Cellularity and the Structure of Pseudo-Trees

Cellularity and the Structure of Pseudo-TreesAuthor(s): Jennifer BrownSource: The Journal of Symbolic Logic, Vol. 72, No. 4 (Dec., 2007), pp. 1093-1107Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/27588593 .

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The Journal of Symbolic Logic Volume 72. Number 4. Dec. 2007

CELLULARITY AND THE STRUCTURE OF PSEUDO-TREES

JENNIFER BROWN

Abstract. Let T be an infinite pseudo-tree. In [2], we showed that the cellularity of the pseudo-tree algebra Treealg(T) was the maximum of four cardinals ct, it, 4>t, and ?iT\ roughly, cj is the "tallness" of J; it is the "width" of T; 0 is the number of "points of finite branching" in T; and ju is the number of "sections of no branching" in T. Here we ask: which inequalities among these four cardinals may be

satisfied, in some sense, by a pseudo-tree? We show that the possible inequalities among ct,it,4>t, and ?uT attained by pseudo-trees T are closely related to the existence of generalized Suslin trees.

?1. Background. A pseudo-tree is a partially ordered set (T<) such that for every nn7\theset{r G T: r < t} is a linear order. The pseudo-tree algebra on T, denoted

Treealg(r), is the subalgebra of ̂(T) generated by the cones {r G T: r > t}, for /in T. For what follows, T is assumed to be an infinite pseudo-tree with a single root. A branch of T is a maximal chain. The cellularity of a Boolean algebra A is c(A)

= sup{|X| : X is a pairwise-disjoint subset of A}. For any sets X and 7,

"X ? 7" means that X is any subset of 7; "X c 7" means that X is a proper subset of 7. See Koppelberg and Monk [8] for background on pseudo-tree algebras.

Most of the tree-related facts and notions we use may be found in Kunen [9]. The cellularity of a pseudo-tree algebra Treealg(T) is characterized in terms of

four cardinal functions on the underlying pseudo-tree. To define these functions, we first need to define some special sorts of elements and chains of T:

Definition 1.1. A fan element of T is an a G T such that there exists a set F = fan (a) with the following properties:

( 1 ) F is a finite set of pairwise-incomparable elements each greater than a, and

|F|>2. (2) For every c > a, c is comparable to a unique b G F.

Definition 1.2. A pure chain in T is a chain C ? T such that the following conditions hold:

(1) |C|>2. (2) If a, b G C, a < b, and a < c G T, then c is comparable to b.

C ? T is a maximal pure chain if C is a pure chain and if whenever C d C, C is not a pure chain.

Note: any two distinct maximal pure chains are disjoint; see [2] for a proof.

Received February 20, 2006.

This paper is based on the Ph.D. thesis of the author, who would like to thank her advisor, J. Donald Monk.

? 2007. Association for Symbolic Logic 0022-4812/07/7204-0002/S2.50

1093



1094 JENNIFER BROWN

Definition 1.3. For any pseudo-tree 7\

The cellularity of T is

cT = sup{c(Intalg(C)) : C is a chain in T} (note that we are using "cellularity" in a special sense here);

The incomparability of T is

iT = sup{|iS| : S is an antichain in T} ("incomparability" is also being used in a special sense);

The number of fan elements of T is

4>t = \{a G T: a is a fan element of T}\\ and The number of maximal pure chains of T is

)?r = |{C ? T: Cisa maximal pure chain}|.

We have shown (see [2]) that:

Theorem 1.4. For any infinite pseudo-tree T,

c(Treealg(r)) =

max{cT,iT,$T,PT}'

We are interested here in asking, which inequalities among the four cardinals

c, i, <j>9 p are attained by a pseudo-tree? To make this question (questions, really) precise, we need some notation. Let o be a permutation on the set of functions

{c, i, <f>9 //}; we will write {o\, cf2,0-3,0-4} =

{c, 1, <t>, //}. Define a general inequality among the four cardinals to be any strict inequality 7 of the form o\ < o2 < 0-3 < a 4 where g is a permutation on {c, z, 4>, p}. We call a the permutation associated with 7. We say that a pseudo-tree T attains the general inequality 7 (or the permutation a) if (g\)t < (oi)t < (o*)t < (0-4)7. Given a permutation a on {c, 1, (?>, ju}, we wish to answer the following Strong Attainment Question. Is it provable in ZFC that for any sequence k\ <

k2 < ^3 < ?4 of infinite cardinals, there exists a pseudo-tree T such that (g?)t = ?/

(for/- 1,2,3,4)?

(Note: for convenience, we will say "The answer to the Strong Attainment Ques tion for an inequality / is ... "to abbreviate "The answer to the Strong Attainment

Question for the permutation a associated with the general inequality / is ... ".) The answer to this Strong Attainment Question depends on the permutation a.

We will show that for 8 of the 24 possible permutations o on {c, 1, cf), //}, the answer is "yes"; and that for the remaining 16 possible permutations, the answer is "no".

If the answer to the Strong Attainment Question is "no" for a given permutation o (and its associated permutation /), we would still like to answer the following two

questions: Weak Attainment Question. Is it consistent that some pseudo-tree T attains

the inequality 7?

Negative Attainment Question. Is it consistent that no pseudo-tree T attains the inequality 7?

Some observations on the Questions:

(1) An answer of "yes" to the Strong Attainment Question implies an answer of "yes" to the Weak Attainment Question and an answer of "no" to the

Negative Attainment Question. (2) An answer of "no" to the Weak Attainment Question, for a given inequal

ity 7, implies an answer of "yes" to the Negative Attainment Question?and,



CELLULARITY AND THE STRUCTURE OF PSEUDO-TREES 1095

in fact, implies that it is provable in ZFC that there is no pseudo-tree T

realizing /.

(3) A "yes" answer to both the Second and Negative Attainment Questions, for a given inequality /, means that the statement "There is a pseudo-tree T

realizing the inequality /" is independent of ZFC.

The reader's intuition may suggest that the maximum of just the two cardinals c and i would seem to be the most likely candidate for the cellularity of Treealg(T). Indeed, we will show that, assuming a weakly compact cardinal, it is not possible to

buildinZFCpseudo-treesrsuchthatmax{(737,/ir} > max{cr, zrjifmaxj^r,/^} is the successor of a regular cardinal. Consistently, however, such pseudo-trees can

be found; for example, let T be a Suslin tree in which every element has exactly two immediate successors. Then ?t

= co\, cj ?

it = co, and jut = 0. For an

example of a pseudo-tree T such that juT > max{cT, iT}, first let S be a Suslin tree in which every element has co-many immediate successors. For each r G S and each immediate successor s of r, replace the empty "link" between r and s with a copy of co. Let T be the resulting pseudo-tree. Then jut

= co\, ct =

it ?

co,

and (j)T = 0. In what follows, we first show that if T is a pseudo-tree such that

max{0r,//r} > max-fc^,^}, then in fact T contains inside of it a generalized Suslin tree. We then address the above three Questions for each of the 24 possible strict inequalities among cj, it, 4>t, and jut- (For a discussion of the 51 possible non-strict inequalities among these four cardinals, see [5].)

?2. The four cardinals and K-Suslin trees. We prove the following: suppose T is a pseudo-tree and k is an infinite cardinal such that maxjc^, ij} < k and either

4>t = ? or jut

= K. Then there exists a K-Suslin tree.

2.1. Fan elements and /c-Suslin trees.

Proposition 2.1. For any pseudo-tree T,

it > sup{| C | : C is a chain of fan elements of T}.

Proof. Let C be a chain in T such that every c G C is a fan element of T. List the elements of C as C =

{ca: a < k}, for some cardinal k. Let B be a branch of T such that C ? B. For each a < k, pick some fa G fan(ca) such that fa ? B. The reader may check that {fa : a < k} is an antichain in T. H

Definition 2.2. Let T be a pseudo-tree. We inductively construct a system of pseudo-levels {Aa : a < 3} (where 3 is an ordinal) of T as follows: let Ao be any maximal antichain in T. Suppose Aa has been constructed. Define

Ba = {a e T : y < a for some y G Aa}.

If Ba = 0, stop; otherwise, let Aa+\ be maximal among antichains contained in Ba. For limit steps: suppose y is a limit ordinal and Aa has been constructed for a < y. Let Cy

= {a G T : for all a < y, there is a y G Aa such that y < a}. If Cy

= 0, stop; otherwise let Ay be maximal among antichains contained in Cy.

It is left to the reader to verify the next lemma.

Lemma 2.3. Let {Aa : a < 3} be a system of pseudo-levels in a pseudo-tree T. If a < ? <3, then for every x G A? there is a y G Aa such that y < x. H

It then follows that:



1096 JENNIFER BROWN

Lemma 2.4. If'{Aa : a < S} is a system of pseudo-levels in 7\ and S = \Ja<? Aa

with the induced order from T, then S is a tree, and Aa = Leva(?). 3

Lemma 2.5. Let k\ < k2 be infinite cardinals. Suppose {Aa : a < K2}isa system of pseudo-levels in T, and suppose max{cr, ij}

< k\ . Then \Ja<Kl Aa is a n2-Suslin tree.

Proof. Let T and {Aa : a < k2} be as above, and denote S = [ja<Kl Aa. By

Lemma 2.4, S is a tree. |jS'| > K2 as each Aa is non-empty and as the Aa's are pairwise-disjoint. |5|

= \[ja<K2Aa\

< Y,a<JM <

Ea<?2 ?i = ?2 ?i =

max{?i,?2} = ?2> since zr < Aq. Thus |S|

= ?2. S has no antichain of size

k2?else T would have an antichain of size k2, contradicting the fact that ij < k\.

Suppose C ? S is a chain of size n2. Since S is a tree, C is well-ordered. But then C is also a well-ordered chain of size k,2 in T, so that cj > k2\ but this is a contradiction. Therefore every chain and antichain in S has size less than k2. 3

Theorem 2.6. Let k\ < k2 be infinite cardinals. Let {Aa : a < S} be a system of pseudo-levels in a pseudo-tree T; and suppose max{cj, it} < k\ while c?r = ^2 Then S =

k2.

Proof. First, ? < n2\ for suppose ? > k2. Then by Lemma 2.4, since AKl =

Le\K2(\Ja<?Aa) ^ 0, T has a well-ordered chain of length k2, contradicting the

assumption that cj < ?i. Now, suppose {Aa : a < ?} have been constructed, for some ? < k2\ we will show

that it is possible to find a non-empty A?. Defined = {/ G T: fis a fan element}.

|^| = n2 by assumption. Let

B = {f G & : for all a < ? there is a y G ̂ a such that y < /}.

It suffices to show that 5^0. Suppose B = 0. We claim that the following holds:

For every / G S? there is a y G Ua<<s ^? such that f < y. (*)

To prove (*): Let / G &~. Then since 5 = 0, there is an a < /? such that for all y G ^4a, y f? /. Let a' be the least such a. If a7 = 0, then f < y ? Ao for some y, by the maximality of Ao. Now suppose a7 > 0. Then for every y < a' there is a y G Ay such that y < f. First suppose a;7 is a successor

ordinal; say a7 = a + 1. Then Aa> was maximal among antichains contained in

Ba = {a e T : y < a for some y G 4}. Now / G Ba by applying the above to

7 = a. So / is comparable with some y G Aa>. Hence f < y, since y it f by the choice of a7. (The case where a' is a limit ordinal is similar.) Therefore (*) holds.

Thus, setting R = Ua</? Aa, we may write

&= \J{f ?& /< y} yeR

Since it < ?i, by Proposition 2.1 no more than ?i-many members of if are below each y G i?, so |{/ G & : f < y}\ < k\ for any y G R. Also since z^ < ?1,

\Aa I < ?1 for each a < ?. With this, and since ? < k2,wq have

?2 = 1*1= \J{f?F:f<y} yeR

< \?\ -Ki <K2,



cellularity and the structure of pseudo-trees 1097

which is a contradiction. Thus B ^ 0 and we are done. H

Theorem 2.7. Let n be an infinite cardinal, and suppose T is a pseudo-tree such that max{c7\ it} < k while <pT

= k. Then there exists a K-Suslin tree.

Proof. Let T be as above. Let {Aa : a < 3} be a system of pseudo-levels of T. Set S =

\Ja<? Aa. By Theorem 2.6, 3 = k. Then by Lemma 2.5, \Ja<KAa is a

K-Suslin tree.

H

2.2. Maximal pure chains and K-Suslin trees. We will show, analogously to The orem 2.7, that if T is a pseudo-tree such that ct^t < k while jut = ?, then there exists a ?-Suslin tree.

Proposition 2.8. For any pseudo-tree 7\ it > sup{|^| : W is a well-ordered or

inversely well-ordered chain of maximal pure chains of T}.

To prove this, we first recall that any two distinct maximal pure chains are disjoint. Proposition 2.8 follows easily from that fact plus the following lemmas, which the reader may verify.

Lemma 2.9. (1) Let T be a pseudo-tree and let C\, C2 be distinct maximal pure chains in T, and suppose that for some branch B ofT, C\, C2 ? B. Then either

Cx < C2orC2 < Ci.

(2) Let C be a pure chain in T, and suppose a G T is comparable to some c G C. Then a is comparable to every element of C.

(3) IfC?T is a maximal pure chain, a,c G C, anda C\ and b _L C2. H

Lemma 2.10. For any pseudo-tree T,

Ct > sup{|^| : & is a chain of maximal pure chains}.

Proof. Suppose W is a family of maximal pure chains contained in some branch B of T. For C G g7, pick xc < yc ? C. Since C is a pure chain, (T ? xc) \ (T ? yc)

= [xc,yc)\ and by Lemma 2.9, for C ^ C/ G f7, [xc,J>c)n[xc', jcO

= 0 Thus {[xc,yc)'- C E ?} is a family of pairwise-disjoint elements of Intalg(^) of size |^|, so cT > c(Intalg(?)) > |^|. H

Theorem 2.11. Lei k\ < k2 be infinite cardinals. Let {Aa : a < 3} be a system of pseudo-levels in a pseudo-tree T\ and suppose max{c^, it}

< k\ while ?ut =

k2

Then 3 = k2.

Proof. First, 3 < k2 as in the beginning of the proof of Theorem 2.6.

Suppose {Aa : a < ?} have been constructed, for some ? < k2, we show that it is possible to construct A?. As it < ?i, |^4a| < k\ for each a < ?; and as /? < ^2,

|Ua<o^4c,| < Av2. Set g7 = {C ? T: C is a maximal pure chain}. By assumption,

1^1 =?2. Define

%? = {C G % : for all a < ?, there is a 7 G Aa so that j < C}.

It is enough to show that %? ^ 0.

Suppose ^ = 0. Similarly to the proof of Theorem 2.6, one may show:

For every C eg7, there is a j; G Ua</i A<* sucn tnat c n (T i y) + 0- (*)



1098 JENNIFER BROWN

(Note that, for C G <? and y G T, if C n (T | y) 7^ 0 then either y G C or

C < y. This follows from part (3) of Lemma 2.9 and by definition of pure chains.) Thus, setting B =

Ua</? ^<*' we may write

% = [J{CG^: Cn(7|y)^0}. VGA

Fix a < ? and y e Aa. Set ̂ = {C G g7: C n (T | y) 7^ 0}. If |^| > co,

then \Qf\ =

\{C e W: C < y}|; so by Lemma 2.10, since cT < k\ it must be that \Qf\ < k\. Thus we will have, by a similar computation to that at the end of Theorem 2.6, that

K2 < ?2, \J{CeW:Cn(T ly)^<b}\ yeB I

which is a contradiction. Thus %> ^ 0. 3

Theorem 2.12. Lei k be an infinite cardinal, and suppose T is a pseudo-tree such that max{cr, it} < k while jut

= k. Then there exists a K-Suslin tree.

Proof. Let T be as above. Let {Aa : a < S} be a system of pseudo-levels of T.

By Theorem 2.11, S = k. Then by Lemma 2.5, S := [ja<K Aa is a tt-Suslin tree. 3

Remark 2A3. We will later show that it is impossible to have a pseudo-tree T in which ct < it < <\>t, it < ct < <1>t> ct < it < Pt, or it < ct < Pt- However, if we allow non-strict inequalities, it is easy to find (consistently) pseudo-trees T for which rnaxjcr, iT} < 4>t or maxjc^, iT} < Pt\ see, for instance, the examples given at the end of Section 1.

?3. Which strict inequalities among the four cardinals are possible? Recall the three Questions that we wish to answer for each of the 24 possible strict inequalities 7

among c, 1, <fi, and//:

Strong Attainment Question. Is it provable in ZFC that for any sequence k\ <

&2 < ?3 < ?4 of infinite cardinals, there exists a pseudo-tree T such that (o?)t = k>?

(for/= 1,2,3,4)? Weak Attainment Question. Is it consistent that some pseudo-tree T attains

the inequality 7?

Negative Attainment Question. Is it consistent that no pseudo-tree T attains the inequality 7?

(Note: the answers to all of these Questions are summarized in the chart at the end of this paper.)

Proposition 3.1. The answer to the Strong Attainment Question is "yes" for the

following eight inequalities:

(j) < I, C < p < (f) < L

(f) < C < p < I, C < <j) < p < I.

(Note that these are exactly the inequalities in which max{0, p} < 1.)



cellularity and the structure of pseudo-trees 1099

Proof. Let ?4, ?3, k2, k\ be distinct infinite cardinals. We prove the theorem by exhibiting a certain type of tree, variations of which will satisfy any of the above

inequalities. For this tree T, we see that cT = ?4, it = max{?i, k2, ?3}, 4>T

= ?1,

and jut = ki- The tree T is constructed as follows:

T ?? times

By varying the values of the kz , it is clear that the tree T may be an example of any of the inequalities listed above. H

Proposition 3.2. Assuming a weakly compact cardinal exists, the answer to the

Strong Attainment Question is "no" for the following 12 inequalities: c < 1 < ? < ju, c < 1 < ju < (j), 1 < c < (f) < ju, 1 < c < ju < (j), ju < c <i < <fr, c < ju < 1 < (/),

c < ? < 1 < ju, (f) < c < 1 < ju, (f) < 1 < c < ju, l < (?) < c < ju, l < JU < C < (f),

JU max{c, 1} or ?1 > max{c, 1}.)

Proof. Suppose that there exists a pseudo-tree T satisfying k = (/)t> maxj/j, cT }. Then by Theorem 2.7, there is a ?-Suslin tree. However, we could choose k to be

weakly compact, so that there are no ?-Suslin trees; so no such pseudo-tree T can be constructed in ZFC. Thus if / is any inequality for which cj> > max{c, 1}, the answer to the Strong Attainment Question is "no" for /.



1100 JENNIFER BROWN

Analogously, Theorem 2.12 is used to show that the answer to the Strong Attain ment Question is "no" for any inequality 7 for which pT > max{cr. iT}. 3

The assumption of a weakly compact cardinal is not necessary to settle the Strong Attainment Question, as we will show next.

The following theorem is due to Monk, as is Theorem 3.6.

Theorem 3.3. Let k be an infinite cardinal. There does not exist a pseudo-tree T such that it < k and (?)t > {2K)+. {In particular, for any pseudo-tree 7\ (?>t < 2lr.)

Proof. Suppose that such a pseudo-tree T exists. Let X be a collection of fan elements of T of size (2K)+. Define a function / : [X]2 ?? 2 by setting, for any x.yeX,

, , J

1 if x and y are comparable,

10 if x and y are incomparable.

By a corollary to the Erd?s-Rado partition theorem ((2*)+ ?> {n+)2K?see Jech [6]),

there is a Y ? X with \Y\ = k+ such that Y is homogeneous for /. The value of/

on [ Y]2 cannot be 0, since ij < k. Thus the value of/ on [ Y]2 is 1 ; i.e., Y is a chain in T. By Theorem 2.1, since Y is a chain of fan elements of T, it > \Y\

= k+ > k, which is a contradiction. 3

By Theorem 3.3, we have:

Corollary 3.4. The answer to the Strong Attainment Question is "no" for all

general inequalities for which i < (/).

Corollary 3.5. The answer to the Negative Attainment Question is "yes" for the following inequalities:

i < c , 1 < C < (?) < //, C < l < p < (j). I < p < (j) < c.

Proof. By Theorem 3.3, under GCH there is no pseudo-tree T realizing any of the above inequalities. 3

Theorem 3.6. Let k be an infinite cardinal. There does not exist a pseudo-tree T such that it < k and pj > (2K)+. {In particular, for any pseudo-tree T, pT < 2lJ '.)

Proof. Suppose T is such a pseudo-tree. Let ? be a collection of //^-many maximal pure chains. Let ^ be a well-order on g7. Define / : [g7]2 ?> 3 by setting, for distinct Q, C2 G? with C\ -< C2,

/({Ci,C2})

1 ifCi<C2, 2 ifC2<Ci, 3 if C\ and C2 are incomparable.

By the Erd?s-Rado theorem, let Qi ? f? be a homogeneous set for / such that

\Q}\ = k+. First suppose that the value of / on Of is 1. Then the pseudo-tree

order < on ?& coincides with the well-order ^, so that QJ is a well-ordered chain of maximal pure chains of T. Then by Theorem 2.8, iT > \2$\

= k+, contradicting the assumption that ij < k. Similarly, if the value of / on QJ is 2, then Of is an

inversely well-ordered chain of maximal pure chains of 7\ and then iT > k+ by



CELLULARITY and the structure of pseudo-trees 1101

Theorem 2.8, which is again a contradiction. Finally, if the value of / on 2 is 3, this is again a contradiction as then it > 2 = ?+. H

By Theorem 3.6, we have

Corollary 3.7. The answer to the Strong Attainment Question is "no" for all

general inequalities for which i < ju.

Corollary 3.8. The answer to the Negative Attainment Question is "yes" for the following inequalities:

l < (j) < c < ju, i < <f> < ju < c.

Proof. The corollary is proved by noting that, by Theorem 3.6, under GCH there is no pseudo-tree satisfying either of the above inequalities. H

Note that this corollary also gives a "yes" answer to the Negative Attainment

Question for any inequality in which i < < ju, but the Negative Attainment

Question for all other such inequalities has already been answered.

Remark 3.9. We can now address the question of whether all four of the cardinals

c, i, (j), and ju are necessary to describe cellularity for pseudo-trees constructed in ZFC. Suppose that Tis a pseudo-tree such that c(Treealg(r))

= 4>t > max{cY,zj}. (The assumption ?iT > maxjcr, z^} is handled similarly, using Theorem 2.12 in

place of Theorem 2.7, and Theorem 3.6 in place of Theorem 3.3.) First suppose that <?>t is a limit cardinal. By Theorem 3.3, <\>T < 2lT. Since 4>T is a

limit, iT < <\>t implies that 0r)+ < <t>T- Thus the existence of such a pseudo-tree T would mean that GCH failed at it

Next suppose that (\>t ? ft+ for some ?. Then by Theorem 2.7, there is a ?+

Suslin tree. Silver showed, in Mitchell's model (see Mitchell [12]), that if? is regular and there exists a weakly compact cardinal above ?, then there is a model of ZFC in which there are no ?+-Aronszajn trees (and hence no ?+-Suslin trees).

Thus, modulo large cardinals, one cannot build in ZFC a pseudotree T such that c(Treealg(T))

= max{0r,/?7} > max{cT,iT} if maxl^,//^} is a limit car

dinal or the successor of a regular cardinal. If max{0r,^r} is the successor of a singular cardinal, the situation is not completely known. If ? is singular and a limit of strongly compact cardinals, then there are no ?+-Aronszajn trees [11]. Assuming large cardinals, it is consistent that there are no tt^+i-Aronszajn trees

tu]. We next wish to answer the Second and Third questions for the inequality cj) <

i < ju < c by constructing, in ZFC, a pseudo-tree T such that 0r < it < jut < cT. We will use the following fact, due to Baumgartner [1]:

Theorem 3.10 (ZFC). There is a dense linear order of size co2 having a dense subset

of size co\.

Proposition 3.11. There is a pseudo-tree T, constructible in ZFC, such that

(/>t = co, it

= co\, jut = 002, and ct = 003.

Proof. By Theorem 3.10, let L be a dense linear order of size co2 having a dense subset TV of size co\. For each n G N, adjoin tola new element bn with the

property that bn > n and for each / G L with / > n, bn _L /. For each k G L\N,



1102 JENNIFER BROWN

replace k with a two-element chain {k\,k2} where k\ <k2. Let S\ be the resulting

pseudo-tree. Adjoin below S\ a copy of C03. To the bottom element of C03 adjoin co-many immediate successors (s? : i < co); and to each s? adjoin two immediate successors tj and tf9 as shown.

Constructing the pseudo-tree T:

k eL\N

n G N

L Si T

One may check that T is as desired. 3

By Proposition 3.11 we have:

Corollary 3.12. The answer to the Weak Attainment Question is "yes" for the

inequality

4> < 1 < p < c. 3

Since the pseudo-tree in Proposition 3.11 was constructed in ZFC, we have

Corollary 3.13. The answer to the Negative Attainment Question is "no" for

the inequality

(f) < 1 < p < c. 3

Proposition 3.14. Suppose there is a dense linear order L having dense subsets M and TV such that \L\

= co3, M n N = 0, \M\ = co\9 and \N\

= coy, and suppose also that there exists an eo2-Suslin tree. Then there is a pseudo-tree T such that

ct = c?4, it = co\9 4>t ?

oj2, and pt = 003.

Proof. We first construct a pseudo-tree T\ similar to the pseudo-tree S\ in Propo sition 3.11: to each m G M, adjoin a new element bm with the property that bm > m

and for each / G L with / > m, bm _L /. For each n G N9 replace n with a two

element chain {n\, n2} where ni < n2. Let T\ be the resulting pseudo-tree. Let T2 be an co2-Suslin tree in which every element has exactly two immediate successors.

Let r be a new element such that r < T\ and r < T2. Finally, adjoin a copy of 0J4

Si

{kuk2}

>bn t2 tx t2 tl Li Li Lo lo C?3




below r, and let T be the resulting pseudo-tree. The reader may verify that T is as

desired. H

It is, in fact, consistent that a pseudo-tree as in Proposition 3.14 exists; we sketch how to show this: Let M be a model of V = L. Then there is an 002-Suslin tree S in M (see Devlin [4]). Extend M to M [G] by Cohen forcing, making the continuum

equal to C03 in the extension. By an exercise in [9] (exercise HI7 in Chapter VII), S remains an <x>2-Suslin tree in M[G]. For the linear order L, we may just take L = R, as we show below. Let P denote R \ Q. We use the following two facts, which can

be proved in ZFC and so are true in M[G]: (i) |P| =

|R|; and (ii) for any a, b G R

witha<b, $a,b)nF\ =

|P|. Lemma 3.15. Suppose |R|

= 003. Then there are dense subsets M, N ofR such that

\M\ = cox, \N\

= C03, and M n TV = 0.

Proof. Let (na : a < 003) be a one-one enumeration of P. Let X = (na: a < co$.

Set M = Q U X and TV = P \ X. Clearly M n N = 0, \M\ = cox, \N\

= co3, and M is dense in R; and by facts (i) and (ii) above, TV is also dense in R. H

Thus we have:

Corollary 3.16. The answer to the Weak Attainment Question is "yes" for the

inequality

z < 4> < ju < c. H

Proposition 3.17. The answer to the Weak Attainment Question is "yes" for the

inequality

JU < l < (j) < c.

Proof. Suppose that ? is an infinite cardinal such that there exists a ?+-Suslin tree. Let S be a ?+-Suslin tree in which every element has exactly two immediate successors. Adjoin below S a copy of ?++. Adjoin to the bottom element of ?++ co

many new immediate successors (tn: n < 00). For each n < co, adjoin to tn exactly one immediate successor rn, as shown below. Let T be the resulting pseudo-tree.

Constructing the pseudo-tree T:

ro

to rx

CO

One may check that jut ?

co, it = ft, (?>t ?

^+, and ct ? ft++.



1104 JENNIFER BROWN

In Proposition 3.2, Theorems 2.7 and 2.12 provided a quick answer of "no" to the Strong Attainment Question for all inequalities 7 in which <j> > max{c, i} or

p > max{c, i}. In fact, a more careful examination of the constructions involved in

proving Theorems 2.7 and 2.12 yields more: there are no pseudo-trees T satisfying any strict inequality 7 in which 0 > max{c, i} or p > max{c, i}.

Proposition 3.18. There is no pseudo-tree T such that cT <it < Pt

Proof. Suppose that such a tree T exists. By Theorem 2.12 and its proof, there is an S ? T such that S, with the induced ordering from T, is a /?r-Suslin tree. Since cT < it < Pt, LevlT(S) ^ 0. Pick x G LevlT(S). Then (S | x) is a well-ordered chain in T of size iT > cT\ but this is a contradiction. 3

Proposition 3.19. There is no pseudo-tree T such that cT <it < (?>t

Proof. This is proved analogously to Proposition 3.18, using Theorem 2.7 in

place of Theorem 2.12. 3

By Propositions 3.19 and 3.18, we have

Corollary 3.20. The answer to the Weak Attainment Question is "no" for the

following inequalities:

C < l < (j) < p, C or c <

i<p.) 3

We will next show that there is no pseudo-tree T for which iT < cT < 4>t or

it < cT < Pt- First we recall some definitions and state, without proof, some facts about trees; the reader is referred to [9].

For what follows, we will mean by a subtree of a tree T, an S ? T that is downwards-closed. (This is the usual meaning, and a stronger meaning than we have been using.) For any regular k, a K-tree is a tree T of height k such that for

every a < k, |Leva(r)| < k. (Thus \T\ =

k.) For any regulars, every ft-Suslin tree is a ft-tree. A well-pruned tt-tree is a tt-tree T so that |Levo(7")|

= 1, and so that for

every x G T and every a, if ht(x, T) < a < k then there exists a y G Leva(7") such that x < y. If k is regular and T is a /^-tree, then T has a well-pruned /^-sub-tree.

Recall that for a regular cardinal k, a K-Aronszajn tree is a K-tree T such that

every chain in T has cardinality less than n. The following fact is Exercise 38,

Chapter II of [9]: Lemma 3.21. Suppose k is a regular cardinal, T is a K-Aronszajn tree, ? < k,

x G T, and \{y G T: y > x}\ = k. Then there is an a > ht(x, T) such that

\{y GLeva(r): y>x}\>L 3

Proposition 3.22. There is no pseudo-tree T such that iT < ct < ?t>

Proof. Suppose that such a tree T exists. By Theorem 2.7 and its proof, there is an S ? T such that S, with the induced ordering from T, is a ^j-Suslin tree

(having 0^-many levels).




First suppose (\>T is regular. Then S has a well-pruned ^-subtree (in the strong sense) U. By Lemma 3.21, for all x G U and X < (?T there is an a > ht(x, T) such that \{y G Leva(c7) \ y > x}\ > ?. Then for every X < (?>T, iT > X. Thus iT > ct,

which is a contradiction. Now suppose 4>t is singular. Then we may choose a cardinal X with cT < X < (\>T

and an x G Lev?(S). But then (S j x) is a well-ordered chain in T of size greater than cT, which is a contradiction. H

Proposition 3.23. There is no pseudo-tree T such that iT < cT < jut

Proof. This is proved analogously to Proposition 3.22, using Theorem 2.12 in stead of Theorem 2.7. H

By Propositions 3.22 and 3.23, we have

Corollary 3.24. The answer to the Weak Attainment Question is "no" for the

following inequalities: l < C < (f) < ju, i < ju < c < </), i < c < ju < 4>, ju < c < ju, (j) t < cT. Set & =

{/ G T: f is a fan element of T}. By the second part of the proof of Theorem 2.6, we may construct a system of pseudo-levels of T, {Aa : a < 3}, such that

(i) S > (j>T, and

(ii) foralla<?,^"nyla ^0.

(Recall that in the proof of Theorem 2.6 we found that, for each ? < ?2 = <\>t, the set B =

{/ G & : for all a < ? there is a y G Aa such that y < /} was non

empty; so we may always choose to include some / G & in the next pseudo-level

A?.) Set S =

[ja<? Aa. Let R = {/ G S : / G y and ht(/, S) < z++}. ? is a tree,

with the ordering inherited from T; we claim that R is an z^+-Suslin tree. \R\ > zj+ by (ii); and

1*1 (J {/GS:/G^andht(/,S)=a}

< ^2\{f e S: f e9- 3Jidht(f9S) =a}\

a<z?+

since each level in S is an antichain in T. Also every antichain in R has size at most it < It+ Finally, every chain in R has size less than zj+: let C be any chain in R. Every / G C is a fan element in 7\ so by Theorem 2.1, \C\ < iT < zj+.



1106 JENNIFER BROWN

Thus R is an z^+-Suslin tree. But then by Lemma 3.21, there is an a < ij+ such that |Leva(i?)|

= zj, and this is a contradiction since Leva(i?) is an antichain in

T. 3

Proposition 3.26. If there is a pseudo-tree T such that iT < </>r, then there is an

z^-Suslin tree.

Proof. Suppose T is a pseudo-tree such that it < <j>t. As in the proof of

Proposition 3.25, there is a system of pseudo-levels of T, {Aa : a < ?}, such that ? >(j)T and for alia <S,^nAa ^ 0. Set S= \Ja<? Aa and set R = {feS: f e& and ht(f,S) < zj}. Then, similarly to the proof of Proposition 3.25, R is an

z^-Suslin tree. 3

Remark 3.27. By Propositions 3.17 and 3.26, the existence of a pseudo-tree re

alizing the inequality p < c is equivalent to the existence of a K+-Suslin tree for some k. Thus the answer to the Negative Attainment Question for the

inequality p < i < (j) < c would be "yes" if there were a model of ZFC in which there were no ^+-Suslin trees for any k\ but it is not known whether there is such a model. Cummings and Foreman [3] have proven, from the existence of infinitely

many supercompact cardinals, that it is consistent that there are no K?-Aronszajn trees for any 2 < n < co.

We summarize the answers to the Strong Attainment Question and the Weak and Negative Attainment Questions for each of the 24 possible general inequalities among c, z, <f), and p in the table on the following page. An entry in square parentheses indicates that the answer follows from previous entries, horizontally, in the table.

references

[I] J. Baumgartner, Almost-disjoint sets, the dense set problem and the partition calculus, Annals of Mathematical Logic, vol. 10 (1976), pp. 401-439.

[2] J. Brown, Cellularity of pseudo-tree algebras, Notre Dame Journal of Formal Logic, vol. 47 (2006), no. 3, pp. 353-359.

[3] J. Cummings and M. Foreman, The tree property, Advances in Mathematics, vol. 133 (1998), no. 1, pp. 1-32.

[4] K. Devlin, Constructibility, Springer-Verlag, 1984.

[5] J. Horne, Cardinal functions on pseudo-tree algebras and a generalization of homogeneous weak

density, Ph.D. dissertation, University of Colorado, 2005.

[6] T. Jech, Set theory, Academic Press, 1978.

[7] S. Koppelberg, General theory of Boolean algebras, Handbook of Boolean algebras (J. D. Monk and R. Bonnet, editors), vol. 1, North-Holland, 1989.

[8] S. Koppelberg and J. D. Monk, Pseudo-trees and Boolean algebras, Order, vol. 8 (1992), pp. 359 374.

[9] K. Kunen, Set theory, Elsevier, 1980.

[10] R. Laver and S. Shelah, The ^-Suslin hypothesis, Transactions of the American Mathematical

Society, vol. 264 (1981), no. 2, pp. 411-417.

[II] M. Magidor and S. Shelah, The tree property at successors of singular cardinals, Archive for Mathematical Logic, vol. 35 (1996), pp. 385-404.

[12] W. Mitchell, Aronszajn trees and the independence of the transfer property, Annals of Mathe matical Logic, vol. 5 (1973), pp. 21-46.




Inequality Strong Attainment Question Weak Attainment Question Negative Attainment Question

c < 9 < JU no (3.4) no (3.20) [yes]

c < c < i < ju < c no (3.7) yes (3.12) no (3.13)

? < ju < c < c < ju

l < < c

JU < c < < C < I

JU < (j) < I < c

JU < I < C < (

?JL < I < (p < C

no (3.4)

no (3.4)

no (3.4)

no (3.4)

yes (3.1)

no (3.4)

yes (3.1)

yes (3.1)

no (3.4)

no (3.4)

no (3.24)

yes (3.16)

no (3.24)

no (3.25)

[yes]

no (3.20)

[yes]

[yes]

no (3.24)

yes (3.17)

[yes (3.8)]

yes (3.8)

[yes (3.5)]

[yes (3.5)]

[no]

[yes]

[yes (3.5)]

' (3.27)

DEPARTMENT OF MATHEMATICS, KENYON COLLEGE. OHIO, USA

E-mail: [email protected]

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Cellularity and the Structure of Pseudo-Trees

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