I is the set of all rules of the form C ⇒ f such that there is some
A ∈ T with f ∈ A and C ⊆ S containing A− f .
We must show with this definition of I, every A in T is I autonomous and there are
no other I autonomous sets.
First, let us show every A in T is I autonomous. By hypothesis, T satisfies (ii),
which implies A can be ordered as ≤′ such that every ≤′ downward closed subset B
of A is in T . It is enough to show ≤′ is an I proof. Take f ∈ A. We must show
{<′ f} ⇒I f . Well, the set B = {≤′ f} is a ≤′ downward closed subset of A and so
is I autonomous by definition of ≤′. We see that {<′ f} = B − f and by definition
of I, we have B − f ⇒I f , hence {<′ f} ⇒I f as desired, therefore ≤′ is an I proof,
which means A is I autonomous.
Second, let us show every I autonomous set is in T . Since T is closed under arbitrary
union by hypothesis, it is enough to show that for every f ∈ A there is B ⊆ A
containing f such that B is in T . So take an I autonomous set A and f ∈ A and we
are looking for such a B. Since A is I autonomous there is an I proof ≤ with domain
A. Since ≤ is an I proof it follows that there is an I rule of inference C ⇒I f such
that C is contained in A− f .
By definition of I, since C ⇒I f in I, it follows that there is a subset B of A containing
f that is in T . This completes the proof.
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7.4.5 Axioms
Axioms are the elements of a proof system you are allowed to infer from nothing at
all.
Definition 7.4.7. Let (S, I) be a preceding set proof system. An I axiom is an
element x of S such that ∅ ⇒I x.
The following lemma is trivial, but we include it here to familiarize the reader with
definitions so far presented.
Lemma 7.4.8. Let (S, I) be a preceding set proof system and x ∈ S. Then the
following are equivalent.
(i) x is an axiom.
(ii) The single element partial order containing just x is a proof.
(iii) {x} is autonomous.
Proof. If x is an axiom then it is immediate from the definition of proof that this
single element partial order is a proof, therefore (i) implies (ii). If the one point
partial order on {x} is a proof then its domain {x} is autonomous by the definition
of autonomous, so (ii) implies (iii). To see that (iii) implies (i), note if {x} is
autonomous then it is the domain of a proof ≤, and {< x} is then ∅. Since ≤P is
a proof, the rule {< x} ⇒ x is in I, hence the rule ∅ ⇒ x is in I, showing x is an
axiom.
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7.4.6 The Information in the Set of Proofs
The set of preceding sets carries strictly more information than the set of proofs,
or equivalently, strictly more information than the set of autonomous sets. In other
words, though the preceding sets uniquely determine the proofs, the proofs do not
uniquely determine the preceding sets. We can find two distinct preceding set proof
systems on S with the same set of proofs.
For example, let S = {1, 2, 3}. We give two distinct preceding set proof systems on
S with the same proofs. For the first proof system, take the rules of inference ∅ ⇒ 1,
1⇒ 2, and 2⇒ 3.
It is understood when a preceding set proof system is described in such a way, we
tend not to bother to say, for instance, that C ⇒ x for all C ⊆ {1, 2, 3} such that C
contains 1. We simply say let 1⇒ 2.
With that understanding, the reader may check that the proofs here are exactly the
downward closed subsets of {1, 2, 3} with the usual ordering of the integers. For
the second proof system, take the same rules of inference as the first, and add the
rule 3 ⇒ 1. The preceding sets in these two proof systems are different, but simple
checking shows that the set of proofs is the same in each case.
The reason that adding the rule yields no new proofs is that there is no chance to use
the rule; inferring 1 from 3 gives nothing when we must prove 1 before ever proving 3.
This suggests the only information lost in passing from the rules of inference to the
set of proofs is the superfluous rules. We make this precise in the following lemma
for finite proof systems, whose simple proof is not included.
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Lemma 7.4.9. Let (S, I) be a finite preceding set proof system. Then there is a
minimum (not just minimal) set J under inclusion of rules of inference on S such
that the set of I proofs and J proofs are equal. In fact, J is the set of all rules T ⇒ x
such that {< x} ⊆ T for some I proof (P,≤).
While a fact that is similar in spirit is true for proof systems in general, it is more
complicated and not central to what we do, so we do not state it here.
Since, in passing to the set of proofs, the only information we lose is (roughly speaking)
that of which superfluous rules had been included, it is plausible, and in fact turns
out to be true, that little of significance is lost and both clarity and efficiency are
gained by “forgetting” the superfluous rules and only “remembering” the proofs,
or equivalently, the autonomous sets. The definition of autonomous system in the
following section accomplishes precisely this.
7.5 Autonomous Systems
7.5.1 Introduction
The following definition is basic to all that follows.
Definition 7.5.1. An autonomous system, or ausys, is a set S together with a set T
of subsets of S, called autonomous sets, satisfying the following two conditions:
(i) T is closed under arbitrary union.
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(ii) (The Order Property) For every autonomous set A there is a total order ≤′ of
A such that every ≤′ downward closed set is autonomous.
Note that since the union of autonomous sets is autonomous, there is a largest au-
tonomous subset S ′ of S. Since all the structure is contained within S ′, we assume
unless otherwise noted that S = S ′.
We call this maximum set S ′ the autonomous part of S. It is convenient to use this
terminology more generally.
Definition 7.5.2. Let S be a subset of an ausys. The autonomous part of S, denoted
by aut(S), is the largest autonomous subset of S under inclusion.
7.5.2 Definition of Proof Revisited
Proposition 7.4.5 characterizes proofs in a preceding set proof system in terms of
autonomous sets. For autonomous systems, in which “autonomous” is the basic
notion, this lemma becomes the definition of proof.
Definition 7.5.3. Let (S, T ) be an ausys. A T proof is a partial order (P,≤P ) of a
subset P of S such that every downward closed set is autonomous.
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7.6 The Canonical Orders
7.6.1 Canonical Order Definition and Basics
This sections formalizes the notion of “y needs x”, or, “in order to prove y, one must
first prove x”, in an autonomous system. However, need is often context dependent.
The element y may need x if only elements from the autonomous set A are allowed,
while not in general needing x. Taking this context dependence into account yields
the following definition, which can be thought of as, “y needs x if all tools used are
from A”, or “y needs x in A”.
Definition 7.6.1. Let A be an autonomous set. The canonical order ≤A is defined
by letting x ≤A y iff x ≤ y for all proofs ≤ with domain A, where x, y ∈ A.
It is clear from the definition that ≤A is in fact a partial order on each autonomous
set A.
For the following lemma, we need the notion of the infimum under lengthening. The
reader may take it as an exercise to show that given a set S, the set of partial orders
on S is itself a partial order under lengthening. We think of ≤1 as less than ≤2 if ≤2
lengthens ≤1. It is also a simple exercise to show that the infimum of every nonempty
set of partial orders on S exists. In fact, given partial orders ≤i on S for i in an index
set I, the infimum ≤ is the unique partial order on S such that for all x, y in S, x ≤ y
iff x ≤i y for all i in I.
We refer to this as the inf under lengthening. We note the supremum of a set of
partial orders on the set S does not always exist as there may in fact be no upper
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bound at all under lengthening. For instance, if x <1 y and y <2 x then ≤1 and ≤2
have no sup under lengthening. However, if an upper bound exists then we may take
the infimum of all such upper bounds to obtain a supremum under lengthening.
With these notions at hand, the following lemma is immediate from the definition.
Proposition 7.6.2. Let A be an autonomous set. Then the canonical order ≤A is
the inf under lengthening of the set of proofs with domain A.
The canonical orders are very often used even in proofs of facts whose statement
makes no mention of them. The following is the most important characterization of
these orders. It is used frequently and without comment.
Proposition 7.6.3. Let A be autonomous. Then x ≤A y iff every autonomous subset
of A containing y also contains x.
Proof. We prove each direction, but it will be more convenient in each case to prove
the contrapositive instead.
First then, we must prove if x 6≤A y then there is an autonomous subset of A con-
taining y and not x. Well, by the definition of the canonical order, since x 6≤A y
there is a proof ≤ with domain A such that x 6≤ y. Consider the set {≤ y}. This set
is a ≤ downward closed subset of the proof ≤, and is therefore autonomous. Since
x 6≤ y, it follows that x is not in this set. Hence we have an autonomous subset of A
containing y and not x.
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For the second half of the proof, we assume there is an autonomous subset of A
containing y and not x and we must prove that x 6≤A y. So let C be such an
autonomous subset of A containing y and not x. C is autonomous so there is a
proof ≤1 with domain C. And A is autonomous by hypothesis so there is a proof ≤2
with domain A. We take the merge ≤m of ≤1 and ≤2 in this order. The merge has
A ∪ C = A as its domain and is a proof since it is in the proof closure of the proofs
≤1 and ≤2. By the definition of the merge, it follows that x 6≤m y. By the definition
of the canonical order and the fact that ≤m is a proof it follows that x 6≤A y.
7.6.2 Descendability
Suppose x is a nonaxiom in an ausys. Then intuitively, x must “need” something in
order to prove it. Formally, we may ask if there is some autonomous set A and y
in A such that y <A x. This is in fact true for finite autonomous systems and even
finitary autonomous systems, as we show below, and it is also true for partial orders.
In general, however, this is not the case.
Let S = {x1, x2, . . .} ∪ {x}. Let each xi be an axiom, and let each infinite subset of
{xi}i≥1 imply x. x is a nonaxiom in this ausys, but suppose there is an autonomous set
A containing x and an element y such that y <A x. Then A is infinite, so A−{x, y} is
an infinite set of axioms. Therefore A−{x, y}∪{x} = A− y is an autonomous subset
of A containing x and not y. Therefore by Proposition 7.6.3, y 6<A x, a contradiction.
This contradiction shows there is no such autonomous set.
The ausys in the previous paragraph is somewhat of an all purpose counterexample,
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so we refer to it by name as the standard nondescendable ausys, or sna. By “de-
scendable”, we vaguely mean various properties that state that in any given context,
there should be a minimal (under inclusion) set from which we are able to prove
x. The word “descendability” comes from the fact that one such notion, strong aut
descendability, is defined as the intersection of every chain of autonomous sets de-
creasing under inclusion itself being autonomous, which may be thought of saying if
autonomous sets are “descended” then the limiting set is autonomous.
The vagueness of the previous paragraph is intentional, as there are many and varied
descendability hypotheses. The descendability axioms are finiteness conditions on
ausyses much as Noetherian ring, well order, compact topological space, et cetera,
are all finiteness conditions on their respective structures.
While many ausys theorems are true in general, many involving canonical orders
require descendability hypotheses. For instance, if x is a nonaxiom then x “should”
need some y in some set A. Typically, any theorem that says certain structural
information “should” follow from knowledge of the canonical orders is very likely to
require an appropriate descendability hypothesis. Descendability hypotheses are also
required for a number of theorems making no reference to canonical orders, such
as the unique lexicographic decomposition theorem for finite ausyses, and forbidden
minor/subdot theorems. The fact that descendability hypotheses are required for the
truth of these theorems suggests the canonical orders are sitting in the background,
and further suggests the importance of these orders for the study of ausyses in general.
With that said, we now prove some basic facts regarding descendability and canonical
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orders. We note the definition of strong aut descendability is particularly important
as the forbidden minor theorems we later prove are proved under this assumption.
Definition 7.6.4. An ausys is strong aut descendable if the intersection of every
chain of autonomous sets under inclusion is also autonomous.
Definition 7.6.5. An ausys is weak aut descendable if for all autonomous sets A and
for all x in A, there is a minimal autonomous subset B of A containing x.
Lemma 7.6.6. Every strong aut descendable ausys S is weak aut descendable.
Proof. Choose an autonomous set A and x in A. We define a sequence of autonomous
sets Aα by transfinite recursion as follows. Define A0 = A. Given an ordinal α such
that Aα is an autonomous subset of A containing x, we do one of two things. If Aα is
minimal with respect to this property, stop. Letting B = Aα, the proof is complete.
If Aα is not minimal with respect to this property, let Aα+1 be a proper autonomous
subset of Aα containing x. If Aα is defined for all α less than a limit ordinal β, let
Aβ =⋂α<β
Aα.
Since there are only 2|S| subsets of S, this induction must stop at some point. But
it was already noted if the induction stops at Aα then the lemma is proved with
B = Aα.
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Note the converse is not true. Take disjoint sets {ai}i≥0 and {xi}i≥0. Let each ai be
an axiom, and for each i let ai ⇒ xi and xi ⇒ xi−1. For each i ≥ 0, let
Ai = {x0, x1, . . .} ∪ {ai, ai+1, . . .}.
The reader may check that each Ai is autonomous, but the intersection {x0, x1, . . .}
of the chain is not, so this ausys is not strong aut descendable. To see that the ausys
is weak aut descendable, note that if an autonomous set A contains a point x then so
does some finite autonomous subset B of x, and we may simply choose B of minimum
cardinality.
Now, we relate this to the canonical orders.
Lemma 7.6.7. Let S be an ausys, let x be in S, and let A be a minimal autonomous
subset of S containing x. Then x is a maximum element of the canonical order ≤A.
Proof. Suppose not. Then there is y is A such that y 6<A x. Therefore there is an
autonomous subset B of A containing x and not y. So A is not a minimal autonomous
set containing x, a contradiction.
Corollary 7.6.8. Let A be an autonomous set in a weak aut descendable ausys S,
and let x be in A. Then there is an autonomous subset B of A containing x such that
x is a maximum element of ≤B.
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Proof. Since S is weak aut descendable, we may choose a minimal autonomous subset
B of A containing x. By Lemma 7.6.7, x is the ≤B maximum element.
As an example of the fact that canonical orders say what they “should” for ausyses
satisfying appropriate descendable hypotheses, we are now in a position to show that
for weak aut descendable ausyses, the canonical orders can in fact be used to uniquely
determine the axioms.
Proposition 7.6.9. Let S be a weak aut descendable ausys and let x be in S. Then
x is a nonaxiom iff there is an autonomous set A and some y in A such that y <A x.
Proof. First, if y <A x for some autonomous set A containing y and x then every au-
tonomous subset of A containing x also contains y. Therefore {x} is not autonomous.
Therefore x is a nonaxiom.
Conversely, suppose x is a nonaxiom. Choose an autonomous set B containing x.
Since S is weak aut descendable, we may shrink B to a minimal autonomous subset
of B containing x. Call this set A. Since x is a nonaxiom, we know {x} is not
autonomous. Therefore A contains some other element y. By Lemma 7.6.7, x is a
≤A maximum element of A. Therefore y <A x.
Corollary 7.6.10. For a weak aut descendable ausys S, the following are equivalent:
(i) Every point is an axiom.
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(ii) Every canonical order of autonomous subsets of S is an antichain.
Proof. It is immediate from the previous lemma that if every canonical order is an
antichain then every point is an axiom. Moreover, y <A x implies x is not an axiom,
so if every point is an axiom then every canonical order is an antichain.
While we have not yet defined partial orders as ausyses, it will be immediate from the
definition that an order theoretic antichain is simply an ausys in which every point
is an axiom. The above corollary is thus thought of more perspicaciously as saying a
weak aut descendable ausys is an antichain iff every canonical order is an antichain.
Note one direction of this corollary does indeed require a descendability assumption,
for the standard nondescendable ausys sna is an ausys such that every canonical order
is an antichain, yet sna is not itself an antichain. The other direction of the corollary
is true for ausyses in general. Similarly, if we look at Proposition 7.6.9, we can see one
direction requires a descendability hypothesis, while the other is true generally. This is
typical for ausys statements that relate the canonical orders to structural information.
The structural information typically implies facts about canonical orders regardless
of the ausys, while the other direction requires a descendability hypothesis.
Since descendability may seem somewhat abstract, we make the discussion more
practical by showing all we have proved for weak aut descendable ausyses applies
also to finitary ausyses.
Definition 7.6.11. An ausys is finitary if every autonomous set is the union of finite
autonomous sets.
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In particular, every finite ausys is finitary. While in mathematical logic, natural proof
systems are always infinite, graphs and partial orders are all examples of ausyses,
so combinatorially, finite ausyses are hardly unnatural, so if a theorem is stated
for finitary ausyses or ausyses satisfying a certain descendability hypothesis, it is
important to realize that in particular, the theorem in question is true for all finite
autonomous systems.
Lemma 7.6.12. Finitary ausyses are weak aut descendable.
Proof. Let A be an autonomous set containing a point x. Since A is the union of
finite autonomous sets, there is a finite autonomous subset B of A containing x. If
B is chosen to have minimum cardinality then B is a minimal autonomous subset of
A containing x.
So every fact we prove assuming weak aut descendability has as corollaries that the
same facts are true for all finitary, and in particular all finite ausyses. We note
that finitary ausyses are not in general strong aut descendable. In fact, the example
already given of a weak but not strong aut descendable ausys was finitary.
7.6.3 Canonical Order Basic Theorems And Examples
There would be no need to define canonical orders for each autonomous set A if ≤A
and ≤B agree on A ∩ B for all A,B. Then we could just use one order. To see
that the orders can nontrivially change as the autonomous set changes, consider the
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following example. Let a and b be axioms, and x a nonaxiom such that a implies
x and b implies x. Then the set {a, x} is autonomous, and since x is a nonaxiom,
{x} is not autonomous, therefore the only autonomous subset of {a, x} containing x
is {a, x} itself. In particular, a <{a,x} x. However, {b, x} is an autonomous subset
of the autonomous set {a, b, x} containing x but not a. Therefore a 6<{a,b,x} x. So x
needs a in {a, x}, but a and x are incomparable in {a, b, x}.
Intuitively, x does not need a in {a, b, x} because x can just use b instead. Restricted
to the set {a, x}, the point x no longer has that choice, and needs x. So the canonical
orders can nontrivially change. There are a number of interesting questions and facts
as to how exactly it may do so.
It is implicit in the discussion of the last paragraph that in going to a larger set in
which we have more tools at our disposal, x may no longer need certain points that
it needs in the smaller set, but in going to a smaller set A under containment, x is
more likely to need any element a in A as there are fewer alternatives. This suggests
the following proposition.
Lemma 7.6.13. Let A ⊆ B be autonomous sets and x, y ∈ A. If x <B y then
x <A y.
Proof. Suppose x <B y. Then every autonomous subset of B containing y also
contains x. Therefore every autonomous subset of A containing y also contains x.
Therefore x <A y.
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The following rewording is useful.
Lemma 7.6.14. Let A ⊆ B be autonomous sets. Then ≤B |A fattens ≤A.
We now ask if there is an ausys with autonomous sets A and B such that
A ∩ B contains two points x, y such that x <A y and y <B x. While this may seem
counterintuitive, a typical ausys does in fact exhibit such behavior, in particular those
arising in mathematical logic and from graphs. Before giving such an example, it is
worth first giving a fallacious proof that no such example exists, as its failure contains
an important point.
We argue by contradiction. So, suppose there is an ausys containing autonomous sets
A and B and x, y in A∩B such that x <A y and y <B x. We know by Lemma 7.6.14
that ≤A∩B lengthens ≤A and ≤B, and we know x <A y and y <B x by hypothesis,
so by lengthening we have x <A∩B y and y <A∩B x, contradicting the fact that ≤A∩B
is a partial order and thus satisfies antisymmetry. So, there are no such autonomous
sets A,B and elements x, y in an ausys.
So where is the cheating? A ∩ B need not be autonomous, and canonical orders are
defined only for autonomous sets.
To see the above behavior can in fact happen, let X = {x1, x2, x3, x4} be the vertices
of a graph theoretic path in that order. Define an ausys P4 on X as follows. x1 and
x4 are axioms, and a implies b whenever a and b are adjacent. The reader may check
that {x1, x2, x3} and {x2, x3, x4} are autonomous sets such that x2 <{x1,x2,x3} x3 and
x3 <{x2,x3,x4} x2.
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We will later see that P4 is in fact the simplest example of this phenomenon in the
sense that every strong aut descendable ausys containing some A,B, x, y such that
x <A y and y <B x in fact contains P4 as a minor. We will define the notion of the
minor of an autonomous system in Definition 7.11.10.
We note in passing that the above construction is a generalization of the definition of
rooted trees in terms of trees. A rooted tree is a partial order (the tree order). Since,
as we will see, partial orders are ausyses, it follows that a rooted tree is an ausys. Its
sole axiom is the root, which we think of as a singleton root set. This construction
can be generalized to rooted graphs and digraphs by choosing a root set to serve as
the set of axioms, and letting a imply b if they are adjacent, or if there is a directed
edge from a to b. The connection between graph connectivity and graph matching
through ausyses makes essential use of these rooted graphs.
We now continue with the basic examples and theorems on the canonical order.
Lemma 7.6.15. Let A be autonomous and B an autonomous subset of A. Then B
is downward closed in the canonical order ≤A.
Proof. Let B be an autonomous subset of A. We have to show B is ≤A downward
closed, so let y ∈ B and x <A y. We must show x ∈ B. Since x <A y, we see every
autonomous subset of A containing y also contains x. In particular, B contains x.
Proposition 7.6.16. Let A be autonomous and S an arbitrary subset of A. Then
S is ≤A downward closed iff it is the (possibly infinite) intersection of autonomous
subsets of A.
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Proof. Let us see that if Si ⊆ A for i ∈ I then⋂i∈I Si is ≤A downward closed. Each
Si is by hypothesis an autonomous subset of A and so is ≤A downward closed by
the previous lemma. Since downward closed subsets of an arbitrary partial order are
closed under intersection, in particular so are the ≤A downward closed subsets. It
follows that⋂i∈I Si is ≤A downward closed.
Now, for the nontrivial direction. We must show every ≤A downward closed set S can
be represented as⋂i∈I Si for some autonomous subsets Si of A. It is enough to show
S is the intersection of all autonomous subsets of A containing it, so let I index all
these sets Si. It is obvious S is a subset of the intersection of all autonomous subsets
of A containing it, so we have only to show the reverse inclusion.
So we have to show the intersection⋂i∈I Si is contained in S, which means we must
show every element of⋂i∈I Si is also an element of S. We show the contrapositive,
that given x ∈ A, if x is not an element of S then x is not an element of⋂i∈I Si.
So take x not in S. To show x is not in⋂i∈I Si is to show there is i ∈ I such that
x is not in Si. Since our Si’s are all the autonomous subsets of A containing S, this
means we have to give an autonomous subset of A containing S but not containing
x. It is enough to give, for each s ∈ S, an autonomous subset B of A containing s
and not x. For then⋃s∈S Bi will be the desired autonomous subset of A containing
S and not x, completing the proof.
So take s ∈ S. How do show there is an autonomous subset of A containing s and
not x? If there were no such autonomous subset of A, then every autonomous subset
of A containing s would also contain x, and therefore we would have x ≤A s. Now s
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is in S and S is ≤A downward closed by hypothesis, which implies x is in S, contrary
to our choice of x as an element not in S. This contradiction proves the lemma.
7.7 Partial Orders As Ausyses
Lemma 7.7.1. Let (P,≤) be a partial order. Then the set T of downward closed sets
is closed under arbitrary union and satisfies the order property.
Proof. To see T is closed under arbitrary union, let Di be downward closed for i in an
index set I. We must show D =⋃i∈I Di is downward closed. Let y ∈ D and x < y.
Since y is in D then y is in some Di. Since Di is downward closed and x < y, we see
that x is in Di. Since Di ⊆ D, we see that x is in D.
To see that T satisfies the order property, let D ∈ T be a downward closed set. We
must show there is a total order ≤′ on D such that every ≤′ downward closed set is in
T . In other words, every ≤′ downward closed set must be ≤ downward closed. This
just says that ≤′ is a total order that lengthens ≤. Such lengthenings are known to
exist for every partial order, in particular for ≤.
Corollary 7.7.2. Let (P,≤) be a partial order and let T be the set of ≤ downward
closed sets. Then (P, T ) is an ausys.
So, given a partial order (P,≤), we may consider it an ausys by taking the downward
closed sets as the autonomous sets. We now characterize the proofs in such a partial
order.
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Lemma 7.7.3. Let (P,≤) be a partial order with set T of ≤ downward closed sets.
Then a partial order (Q,≤′) of a subset Q of P is a (P, T ) proof iff Q is ≤ downward
closed and ≤′ lengthens ≤ |Q.
Proof. First, suppose ≤′ is a proof. We must show Q is ≤ downward closed and
that ≤′ lengthens ≤ |Q. Now, ≤′ is a proof so the domain Q is autonomous. The
autonomous sets here are just the ≤ downward closed sets, so Q is ≤ downward closed
as needed. To show ≤′ lengthens ≤ |Q, suppose x < y. Then every ≤ downward
closed set containing y also contains x. Since the autonomous sets in (P, T ) are
exactly the ≤ downward closed sets, this means every autonomous set containing y
also contains x. In particular, since ≤′ is a proof, the set {≤′ y} is autonomous and
therefore contains x. This means x <′ y. Since x and y were chosen arbitrarily in Q,
we see that ≤′ lengthens ≤ |Q as claimed.
For the converse, suppose Q is ≤ downward closed and ≤′ lengthens ≤ |Q. We must
show ≤′ is a proof. By Proposition 7.4.5, it is enough to show every ≤′ downward
closed set is autonomous. This means we must show every ≤′ downward closed set
is ≤ downward closed. Let D be a ≤′ downward closed set, let y be in D, and let
x < y. It is enough to show x is in D. Since x < y and Q is ≤ downward closed, we
see that x is in Q. Since ≤′ lengthens ≤ |Q, the points x, y are in Q, and x < y, we
see that x <′ y. Since D is a ≤′ downward closed set containing y by assumption, we
see that x is in D. This completes the proof.
We now describe the canonical orders in a partial order.
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Lemma 7.7.4. Let A and B be autonomous sets in a poset (P,≤) and let x, y ∈ A∩B.
Then x ≤A y iff x ≤B y.
Proof. Let T be the set of ≤ downward closed sets. We see that x ≤A y iff every
T autonomous subset of A containing y also contains x. Since T is the set of ≤
downward closed sets, this means x ≤A y iff every ≤ downward closed subset of A
containing y also contains x. Since x is in A, this is equivalent to saying x ≤ y.
Similarly, x ≤B y iff x ≤ y. Therefore x ≤B y iff x ≤A y as claimed.
Corollary 7.7.5. Let (P,≤) be a poset, T the set of ≤ downward closed subsets of
P , and A in T . In the ausys (P, T ), the canonical order ≤A of A is the restriction
≤P |A of the canonical order of P .
We may ask if Corollary 7.7.5 has a converse. So suppose (P, T ) is an ausys such that
for all autonomous A,B and for all x, y ∈ A ∩ B we have x ≤A y iff x ≤B y. Does
this imply (P, T ) is a poset?
The answer is no. We can take the standard nondescendable ausys sna as our coun-
terexample. In the standard nondescendable ausys every canonical order is an an-
tichain, which implies x ≤A y iff x ≤B y whenever x and y are in A ∩ B, but sna is
not an antichain.
However, our next proposition states Corollary 7.7.5 does in fact have a converse
under the assumption of weak aut descendability.
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Proposition 7.7.6. Let (P, T ) be a weak aut descendable ausys such that for all
autonomous A and B and for all x, y in A ∩ B, x ≤A y iff x ≤B y. Then P is a
poset.
Proof. Choose x in P . P itself is an autonomous set in the weak aut descendable
ausys P , so we may choose a minimal autonomous subset A of P containing x. We
wish to show A is the unique minimal autonomous set containing x. If A = {x}, we
are done. Suppose not. Then we may choose y in A − x. By minimality of A and
Lemma 7.6.7, we see that y <A x. Therefore y <B x for all autonomous sets B by
hypothesis. Therefore every autonomous set containing x also contains y. Since y
was an arbitrary member of A − x, we see that every autonomous set containing x
also contains A, so A is the unique minimal autonomous set containing x as claimed.
As x was an arbitrary point in P , we see that for every x in P there is a minimum
autonomous set Mx containing x.
Define a binary relation ≤ on P by letting x ≤ y iff x ∈ My. It is simple to check
≤ is in fact a partial order on P . To prove the proposition, it is enough to show the
T autonomous sets are exactly the ≤ downward closed sets. Note that by definition,
{≤ x} = Mx.
If a set A is T autonomous then it contains Mx for all x in A, so A is the union of
the ≤ downward closed sets Mx. Therefore A is ≤ downward closed. Conversely, if a
set A is ≤ downward closed then for all x in A, the set A contains the ≤ downward
closure Mx of x. Therefore A is the union of T autonomous sets Mx. Therefore A is
T autonomous .
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We state another proposition that holds only under the assumption of weak aut
descendability. This will be used in the proof of the forbidden subdot characterization
of posets.
Proposition 7.7.7. Let (P, T ) be a weak aut descendable ausys such that A ∩ B is
autonomous for all autonomous sets A and B. Then P is a poset.
Proof. By Proposition 7.7.6, it is enough to show that for all autonomous sets A and
B and for all x, y in A ∩ B, x ≤A y iff x ≤B y. By symmetry, it is enough to show
that if x ≤A y then x ≤B y.
Suppose this is not always the case. Take autonomous sets A and B and x and
y in their intersection such that x ≤A y but x 6≤B y. Since x 6≤B y, there is an
autonomous subset B′ of B containing y and not x. Since A and B′ are autonomous,
so is A ∩ B′ = C. Note that C is an autonomous subset of A containing y and not
x. Therefore x 6≤A y, contrary to assumption. This contradiction shows x ≤B y as
claimed.
Since all partial orders are ausyses, but the converse is not true, we now wish to know
which ausyses are partial orders. There are a number of such characterizations of the
class of partial orders, some holding for ausyses in general, and some holding only
under an appropriate assumption of descendability. The number and the variety of
characterizations further suggests the naturality of the class as a subclass of ausyses.
The next few lemmas give some such characterizations.
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We first characterize posets in terms of proofs and canonical orders.
Lemma 7.7.8. Let (P, T ) be an ausys given by autonomous sets. Then the following
are equivalent.
(i) (P, T ) is a poset.
(ii) Letting ≤P be the canonical order of the whole ausys P , the proof closure of
{≤P} is the set of all T proofs.
(iii) There is a T proof ≤ such that the proof closure of {≤} is the set of all T proofs.
(iv) There is a partial order ≤ such that the proof closure of {≤} is the set of all T
proofs.
(v) The canonical order ≤P of the whole ausys is a T proof.
Proof. (i)⇒ (ii): We are assuming (P, T ) is a poset, which means there is a partial
order ≤ on P such that the corresponding poset is (P, T ). By Lemma 7.7.3, it
follows that the set of T proofs is exactly the set of lengthenings of ≤ restricted to
≤ downward closed subsets of P . It is an exercise in partial orders, needing no ausys
notions, to show that the proof closure of a single partial order ≤ on a set P is the
set of all lengthenings of ≤ on ≤ downward closed subsets Q of P . Therefore the set
of T proofs is exactly the proof closure of {≤}. Since (P,≤) is a poset by hypothesis,
we know that ≤ and ≤P are in fact the same order. We have thus shown (i) implies
(ii).
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(ii) ⇒ (iii): Assuming the canonical order ≤P of P has the set of all proofs as its
proof closure, we must show there is a proof having the set of all proofs as its proof
closure. It is enough to show the canonical order ≤P is in fact a proof. But this is
trivial. ≤P has the set of all proofs as its proof closure. And ≤P is of course in the
proof closure of ≤P . So ≤P is in the set of all proofs. In other words, the canonical
order ≤P is a proof.
(iii)⇒ (iv): Immediate.
(iv)⇒ (v): So, assume there is a partial order ≤ such that the proof closure of {≤}
is the set of all proofs. We must show the canonical order ≤P of the whole ausys
is a proof. Note that ≤ is of course in the proof closure of {≤} and so is a proof,
therefore it is enough to show that ≤ is in fact the same partial order as ≤P , for then
≤P is a proof as desired.
So let us show ≤ and ≤P are the same partial order. ≤ is a proof and the canonical
order ≤P fattens all proofs therefore ≤P fattens ≤. We have only to show that ≤
fattens ≤P . First, we need the fact that the only partial orders with domain P in
the proof closure of {≤} are the partial orders with domain P lengthening ≤. This
is easy from the definition of proof closure and is left as an exercise. Assuming this
fact, recall that the canonical order ≤P of P is the inf under lengthening of all proofs
with domain P . By the exercise just mentioned, that means the canonical order ≤P
is the inf of all the partial orders with domain P lengthening ≤. The inf of a set of
partial orders lengthening ≤ also lengthens ≤, therefore ≤P lengthens ≤ as claimed,
thus showing ≤ is the same order as ≤P , thus completing the proof that (iv)⇒ (v).
(v)⇒ (i): To complete the proof of the lemma, showing the equivalence of these 5
128
conditions, we must prove finally that if the canonical order ≤P of the whole ausys
is a proof then (P, T ) is in fact a poset.
To show that (P, T ) is a poset, we show it is in fact the poset (P,≤P ). That is, we
show a set is in T iff it is ≤P downward closed.
First, we must show every T autonomous set is ≤P downward closed. This just says
the autonomous sets are downward closed in the canonical order, which is exactly
Lemma 7.6.15. Second, we must show every≤P downward closed set is T autonomous.
By assumption, ≤P is a T proof. Since every downward closed subset of a proof is
autonomous, it follows that every ≤P downward closed set is T autonomous.
The following lemma characterizes posets in terms of autonomous sets.
Proposition 7.7.9. Let (P, T ) be an ausys given by autonomous sets. Then the
following are equivalent:
(i) (P, T ) is a poset.
(ii) The T autonomous sets are closed under arbitrary intersection.
Proof. (i)⇒ (ii): If (P, T ) is a poset then we may take ≤ on P such that the sets
in T are exactly the ≤ downward closed sets. Since the downward closed sets of a
partial order are closed under arbitrary intersection, we see the autonomous sets of
(P, T ) are as well.
(ii) ⇒ (i): For the converse, we assume the T autonomous sets are closed under
arbitrary intersection. By Proposition 7.6.16, the ≤P downward closed subsets of
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P are exactly the intersections of autonomous subsets of P . Since we are assuming
the arbitrary intersection of autonomous sets is autonomous, this implies the ≤P
downward closed sets are exactly the T autonomous sets. Therefore (P, T ) is a partial
order as claimed.
Lemma 7.7.10. Let (P, T ) be an arbitrary ausys. Then the following are equivalent.
(i) For all autonomous sets A,B and for all x, y in A ∩B, x ≤A y ⇔ x ≤B y.
(ii) The canonical orders never fatten. In other words, for all autonomous sets
A ⊆ B the orders ≤A and ≤B |A are the same.
Proof. (i) ⇒ (ii) should be obvious, so let us prove (ii) ⇒ (i). We prove the
contrapositive instead, so assume (i) fails. Then there are autonomous sets A,B and
x, y ∈ A ∩ B such that x ≤A y and x 6≤B y. Now B ⊆ A ∪ B and so ≤A∪B fattens
≤B. Since x 6≤B y this implies x 6≤A∪B y. So we have x ≤A y and x 6≤A∪B y, and A is
a subset of A∪B. Hence the canonical orders fatten from A to A∪B, and (ii) fails.
7.8 Well Founded Autonomous Systems
While almost all of the general theory goes through without the assumption of well
foundedness, it is still a particularly important class of autonomous systems, so a few
words are in order. Moreover, when we define blocking and see it carries the same
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information as the proofs and autonomous sets, we naturally attempt to axiomatize
autonomous sets in terms of blocking, but at present, the only blocking axiomatization
known requires well foundedness.
Definition 7.8.1. An ausys is well founded if each proof is in the proof closure of
the set of well ordered proofs.
The reader may show the above definition remains unchanged if we say well founded
proofs instead of well ordered proofs. We now wish to characterize well foundedness
in terms of autonomous sets. In the following lemma, well orders can also be replaced
with well founded partial orders with no effect.
Lemma 7.8.2. An autonomous system P is well founded iff for every autonomous set
A there is a well order ≤′ of A such that every ≤′ downward closed set is autonomous.
Proof. Suppose P is well founded. Choose an autonomous set A. We know by the
order property that there is a total order ≤ of A such that every ≤ downward closed
set is autonomous. In other words, ≤ is a totally ordered proof of A. Since P is well
founded, the proof ≤ is in the proof closure of some set {≤i}i∈I of well ordered P
proofs. Merging these proofs in any order yields a well ordered proof ≤m of A.
For the converse, suppose for every autonomous set A there is a well order ≤′ of A
such that every ≤′ downward closed set is autonomous. We must show the set of
proofs is the proof closure of a set of well founded partial orders. Let ≤ be a proof
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with domain A and let x be in A. Then {≤ x} is autonomous. Therefore there
is a well order ≤′ with domain {≤ x} such that every ≤′ downward closed set is
autonomous. Therefore ≤′ is a well ordered proof such that
{<′ x} ⊆ {< x}.
Since the proof ≤ was arbitrary and x was an arbitrary point in its domain, it follows
that each proof is in the proof closure of a set of well founded proofs, as was to be
shown.
Our choice of wording of the previous lemma was intentional, in order to show how
one may define ausys well foundedness directly in terms of the autonomous sets.
However, it is unnecessarily verbose for practical purposes. More simply, it states
every autonomous set has a well ordered proof.
This shows, moreover, that our definition of ausys well foundedness agrees with the
notion of partial order well foundedness in the case that an ausys P is a partial order,
for proofs in partial orders are just lengthenings of the order and the autonomous
sets are just downward closed sets, so ausys well foundedness for a partial order
simply states that the restriction of the order to every downward closed set can be
lengthened to a well order. A partial order satisfies this property iff it is partial order
theoretically well founded.
We now state a characterization of well foundedness in terms of autonomous sets that
should be considered the well founded version of the order property.
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Corollary 7.8.3. An ausys P is well founded iff for all autonomous sets A and all
autonomous sets B properly containing A, there is x in B − A such that A ∪ {x} is
autonomous.
Proof. Suppose P is well founded. Let A and B be autonomous sets such that B
properly contains A. Since P is well founded, we may choose well ordered proofs ≤1
of A and ≤2 of B. Merge the proof ≤1 and ≤2 in that order to obtain a well ordered
proof ≤m of B containing A as an initial segment. Since B properly contains A, there
is x in B such that y <m x for all y in A. Since ≤m is a well order, we may assume
without loss of generality that we choose the minimum such x. Then A ∪ {x} is an
initial segment of the proof ≤m. Therefore A ∪ {x} is autonomous as needed.
For the converse, suppose for all autonomous sets A and all autonomous sets B
properly containing A, there is x in B − A such that A ∪ {x} is autonomous. It
is enough to show every autonomous set has a well ordered proof. This is true by
transfinite recursion, noting the null set has a well ordered proof (them empty partial
order is well founded) for α = 0, using the fact that autonomous sets are closed
under arbitrary union for the limit ordinals, and adding an x as in the statement of
the lemma for the successor ordinals.
The previous corollary can be used to show that the well founded ausyses can be
axiomatized by their autonomous sets as in the following proposition. Details are left
to the reader.
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Proposition 7.8.4. Let P be a set and T be a set of subsets of P . Then (P, T ) is a
well founded ausys iff the following conditions hold:
(i) T is closed under arbitrary union.
(ii) The null set is in T .
(iii) Given a set A in T and a set B in T properly containing A, there is x in B−A
such that A ∪ {x} is in T .
Now that we have an ausys definition of well foundedness, we wish to address the
potential problem that if autonomous systems allow nonwell founded proofs then for
proof systems arising in mathematical logic, partial orders that “should not be” proofs
are, and this actually leads to unprovable formulas being provable in the ausys. The
solution is to define the well founded part of an ausys. First, we need a lemma that
will justify the definition.
Lemma 7.8.5. Let (P, T ) be an ausys, and let W be the set of autonomous sets with
a well ordered proof. Then (P,W ) is a well founded ausys.
Proof. For closure under arbitrary union, if sets Ai have well ordered proofs ≤i for
i in some index set I then the proofs can be merged in any order to obtain a well
ordered proof of⋃i∈I Ai. For the order property, it is trivial that every member of
W can be well ordered such that every initial segment is in W .
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We call the ausys of the previous lemma the well founded part of an autonomous
system. Given a preceding set proof system arising in mathematical logic, one may
define an ausys (P, T ) from it. While it is true that many of the T proofs “should not
be” proofs, and unprovable formulas in general become provable, this ceases to occur
on passing to the well founded part. The well founded part has exactly the proofs
and provable elements one would expect.
7.9 Blocking
We have so far met two important axiomatizations of ausyses, one by proofs and
the other by autonomous sets. The third and final important axiomatization is by
blocking, as we will see that blocking gives the same information as the proofs and
the autonomous sets.
Definition 7.9.1. Let (P, T ) be an ausys, S an arbitrary subset of P , and x an
element of P . Then S blocks x iff for all proofs ≤ containing x, there is s in S such
that s < x.
In particular, s must be in the domain of ≤. If S is the singleton {s}, we will also
say s blocks x. We now give some properties equivalent to S blocking x. The simple
proof is left to the reader.
Lemma 7.9.2. Let (P, T ) be an ausys, S a subset of P , and x in P . Then the
following are equivalent.
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(i) S blocks x.
(ii) For every proof ≤ containing x, ≤ also contains an element of S other than x.
(iii) For every proof ≤ containing x, the set S ∩ {< x} is nonempty.
(iv) Every autonomous set A containing x also contains an element of S other than
x.
(v) For every autonomous set A containing x, the set A ∩ S − x is nonempty.
We have just given equivalent definitions of blocking in terms of proofs and in terms
of autonomous sets. Not surprisingly, a characterization of blocking in terms of
canonical orders requires a descendability hypothesis.
Lemma 7.9.3. Let P be a weak aut descendable ausys, S a subset of P , and x an
element of P . Then S blocks x iff for all autonomous sets A containing x there is an
autonomous subset B of A containing x such that the set {<B x} ∩ S is nonempty.
Proof. First, suppose S blocks x, and choose an autonomous set A containing x. Since
P is weak aut descendable, we know by Corollary 7.6.8 that there is an autonomous
subset B of A containing x such that x is ≤B maximum. Since S blocks x, the set
(B − x) ∩ S = {<B x} ∩ S
is nonempty as claimed.
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For the converse, note that if S does not block x then there is an autonomous set A
containing x and disjoint to S. Then every autonomous subset B of A containing x
is disjoint to S as well. In particular, {<B x} ∩ S is empty.
We claimed blocking carries the same information as the autonomous sets and as the
proofs. Let us see that this is so. By the definition of blocking, the proofs obviously
uniquely determine blocking. We know the autonomous sets also uniquely determine
blocking by the last lemma, or by the fact that the autonomous sets uniquely de-
termine the proofs. The next two lemmas finish showing blocking carries the same
information by characterizing the proofs and the autonomous sets in terms of block-
ing.
Proposition 7.9.4. Let (P, T ) be an ausys and A a subset of P . Then A is au-
tonomous iff P − A does not block any element of A.
Proof. First, suppose A is autonomous and let x in A. We must show P − A does
not block x. By lemma Lemma 7.9.2, it is enough to show there is an autonomous
subset of P containing x and no element of P −A other than x. Since x is in A, the
“other than x” makes no difference, and so we are looking for an autonomous subset
of P containing x and no element of P −A. Of course A is such an autonomous set,
thus P − A does not block x, and this direction of the lemma is proved.
For the converse, let A be an arbitrary subset of P such that P − A does not block
any element of A. We must show A is autonomous. We know there is a largest
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autonomous subset A′ of A, the autonomous part of A. A is autonomous iff A = A′.
Suppose A 6= A′. Since A′ is a subset of A, this implies A′ is a proper subset of A, so
take y ∈ A−A′. By hypothesis, P −A does not block y. By lemma ??, this implies
there is an autonomous subset B of A containing y and no element of P − A other
than y. You can check this just means B is an autonomous subset of A containing y.
Now consider B ∪A′. B ∪A′ is an autonomous subset of A, and since y is in B ∪A′
but not in A′, we know B ∪A′ properly contains A′, contrary to the assumption that
A is the largest autonomous subset of A.
Proposition 7.9.5. Let (P, T ) be an ausys and ≤ a partial order of a subset of P .
Then ≤ is a proof iff for all x in the domain of ≤, the set P − {≤ x} does not block
x.
Proof. First, suppose ≤ is a proof and take x in the domain of ≤. We will show that
P − {≤ x} does not block x. It is enough to show there is an autonomous subset B
of P containing x and no element of P − {≤ x} other than x. Just set B = {≤ x}.
This completes one direction.
For the other direction, let ≤ be a partial order such that for all x in the domain of ≤
the set P − {≤ x} does not block x, and we must show ≤ is a proof. It is enough to
show every ≤ downward closed set is autonomous. Since every ≤ downward closed
set is a union of sets of the form {≤ x}, and since the ≤ downward closed sets are
closed under arbitrary union, it is enough to check that all subsets of the domain of
≤ of the form {≤ x} are autonomous.
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So take such a set of the form C = {≤ x}. By the previous lemma, it is enough to
show P−C does not block any element c of C. Take c in C. By hypothesis, P−{≤ c}
does not block c. And c ∈ C = {≤ x} so c ≤ x which implies {≤ c} ⊆ {≤ x}. Thus
P − {≤ x} ⊆ P − {≤ c}. So P − {≤ x} ⊆ P − {≤ c} and P − {≤ c} does not block
c, which implies P − {≤ x} does not block c. In other words, P − C does not block
c. Since c was arbitrary, the proof is complete.
Now that we know blocking carries the same information as the proofs and as the
autonomous sets, we want to know the axiomatization of ausyses in terms of blocking.
Currently, the only known axiomatization by blocking requires the assumption of well
foundedness of the autonomous system. We should note there is no intrinsic reason
an axiomatization by blocking should be limited to well founded partial orders while
the rest of the theory goes through in general. It is the author’s belief that a natural
axiomatization of general autonomous systems exists, and simply has not been found
yet.
First, we will state two lemmas, each one showing that blocking in a well founded
ausys satisfies a certain property. Next, we prove these properties are in fact sufficient
to guarantee that the blocking arises from a well founded ausys.
Lemma 7.9.6. Let P be an ausys, S1 ⊆ S2 ⊆ P , and x ∈ P . If S1 blocks x then S2
blocks x also.
Proof. If S1 blocks x then in every proof ≤ containing x, there is some element s < x
in S1. This x is also in S2, and thus S2 blocks x as ≤ was arbitrary.
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Lemma 7.9.7. Let P be a well founded ausys and S ⊆ P , and let S ′ be the set of all
elements s in S such that S − {s} blocks s. Then S − S ′ blocks every element of S ′.
Proof. Let t in S ′. We must show S − S ′ blocks t. That’s equivalent to showing in
every proof ≤ containing t there is some element of S − S ′ that is < t. So take any
proof ≤ containing t. The element t is in S ′ so by definition of S ′, we know S − {t}
blocks t. That is, there is some element c < t with c ∈ S − {t}. If c is in S − S ′ then
we are done, so suppose c is in S ′. Then ≤ is a well founded partial order containing
at least one element of S ′ that is ≤ c, namely c itself, so take a ≤ minimal such
element c′ ≤ c. c′ is in S ′ by choice, so S−{c′} blocks c′, and there is thus an element
c′′ < c′ with c′′ ∈ S −{c}. c′′ can not be in S ′ by choice of c′ as minimal, and so c′′ is
in S − S ′. We thus have c′′ < c′ ≤ c < t and so c′′ is an element of S − S ′ that is < t.
As ≤ was arbitrary, there is always such a c′′, and hence S −S ′ blocks every element
of S ′ as claimed.
Now, we show the last two properties are in fact enough to guarantee the blocking
arises in a well founded ausys.
Proposition 7.9.8. Let P be a set and B be a set of ordered pairs (S, x) with S ⊆ P
and x ∈ P . Then B is the set of pairs of the form S blocks x for some well founded
ausys on P iff the following two conditions hold:
(i) If (S1, x) is in B and S1 ⊆ S2 ⊆ P then (S2, x) is in B.
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(ii) For every S ⊆ P , if S ′ is defined as the set of all s ∈ S such that the pair
(S − {s}, s) is in B, then the pair (S − S ′, t) is in B for every t in S ′.
Proof. We have obviously shown in the last two lemmas that the two conditions are
necessary in order for B to be blocking in a well founded ausys. We need only show
sufficiency. Throughout the lemma, since we do not know B actually gives blocking
yet but would like to speak of blocking as opposed to ordered pairs, we will say “S
B-blocks x” to mean the pair (S, x) is in B.
So suppose the two conditions are satisfied. We must show B is the set of pairs of the
form S blocks x for some ausys on P . We take ausyses to be given by autonomous
sets. So we have to show if B satisfies the two conditions, then the corresponding
autonomous sets are the autonomous sets of an ausys. What do we mean by “corre-
sponding” autonomous sets? We know how to characterize the autonomous sets in
terms of blocking, so we take this as our definition of corresponding. More rigorously,
given our set of pairs B as above, we define T to be the set of all subsets A of P such
that P − A B-blocks no element of A.
How do we show T is the set of autonomous sets of a well founded ausys? It is enough
to show three things. First, that the null set is in T . Second, that T is closed under
arbitrary union. And third, that if C and D are both in T with C a proper subset of
D, then there is x ∈ D − C such that C ∪ {x} is in T .
First, let us see that the null set is in T . By definition of T , we must show that P −∅
B-blocks no element of ∅. This is vacuously true.
Second, let us see that T is closed under arbitrary union. So let Ai be in T for i ∈ I
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and let A =⋃i∈I Ai be their union. We must show A is in T . We know that for each
i ∈ I, the set P − Ai B-blocks no element of Ai. By (i), which we are assuming B
satisfies, and the fact that P −A ⊆ P −Ai, it follows that P −A B-blocks no element
of Ai. Every element of A is in some Ai, hence it follows that P − A B-blocks no
element of A. Therefore A is in T by definition of A.
Third, and finally, we must show that if C and D are both in T with C a proper
subset of D, then there is x ∈ D−C such that C∪{x} is in T . So take such C and D,
and we have to find an x ∈ D−C such that that C∪{x} is in T , or equivalently such
that P − (C ∪ {x}) B-blocks no element of C ∪ {x}. Since C is in T we know P −C
B-blocks no element of C. P − (C∪{x}) is a subset of P −C so by (i), P − (C∪{x})
does not B-block any element of C either. So saying P − (C ∪ {x}) B-blocks no
element of C ∪ {x} is equivalent to saying P − (C ∪ {x}) does not B-block x. Thus,
our problem is reduced to showing there is x ∈ D −C such that P − (C ∪ {x}) does
not B-block x.
So suppose there is no such x. Then the set P − (C∪x) B-blocks x for all x ∈ D−C.
Let S = P−C. Then what we know is that the set S−x B-blocks x for all x in D−C.
Defining S ′ in terms of S as above as the set of all s in S such that S−{s} B-blocks
s, we see this just means D − C is a subset of S ′. By (ii), the set S − S ′ B-blocks
every element of S ′, in particular every element of D − C. So S − S ′ B-blocks every
element of D−C, and it is obvious via diagram or simple boolean computation that
P − D contains S − S ′, so it follows from (i) that P − D B-blocks every element
of D − C. In particular, the set P − D B-blocks x. This is a contradiction, for we
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assumed D is in T , which by definition of T means P −D B-blocks no element of D.
7.9.1 The Blocking Order
The second axiom in our blocking axiomatization says that for every subset S of the
ausys, if we define S ′ as before, then S − S ′ blocks every element of S ′. Rather than
saying this, it is more convenient simply to say S−S ′ blocks S ′. As we will soon see,
blocking then becomes a well founded partial order on the subsets of of every well
founded ausys.
Definition 7.9.9. Let P be an ausys and let C and D be subsets of P . We say C
blocks D if C blocks every element of D.
Proposition 7.9.10. Let P be a well founded ausys. With the definition above, the
set of nonempty subsets of P is a strict partial order under blocking.
Proof. We have to show blocking is an irreflexive, antisymmetric, transitive relation
on the set of subsets of P .
For irreflexivity, take an arbitrary nonempty subset C of P and we must show C does
not block C, so we must show there is an element c of C such that C does not block
c. Take a well founded proof ≤ of the whole ausys. C is nonempty so there is some
element of C in the domain of ≤. Since ≤ is well founded there is a minimal such.
We take this as our c. If C blocked c, then there would be c′ < c in the domain of
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≤, which there is not by our choice of c as minimal. So C does not block c and we
are done.
For antisymmetry, we must show there are no nonempty sets C and D such that C
blocks D and D blocks C. Suppose there are such C and D. Take a proof ≤ of the
whole ausys. Take a ≤ minimal element b of C ∪D. b is in at least one of C or D,
so without loss of generality b is in C. Since b is ≤ minimal there is no element of D
that is < b, hence D does not block b, and b is in C, therefore D does not block C.
So there are no nonempty sets C and D that both block one another, as was to be
shown.
For transitivity, let C, D, and E be nonempty sets such that C blocks D and D
blocks E. We need to show C blocks E, which is to show C blocks every element e
of E. So take e in E. It is enough to show every proof ≤ containing e contains an
element c ∈ C such that c < e.. Take any proof ≤ containing e. By assumption,
D blocks E and so D blocks e also. Thus there must be d ∈ D such that d < e.
Similarly, by assumption we know C blocks D and in particular C blocks d. Hence
there is c ∈ C such that c < d. We then have c < d < e so c < e and we are done.
We now show the blocking order on a well founded ausys is in fact a well founded
partial order.
Proposition 7.9.11. Let P be a well founded ausys. Then blocking is a well founded
partial order on the nonempty subsets of P .
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Proof. Suppose not. Then there is an infinite sequence of subsets Cn for positive
integers n such that Cn+1 blocks Cn for all n. Take a proof ≤ of the whole ausys. ≤
is well founded so there is a ≤ minimal element c of⋃Cn. Since c is in the union,
it must be in some Ci. Ci+1 blocks Ci by assumption, and in particular Ci+1 blocks
c. Hence there is c′ ∈ Ci+1 such that c′ < c, contrary to ≤ minimality of c. This
contradiction completes the proof.
We note that while C blocking D is defined for arbitrary ausyses, blocking is only a
preorder for ausyses in general. Moreover, the corresponding quotient order is not in
general well founded. We leave details to the reader.
We also note that the canonical order of the entire ausys is the restriction of the
blocking order to the singletons. In other words, given points x and y in an ausys P ,
the singleton {x} blocks {y} iff x <P y.
7.9.2 Blocking In Posets
We now discuss blocking in partial orders. It is worth noting that no statement in
this section requires the assumption of well foundedness.
Lemma 7.9.12. Let (P,≤) be a poset, S a subset of P , and x an element of P . Then
S blocks x iff there is s ∈ S such that s < x.
Equivalently,
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Lemma 7.9.13. Let (P,≤) be a poset, S a subset of P , and x an element of P . Then
S blocks x iff S ∩ {< x} is nonempty.
Corollary 7.9.14. Let (P,≤) be a poset and x, y ∈ P . Then x blocks y iff x < y.
The above corollary can also be proved using the fact that the restriction of blocking
to the singletons is the canonical order of P , together with the fact that the canonical
order ≤P of a poset (P,≤) is just ≤ itself.
We now characterize partial orders by blocking.
Lemma 7.9.15. Let P be an ausys. Then P is a poset iff whenever S blocks x, there
is s ∈ S such that s blocks x.
Proof. That posets satisfy the stated condition follows trivially from what has been
said. We prove the less trivial direction.
For the less trivial direction, we have an ausys P such that whenever S blocks x,
there is s in S such that s blocks x. We have to show P is a poset, which means
showing there is a partial order ≤ on P such that the ≤ downward closed sets are
exactly the autonomous sets in P . It is enough to show a subset of P is autonomous
iff it is downward closed in the canonical order ≤P .
We know by Lemma 7.6.15 that every autonomous subset of P is ≤P downward
closed. We just have to show every ≤P downward closed subset of P is autonomous.
Take a ≤P downward closed subset A of P . To show A is autonomous it is enough
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to show that P −A blocks no element of A. Suppose P −A does block some element
a of A. By hypothesis, this means there is some s in P − A such that s blocks a.
We know the blocking order on the singletons is just the canonical order ≤P , so s
blocks a just means s <P a. A is autonomous and therefore ≤P downward closed ,
and a ∈ A so s must be in A also. This contradicts our choice of s in P − A. This
contradiction completes the proof.
7.10 Ausys Lexicographic Sum
In this section, we define the lexicographic sum of autonomous systems that gener-
alizes the partial order definition of lexicographic sum. The lexicographic sum may
roughly be thought of as thinking of an ausys as an “outer” ausys, and for each point
of the outer ausys, giving it the “inner” structure of another ausys, which we call an
inner ausys.
Before defining ausys lexicographic sum, we must define the notion of lexicographic
proof, which uses the notion of a lexicographic partition of a poset.
Definition 7.10.1. Let (P,≤) be a partial order. A partition τ of P is called lexi-
cographic if for all cells C and D of τ , for all x, x′ in C, and for all y, y′ in D, you
have x ≤ y iff x′ ≤ y′.
Definition 7.10.2. Let O be an ausys and let {Po : o ∈ O} be a set of pairwise
disjoint ausyses indexed by O. A lexicographic proof is a partial order (P,≤) with
domain⋃o∈O Po such that the partition into cells Po is lexicographic.
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For clarity, it is worth pointing out that the lexicographic “proofs” defined in the
previous lemma are, at this point, not proofs of anything at all, as we have not
yet defined an ausys in which they are proofs, and “lexicographic proof” is just
a suggestive name for a certain type of partial order. The name comes from the
following definition.
Definition 7.10.3. Let O be an ausys and let {Po}o∈O be a set of disjoint ausyses
indexed by O. Then the lexicographic sum⊕
O Po is defined as the unique ausys on⋃o∈O Po whose set of proofs is the proof closure of the set of lexicographic proofs.
Note that now, the lexicographic proofs are indeed proofs, as we have just defined
the ausys lexicographic sum they are proofs in.
We must fix some terminology for efficiency of presentation. We call O the outer
ausys and each Po an inner ausys. Given a subset S of⊕
O Po, for each o we call
S ∩Po the o part of S. We call Po a full inner cell of S if Po = S ∩Po. Otherwise, Po
is a partial or nonfull cell of S. We define the full part of S to be the union of the
full inner cells. We define NO(S) and FO(S) to be the set of points o in O such that
the o part of S is nonempty, or full, respectively. An O proof is sometimes called an
outer proof. Pi proofs may be called inner proofs.
Given an ausys (P, T ) and a partition τ of P , we say that τ is a lexicographic partition
if there is an ausys (O, T ′) and ausyses (Po, To) for o in O such that
⊕O
Po = P.
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Note that since we say equal, not isomorphic, in particular, each Po is an ausys on
a τ cell. We call a subset S of an ausys P an inner cell of P if S is a cell of some
lexicographic partition of P . Note such lexicographic partitions are generally not
unique. We will later see there is in a certain sense a most natural lexicographic
partition of each finite ausys.
The reader may check that the ausys theoretic and order theoretic definitions of
lexicographic sum and lexicographic partition coincide in the case that all ausyses
are partial orders, so that we may use these terms without risk of ambiguity. We
now prove some other basic properties, including characterizing the autonomous sets,
blocking, and canonical orders in the lexicographic sum.
First, we need a lemma regarding the autonomous sets and proof closure. A spanning
set of proofs in an ausys is a set of proofs whose proof closure is the set of all proofs.
Lemma 7.10.4. Let P be an ausys and let U be a spanning set of proofs for P .
Then a subset A of P is autonomous iff there are proofs {(Ai,≤i)}i∈I in U and ≤i
downward closed sets Di for i in the index set I such that
A =⋃i∈I
Di.
Proof. First, let A be autonomous. Then there is a proof ≤ with domain A. Let
x ∈ A. Since U is a spanning set of proofs, there is a proof (Bx,≤x) in U such that
{≤x x} ⊆ {≤ x} ⊆ A.
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Let Ax = {≤x x}. Then Ax is a ≤x downward closed subset of A containing x, so
A =⋃x∈A
Dx
as required.
For the converse, suppose
A =⋃i∈I
Di
for some proofs ≤i in U and ≤i downward closed sets Di. Each Di is a downward
closed subset of a proof and therefore autonomous. Therefore their union A is au-
tonomous as well.
In characterizing the autonomous sets of the lexicographic sum, we make use of the
following fact, whose proof is left to the reader.
Lemma 7.10.5. Given sets X ⊆ Y in an ausys P , the following conditions are
equivalent:
(i) There is a proof of Y in which every element of Y −X is maximal.
(ii) X is autonomous and X ∪ y is autonomous for all y in Y −X.
(iii) For each y in Y there is a proof ≤ containing y such that {< y} ⊆ X.
Proposition 7.10.6. Let O be an ausys and let {Po : o ∈ O} be a set of pairwise
disjoint ausyses indexed by O. A subset S of the lexicographic sum P =⊕
O Po is
autonomous iff the following conditions hold:
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(i) For all o ∈ O, the o part of S is autonomous in Po.
(ii) There is an outer proof ≤ of NO(S) such that every partial cell of S is ≤
maximal.
Proof. First, suppose that for all o ∈ O, the o part of S is autonomous in Po and that
there is an outer proof ≤ such that the partial cells of S are all ≤ maximal. Let So
be the (possibly empty) o part of S for each o in O. Then for each o the ausys Po
has a proof ≤o such that So is ≤o downward closed. Then the lexicographic sum
⊕≤
≤o
is a lexicographic proof in which S is downward closed. Therefore S is autonomous.
For the converse, suppose S is autonomous. We have two things to show. First, we
show the o part S∩Po of S is autonomous in Po by applying Lemma 7.10.4 with U the
set of lexicographic proofs. By Lemma 7.10.4, S can be written as⋃i∈I Di for some
lexicographic proofs ≤i of P and some ≤i downward closed sets Di for i in an index
set I. Since Di is ≤i downward closed, we see in particular that the o part Di ∩Po is
≤i |Po downward closed for each o. Since ≤i is a lexicographic proof, it follows that
≤i |Po is a Po proof, so the ≤i |Po downward closed set Di ∩ Po is autonomous in Po.
Since this is true for each i, we see that
⋃i∈I
(Di ∩ Po) =
(⋃i∈I
Di
)∩ Po = S ∩ Po
is autonomous as claimed.
151
Now we show there is an outer proof ≤ such that the partial cells of S are all ≤
maximal. By the previous lemma, with X = FO(S) and Y = NO(S), it is enough to
show for each o in NO(S) there is an outer proof ≤′ containing o such that {<′ o} ⊆
FO(S).
Similar to before, we write S as⋃i∈I Di for some index set I, lexicographic proofs ≤i
defined as ⊕≤O,i
≤o,i
for some outer proof ≤O,i and inner proofs ≤o,i, and ≤i downward closed subsets Di
of S. Choose o in NO(S). Since o is in NO(S), we may choose x in Po ∩ S. Then
there is some i such that x is in Di. Suppose p <O,i o. Then every y in Pp satisfies
y <i x. Since Di is a ≤i downward closed subset of S, we see that Pp ⊆ S. Therefore
p is in FO(S). Since p was an arbitrary member of {<O,i o}, it follows that ≤O,i is an
outer proof such that {<O,i o} ⊆ FO(S), thus completing the proof.
We leave the verification of the following characterizations of blocking and canonical
orders in the lexicographic sum to the reader.
Proposition 7.10.7. S blocks x in the lexicographic sum iff one of the following
conditions holds:
(i) S contains a subset T of the inner cell of x such that T blocks x in the inner
cell.
(ii) There is a subset T of the outer ausys such that T blocks the inner cell of x in
the outer ausys and for all t in T , the t part of S is nonempty.
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Proposition 7.10.8. If A is autonomous in the lexicographic sum and x, y ∈ A then
x <A y iff one of the following conditions holds:
(i) x and y belong to distinct inner cells, and the inner cell of x is less than the
inner cell of y in the canonical order of the outer ausys.
(ii) x and y belong to the same inner cell Po and x is less than y in the canonical
order of A ∩ Po in the ausys Po.
7.11 Subausys, Dot, Homomorphisms, and Minors
We now make two definitions that are both, intuitively, notions of “subausys”. The
first, which we in fact call subausys, is analogous to the definition of subspace of a
topological space. The second, which we simply call “P dot A” for notation and lack
of a better name, is instead obtained by restricting the domain but keeping the same
autonomous sets. We make these statements precise in the following definitions.
Definition 7.11.1. Let (P, T ) be an ausys and S a subset of P . Then the restriction
P |S of P to S, or the subausys of P with domain S, is defined as the ausys (S, T ′) with
domain S and set of autonomous sets T ′ = {B ⊆ S : B = A ∩ S for some A in T}.
Definition 7.11.2. Let (P, T ) be an ausys and A an autonomous subset of P . Then
P.A, read “P dot A”, is defined as the ausys (A, T ′) with domain A and set of
autonomous sets T ′ = {B : B ⊆ A and B ∈ T}.
153
Note P |S is defined for all subsets S of P , but we define P.A for autonomous A only.
Definition 7.11.3. Let P and Q be ausyses. We say that Q is a subdot of P if there
is a sequence P = P1, . . . , Pn = Q of ausyses such that for each i with 1 ≤ i < n, one
of the following conditions holds:
(i) Pi+1 = Pi|S for some subset S of Pi.
(ii) Pi+1 = Pi.A for some autonomous subset A of Pi.
Definition 7.11.4. An ausys homomorphism is a function f from an autonomous
system P to an autonomous system Q such that f−1(A) is autonomous in P for all
autonomous A ⊆ Q.
We say an equivalence relation ∼ on an ausys (P, TP ) is homomorphism induced if
there is an ausys (Q, TQ) and a homomorphism f : P → Q such that for all x, y in
P , x ∼ y if and only if f(x) = f(y). In other words, when there is a homomorphism
f for which the ∼ classes are exactly the inverse images of points under f . We call a
partition of P homomorphism induced if the associated equivalence relation is also.
We note that ∼ is an equivalence relation induced by a homomorphism iff it is an
equivalence relation induced by some onto homomorphism, a fact we will sometimes
make use of. To see this, the reader may verify f : P → Q is a homomorphism iff
f : P → Q|f(P ) is an onto homomorphism. Note there is in fact something to check
here. For instance, the iff statement would not hold with Q.f(P ) used in place of
154
Q|f(P ). While surjectivity is trivial, it must be checked that the inverse images of
Q|f(P ) autonomous sets are P autonomous.
We wish to characterize the homomorphism induced equivalence relations, as we make
use of this characterization in later proofs. First, we need the following lemma.
Lemma 7.11.5. Let ∼ be an equivalence relation on an ausys P . Let ≤ be a total
order of the set τ of ∼ classes. Suppose for each ∼ class C in τ , the set
AC :=⋃
B∈{≤C}
B
is autonomous in P . Then there is a totally ordered proof ≤′ of P such that every ∼
class is a ≤′ interval.
Proof. Since each AC is autonomous, the set
A<C :=⋃C′<C
AC′
is the union of autonomous sets AC′ and is therefore autonomous as well. So A<C is
an autonomous subset of the autonomous set AC . Therefore there is a proof ≤C of
AC such that A<C is a ≤C initial segment.
Define ≤′ as follows. Given x and y in P , suppose x is in the ∼ class C and y is in
the ∼ class D. If C = D then we let x ≤′ y iff x ≤C y. If C 6= D then C < D or
C > D. We let x <′ y if C < D and y <′ x if D < C.
The reader may check ≤′ is a total order on P that is in the proof closure of the
proofs of the form ≤C for ∼ classes C. Therefore ≤′ is itself a proof. Each ∼ class is
a ≤′ interval by definition of ≤′.
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Proposition 7.11.6. Let (P, TP ) be an ausys and ∼ an equivalence relation on P .
Then ∼ is homomorphism induced iff P has a linearly ordered proof ≤ such that each
∼ class is a ≤ interval.
Proof. Suppose ∼ is induced by some homomorphism f : P → Q. Since Q is au-
tonomous, it has a linearly ordered proof ≤1. Since ≤1 is a proof, each initial segment
D of ≤1 is autonomous. Since f is a homomorphism, the inverse image f−1(D) is au-
tonomous for each such ≤1 initial segment. Thus ≤1 induces a total order ≤2 on the
∼ classes such that the union of any ≤2 initial segment of ∼ classes is autonomous in
P . We may therefore apply the previous lemma to obtain the desired totally ordered
proof ≤ of P in which each ∼ class is an interval.
For the converse, suppose there is a linearly ordered proof ≤ of P such that each ∼
class is an interval. Then the quotient linear order (P/ ∼,≤′) is uniquely defined by
letting X ≤′ Y if x ≤ y for some x in X and Y in y. This linear order is in particular
an ausys whose autonomous sets are the initial segments of the order. Trivial checking
shows the function taking each x to its ∼ class is the desired homomorphism.
In many branches of mathematics, the bijective homomorphisms are exactly the iso-
morphisms. This is far from true for ausys homomorphisms. For example, let P be
an antichain, let Q be any ausys at all, and let f be any function at all from P to
Q. Then f is automatically a homomorphism because every single subset of P is
autonomous.
156
In particular, every total order is the homomorphic image of each antichain of greater
or equal cardinality. While this definition is useful for some purposes, for others, it is
not. For other purposes, we want homomorphic images to be quotients of sorts, and
a chain is in no sense a quotient of an antichain.
The solution is to factor each surjective homomorphism into two maps, the first map
being what we call a pure contraction and the second a bijective homomorphism. We
formalize this in the following definitions and lemmas.
Definition 7.11.7. Let (P, TP ) and (Q, TQ) be ausyses and f : P → Q an onto ausys
homomorphism. Then f is said to be a pure contraction if TQ is exactly the set of all
subsets A of Q that can be linearly ordered as (A,≤) in such a way that f−1(D) is in
TP for every ≤ downward closed subset of A. We say that Q is a pure contraction of
P if there is some pure contraction f : P → Q.
Intuitively, f is a pure contraction if TQ is the maximum possible set of autonomous
subsets of Q that still lets f be a homomorphism. The following lemma makes this
precise.
Lemma 7.11.8. Let (P, TP ) and (Q, TQ) be ausyses, and let f : (P, TP ) →(Q,TQ)
be a pure contraction. Let (Q, T ′) be another ausys with domain Q. If f : (P, TP )→
(Q, T ′) is a homomorphism then T ′ ⊆ TQ.
Proof. Take A in T ′. We must show A is in TQ. By definition of pure contraction,
this is equivalent to showing that A can be linearly ordered as (A,≤) in such a way
157
that f−1(D) is in TP for every ≤ downward closed subset D of A. By assumption, A
is autonomous in (Q, T ′), so let ≤ be any totally ordered (Q, T ′) proof of A. Given a
≤ downward closed set D, we know D is in T ′ since every downward closed subset of
a proof is autonomous. Since D is in T ′ and f is a homomorphism from (P, TP ) to
(Q, T ′), it follows that f−1(D) is in TP , as needed.
Proposition 7.11.9. Let (P, TP ) and (Q, T ′) be ausyses and f : (P, TP )→ (Q, T ′) an
onto ausys homomorphism. Then there is an ausys (Q, TQ) on Q and homomorphisms
g : (P, TP )→ (Q, TQ) and h : (Q, TQ)→ (Q, T ′) such that g is a pure contraction, h
is a bijection, and f = h ◦ g.
Proof. Define TQ as the set of all subsets A of Q that can be linearly ordered as (A,≤)
in such a way that f−1(D) is in TP for every ≤ downward closed subset of D. Then
f : (P, TP )→ (Q, TQ) is a pure contraction by definition. So set theoretically, we let
g = f , the only difference between the maps being that as morphisms, the codomain
of g is (Q, TQ) while the codomain of f is (Q, T ′). Since g is a pure contraction,
we know by Lemma 7.11.8 that T ′ ⊂ TQ, so if we let h : (Q, TQ) → (Q, T ′) be the
identity function on Q then h−1(A) is in TQ for all T in A. Therefore h is a bijective
homomorphism. Trivially, f = h ◦ g.
With the definitions of subausys and dot made, and the basic definitions and lemmas
related to ausys homomorphisms and pure contractions given, we are finally ready to
define the notion of the minor of an ausys.
158
Definition 7.11.10. Let P and Q be ausyses. We say that Q is a minor of P if there
is a sequence P = P1, . . . , Pn = Q of ausyses such that for each i with 1 ≤ i < n, one
of the following conditions holds:
(i) Pi+1 = Pi|S for some subset S of Pi.
(ii) Pi+1 = Pi.A for some autonomous subset A of Pi.
(iii) Pi+1 is a pure contraction of Pi.
Numerous theorems in combinatorics, Kuratowski’s theorem characterizing planar
graphs being the most famous, characterize some well known class of objects in terms
of certain “forbidden” substructures the object is not allowed to “contain”. For in-
stance, Kuratowski’s Theorem states the planar graphs are exactly the graphs “con-
taining” no K5 or K3,3 in the sense of not having either graph as a graph theoretic
minor. The notions of containment are varied: the minor relation for graphs and
matroids, the subgraph and induced subgraph relation for graphs and its variants for
degree sequences, the subposet relation for partial orders, and so on.
The above definition of the minor relation for ausyses allows us to ask such questions
for natural classes of ausyses as well. When studying ausyses, the most frequently and
naturally arising class is the class of partial orders. Naturally, we wish to characterize
this class. It turns out this class has exactly one forbidden minor, an ausys we call
2-choice, or Ch2.
Definition 7.11.11. We let Ch2 denote the three point ausys with axioms a, b and
nonaxiom x such that a⇒ x and b⇒ x.
159
We will prove a finite ausys P is a partial order iff P has no Ch2 minor. First, we
prove a minor of a partial order is in fact a partial order. It is enough to show that
dots, subausyses, and pure contractions of partial orders are partial orders, which we
do in three separate lemmas below.
Lemma 7.11.12. If the ausys P is a poset and A is an autonomous subset of P then
P.A is a partial order. In fact, it is the partial order theoretic restriction to A.
Proof. A set B is autonomous in P.A iff it is a subset of A that is autonomous in the
poset (P,≤P ). A subset of a poset is autonomous iff it is downward closed, so the
autonomous sets of P.A are the ≤P downward closed subsets of A. These are exactly
the autonomous subsets of the poset (A,≤P |A) when thought of as an ausys.
Lemma 7.11.13. Let (P,≤) be a poset and S a subset of P . Then P |S is a partial
order. In fact, it is the partial order theoretic restriction (S,≤ |S).
Proof. Take an autonomous subset B of (S,≤ |S). In other words, B is a ≤ |S
downward closed subset of S. Let B′ be the ≤ downward closure of B. Then B′ is a
(P,≤) autonomous set such that B′ ∩ S = B.
Conversely, suppose a subset B of S has the form B′∩S for some (P,≤) autonomous
set B′. Then B′ is ≤ downward closed so B = B′ ∩ S is ≤ |S downward closed.
Therefore B′ is (S,≤ |S) autonomous.
160
Together, we see that a subset B of S is (S,≤ |S) autonomous iff B has the form
B′ ∩ S for some (P,≤) autonomous set B′. Therefore the ausys P |S is in fact the
poset (S,≤ |S) as claimed.
Lemma 7.11.14. Let (P, TP ) be a poset and (Q, TQ) a pure contraction of P . Then
Q is a partial order.
Proof. By Proposition 7.7.9, we need only show TQ is closed under arbitrary intersec-
tion. Since Q is a pure contraction of P , by definition, there is an onto homomorphism
f : P → Q such that TQ is exactly the set of all subsets A of Q that can be linearly
ordered as (A,≤) in such a way that f−1(D) is in TP for every ≤ downward closed
subset of A.
So take sets Ai for i in an index set I such that for each i, the set Ai can be linearly
ordered as (Ai,≤i) in such a way that f−1(D) is in TP for every ≤i downward closed
subset of Ai. We must show A =⋂i∈I Ai can be linearly ordered as (A,≤) in such a
way that f−1(D) is in TP for every ≤ downward closed subset of A. As ≤, we choose
a j in I arbitrarily and let ≤ be ≤j |A.
To show ≤ is the desired linear order, choose a ≤ downward closed subset D of A.
We must show f−1(D) is in TP . Well, since D is downward closed in
≤j |
(⋂i∈I
Ai
),
it follows that
D = D′ ∩
(⋂i∈I
Ai
)
161
for some ≤j downward closed set D′. Since D′ is downward closed in ≤j, we see by
choice of ≤j that f−1(D′) is autonomous. Also, f−1(Ai) is autonomous for each i
since f is a homomorphism. Since P is a poset, we therefore see that
f−1(D′) ∩
(⋂i∈I
f−1(Ai)
)= f−1
(D′ ∩
(⋂i∈I
Ai
))= f−1(D)
is autonomous.
Corollary 7.11.15. Let (P,≤) be a poset and (Q, TQ) a minor of P . Then Q is a
partial order.
Proof. Immediate from the fact that the dot, restriction, and pure contraction of a
poset is a poset.
Theorem 7.11.16. Let P be a strong aut descendable ausys with autonomous sets
A and B such that A ∩B is not autonomous. Then P contains Ch2 as a subdot.
Proof. By dotting to A ∪ B if necessary, we may assume P = A ∪ B. A ∩ B is
not autonomous, so we may choose x in the nonautonomous part of A ∩ B. Let
A′ = (A−B) ∪ {x}, B′ = (B −A) ∪ {x}, let S = (A′ ∪B′ ∪ {x}), and let P ′ = P |S.
Since A′ = A ∩ S and B′ = B ∩ S, we see by definition of P ′ that A′ and B′ are
autonomous in P ′. If {x} = A′ ∩ B′ is autonomous in P ′ then by definition of P ′ as
a restriction, P must contain an autonomous subset of A ∩B containing x, contrary
162
to choice of x in the nonautonomous part. This contradiction shows {x} = A′ ∩B′ is
not autonomous in P ′.
Since P ′ is strong aut descendable, we may choose a minimal P ′ autonomous subset
A′′ of A′ containing x. Similarly, we may choose a minimal P ′ autonomous subset B′′
of B′ containing x. A′′ ∪B′′ is P ′ autonomous. Let P ′′ = P.(A′′ ∪B′′). Then A′′ and
B′′ are P ′′ autonomous sets such that A′′ ∩B′′ = {x} is not P ′′ autonomous.
Note that since A′′ is a minimal P ′ autonomous subset of A′ containing x, it follows
that x is a ≤A′′ maximum element of A′′ in the ausys P ′. Since P ′ and P ′′ have the
same autonomous subsets of A′′, it follows that x is a ≤A′′ maximum element of A′′ in
the ausys P ′′ as well. Since {x} is not autonomous in P ′′, it follows that there is a in
A′′ such that a <A′′ x in the ausys P ′′. Similarly, there is b in B′′ such that b <B′′ x
in the ausys P ′′.
Let P ′′′ = P ′′|{a, b, x}. Simple checking of the autonomous sets of P ′′′ shows that P ′′′
is isomorphic to Ch2. Obviously, P ′′′ is a subdot of P .
Corollary 7.11.17. A strong aut descendable ausys is a poset iff it has no Ch2 subdot.
Proof. We know the subdot of a poset is a poset, which is therefore not Ch2. Inversely,
if a strong aut descendable ausys is not a poset then by Proposition 7.7.7 there are
autonomous sets A and B such that A∩B is not autonomous. By Theorem 7.11.16,
it follows the ausys has a Ch2 subdot.
163
Corollary 7.11.18. A strong aut descendable ausys is a poset iff it has no Ch2 minor.
Proof. Immediate from the fact that a subdot is a minor and the fact that the pure
contraction of a poset is a poset.
Now consider the four point ausys with axioms a and b and nonaxioms x and y such
that a implies x, b implies y, x implies y, and y implies x. We call this ausys P4 since
it may be thought of as a rooted graph, the roots (axioms) being the endpoints. We
have previously noted that x <{a,x,y} y and y <{b,y,x} x. In fact, P4 is the simplest
ausys exhibiting such behavior in the sense that every strong aut descendable ausys
containing points x and y and sets A and B such that x <A y and y <B x contains
a P4 minor. We prove this claim now.
Theorem 7.11.19. Let P be a strong aut descendable ausys with autonomous sets
A and B and points x and y such that x <A y and y <B x. Then P contains a P4
minor.
Proof. The proof is similar in spirit to that of Theorem 7.11.16.
By dotting to A∪B if necessary, we may assume P = A∪B. Given such A, B, x, and
y, let A′ = (A−B)∪{x, y}, B′ = (B−A)∪{x, y}, let S = (A′∪B′∪{x, y}), and let
P ′ = P |S. Since A′ = A ∩ S and B′ = B ∩ S, we see by definition of P ′ that A′ and
B′ are both autonomous sets in P ′ containing x and y. If there is a P ′ autonomous
subset C of A′ containing y and not x then C = D ∩ S for some P autonomous set
164
D. But then D ⊆ A, contrary to the fact that there is no P autonomous subset of A
containing y and not x. This contradiction shows that every P ′ autonomous subset
of A′ containing y also contains x. Therefore x <A′ y in the ausys P ′. Similarly,
y <B′ x in the ausys P ′.
Since P ′ is strong aut descendable, we may choose a minimal P ′ autonomous subset
A′′ of A′ containing y. Similarly, we may choose a minimal P ′ autonomous subset B′′
of B′ containing x. A′′ ∪B′′ is P ′ autonomous. Let P ′′ = P.(A′′ ∪B′′). Then A′′ and
B′′ are P ′′ autonomous sets such that A′′ ∩B′′ = {x, y}.
Note that since A′′ is a minimal P ′ autonomous subset of A′ containing y, it follows
that y is a ≤A′′ maximum element of A′′ in the ausys P ′. In particular, x <A′′ y in
the ausys P ′. Since P ′ and P ′′ contain the same autonomous subsets of A′′, we see
that x <A′′ y in the ausys P ′′ as well. Similarly, y <B′′ x in the ausys P ′′. Since
x <A′′ y in P ′′, we see that y is not a P ′′ axiom. Therefore B′′ − {x, y} is nonempty.
Similarly, A′′ − {x, y} is nonempty.
The reader may check that the partition of P ′′ with cells A′′ − {x, y}, B′′ − {x, y},
{x}, and {y} is homomorphism induced. Let P ′′′ be the pure contraction of P ′′ with
respect to this partition. The reader may check that this ausys is isomorphic to P4.
We note that while our forbidden minor theorem for partial orders can also be seen as
a forbidden subdot theorem, the same is not true for the theorem just stated. Let P
be the six point ausys with axioms a1, a2, b1, b2 and nonaxioms x and y such that a1
implies x, a2 implies x, {a1, a2, x} implies y, b1 implies y, b2 implies y, and {b1, b2, y}
165
implies x. Let A = {a1, a2, x, y} and B = {b1, b2, x, y}. The reader may show that
x <A y and y <B x yet P has no P4 subdot.
7.12 Relations to Matching and Connectivity
In this section, we use ausyses to give a problem related to both graph matching and
graph connectivity. We recall Konig’s Duality Theorem and then Menger’s Theorem.
The reader unfamiliar with the relevant definitions may consult [3].
In this section, usual graph theoretic conventions apply. In particular, subgraphs are
not necessarily induced.
Theorem 7.12.1. Let G be a finite, bipartite graph. Then the maximum cardinality
of a matching is the minimum cardinality of a vertex cover.
Theorem 7.12.2. Let G be a finite graph and let x and y be distinct vertices of G.
Then the maximum number of disjoint paths from x to y is equal to the minimum
number of vertices separating x and y.
We note that in appropriately defined ausyses, both these theorems say that in this
ausys, the maximum number of disjoint autonomous sets containing x equals the min-
imum size of a set S such that S blocks x. Here, by disjoint we mean the autonomous
sets only intersect in x. We make these notions precise as follows.
First, let G be a graph. Let x be a point not in G. We may define an ausys PG such
that every v in G is an axiom, x is a nonaxiom, and {v, w} implies x iff v is adjacent
166
to w. We call such an ausys a graph of axioms. A matching in a graph of axioms
exactly corresponds to a set of disjoint autonomous sets containing x. A vertex cover
in a graph of axioms is a set of vertices incident with every edge. In this ausys, a
vertex cover is then exactly a set S of points that blocks x. Konig’s Duality Theorem
therefore states that in a bipartite graph of axioms, the maximum number of disjoint
autonomous sets containing the nonaxiom x equals the minimum cardinality of a set
blocking the nonaxiom x.
Second, let G be a graph and let y and x be distinct vertices in G. We let QG be
the ausys with domain G−{y} such that the QG autonomous sets are exactly sets of
the form S − {y}, where S is a connected set in G containing y. In this ausys, a set
of disjoint paths corresponds exactly to a set of disjoint autonomous sets containing
x. A set separating x and y in G corresponds exactly to a set S blocking x in QG.
Therefore Menger’s Theorem says exactly that for all such G, x, and y, the maximum
number of disjoint autonomous sets containing x equals the minimum cardinality of
a set blocking x. We note this discussion applies equally well to the infinite version
of Menger’s Theorem.
This naturally leads us to ask for which finite ausyses P and points x in P is the
maximum number of disjoint autonomous sets containing x equal the minimum size of
a set blocking x. We note this is certainly not true for all ausyses P and x contained
in P . That would imply that Konig’s Theorem holds for all finite graphs, which of
course it does not.
167
BIBLIOGRAPHY
[1] M. Chudnovsky, P. Seymour, The proof of Rao’s Conjecture on degreesequences, In Preparation.
[2] V. Chvatal, W. Lenhart, N. Sbihi, Two-colourings that decomposeperfect graphs, J. Combin. Theory Ser. B 49 (1990), pp. 1-9.
[3] R. Diestel, Graph Theory, 2nd ed., Springer-Verlag, New York, 2000.
[4] P. Erdos, T. Gallai, Graphs with Prescribed Degrees of Vertices, Mat.Lapok. 11 (1960), pp. 264–274 [In Hungarian].
[5] S. Foldes, P. Hammer, Split Graphs, Proceedings of the Eighth South-eastern Conference on Combinatorics, Graph Theory and Computing, 1977.
[6] D. Fulkerson, A. Hoffman, M. McAndrew, Some Properties ofGraphs With Multiple Edges, Canad. J. Math. 17 (1965) pp. 166–177.
[7] P. Grillet, Commutative Semigroups, Kluwer Academic Publishers, Dor-drecht, The Netherlands, 2001.
[8] P. Halmos, Naive Set Theory, Springer-Verlag, New York, 1974.
[9] K. Kunen, Set Theory: An Introduction to Independence Proofs, Elsevier,San Diego, 1980.
[10] S. Rao, (A private communication to Neil Robertson), 1971.
[11] S. Rao, A survey of the theory of potentially P-graphic and forcibly P-graphic degree sequences, Lecture Notes in Mathematics 885, Springer Berlin/Heidelberg,(1981) pp. 417–440.
[12] S. Rao, Towards a theory of forcibly hereditary P-graphic degree sequences,Lecture Notes in Mathematics 885, Springer Berlin/Heidelberg, (1981) pp.441–458.
168
[13] H. Rose, A Course in Number Theory, Oxford Science Publications, 1994.
[14] J. Rosenstein, Linear Orderings, Academic Press, New York, 1982.
[15] B. Schroder, Ordered Sets: An Introduction, Birkhauser, Boston, 2003.
[16] R. Tyshkevich, Decomposition of graphical sequences and unigraphs, Dis-crete Math 220 (1-3) (2000) pp. 201–238.
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