Lecture 2Plan:

1. Automatic Boolean Algebras

2. Automatic Linear Orders

3. Automatic Trees

4. Automatic Versions of König’s lemma

5. Intrinsic Regularity and Definability:

a) Decidability Theorem III.

b) Intrinsic Regularity in (,S)

Automatic Boolean AlgebrasA Boolean Algebra is (B,,∩, /, 0,1), where

the operations ∩,, and / satisfy the usual

properties of the set-theoretic operations, 0 is

the minimal element and 1 is the maximal

element.

P(A) is an example of a Boolean Algebra.

Automatic Boolean Algebras

Let L be a linear order. An interval is the set:[a,b)={x | ax<b}.

Consider the set BL consisting of all finite unions of all intervals.

Claim 1: BL is a Boolean algebra.Claim 2: Any Boolean algebra is isomorphic

to BL for some linear order L.

Automatic Boolean Algebras

Examples: Bω, Biω, Bη are Boolean algebras.

Lemma 1. The Boolean algebra Biω has an automatic presentation.

Lemma 2. The Boolean algebra Bη does not have an automatic presentation.

Proof. Assume Bη is automatic.

Automatic Boolean Algebras

For each binary string σ construct an element

bσ as follows.

bλ =1.

Assume bσ has been constructed. Find the

length-lexicographically first x such that

bσ∩x≠0 and bσ∩(1/x) ≠0 . Set:

bσ0 = bσ∩x.

bσ1 = bσ∩ (1/x).

Automatic Boolean Algebras

Claim 1: There is a C1such that

|bσα| |bσ| +C1.

Claim 2: Let Xn={bσ | |σ|=n}. There is a C2

such that for all x in Xn we have |x| C2n.

Claim 3: There is a C3 such that for all y

generated from Xn we have |y| C3n.

Automatic Boolean Algebras

Thus, the set of all elements of the Boolean

algebra generated by Xn is a subset of

{0,1}O(n). However, the number of elements

of the the Boolean algebra generated by Xn

is 2 to the power of 2n. Contradiction.

Thus, the atomless Boolean algebra does not

have an automatic copy.

Automatic Boolean Algebras

The technique can now be used to prove the following

Theorem [Characterization of Automatic BA] (Khoussainov, Nies, Rubin, Stephan. 2003)A Boolean algebra has an automatic copy if

and only if it is isomorphic to Biω for some i.

Proof. One direction is done.

Automatic Boolean Algebras

Assume B is automatic but not of the desired

form. An element b in B is infinitary if it has

infinitely many atoms below it. An element b

is large if its image in the factor algebra by

the finitary ideal is not a union of finitely

many atoms.

By assumption, the element 1 is large.

Automatic Boolean Algebras

Now, one constructs a tree Tn with the

following properties:

1. The number of leaves is at least n2.

2. There are at least n-1 leaves that are infinitary.

3. There is at least one leave that is large.

4. The length of each element is O(n).

Automatic Boolean Algebras

Thus, the sub-algebra generated by leaves has

2 to the n2 elements. But the length of each

element in the sub-algebra is O(n). Hence

The number of elements of the subalgebra

is a subset of {0,1}O(n).

Again, we have a contradiction.

The theorem is proved.

Automatic Linear Orders

Examples:

1. , i called small ordinals.

2. The order of rational numbers η.

3. The sum and products of automatic linear orders.

4. Σ(η+f(n)), where f(n)=2an+b or f(n) is a polynomial with positive coefficients.

Automatic Linear Orders

Let (L,) be a linear order. Elements a,b are

equivalent if there are finitely many elements

between them.

Factorize (L,) w.r.t the equivalence

relation, and get the linear order (L1,1).

Apply the process to (L1,1) and get (L2,2).

Automatic Linear Oders

Continue on. The first point (ordinal) at which

(Ln,) equals (Ln+1,n+1) is called the Cantor-

Bendixson rank of the linear order.

Example. The CB rank of i is i.

Note: if (Ln,) = (Ln+1,n+1) then either (Ln,)

is singleton or (Ln,) is isomorphic to η

Automatic Linear Orders

Theorem [Ranks of Automatic Linear Orders] (Khoussainov, Nies, Rubin, Stephan. 2003)If (L,) is an automatic linear order then its Cantor-Bendixson rank is finite.

Proof. We provide our proof when the linear order is an ordinal. This is due to Goranko, Delhomme and Knapik (2000). Ordinals of finite rank are small ordinals. All are automatic.

Automatic Linear Orders

If ordinal α is automatic and β α then β is

automatic. So it suffices to prove that is

not automatic.

Picture of : ω+ω2+ω3+ω4+……

Assume is automatic.

Automatic Linear Orders

Let M be an automaton recognizing the order relation in . Let D be an automaton recognizing the domain of .

Pairs (u,v) and (u1,v1) with |u|=|v| and |u1|=|v1| are equivalent if:

1. v, v1 are accepted by D.

2. D does not distinguish u and u1

3. M does not distinguish (u,v) and (u1,v1) .

Automatic Linear Orders

Claim 1.

The number of equivalence classes |D| |M|.

Claim 2.

If (u,v) and (u1,v1) are equivalent then

{uw | uw is in D & uw v} is isomorphic to

{u1w | u1w is in D & u1w v1}

Automatic Linear Orders

Consider v1 v2 v3 ….. such that

vi={x | x vi} is isomorphic to i.

With each vi associate the set Char(vi)

consisting of all (Du, M(u,vi)), where |u|=|vi|.

Consider the sequence:

Char(v1), Char(v2) , Char(v3) ,……..

Fact 3. i j ( i < j & Char(vi)=Char(vi)).

Automatic Linear Orders

Decompose vj into the union of sets of the

type {uw | uw is in D & uw vj & |u|=|vj|}.

Thus, vj =X1 X2 …..Xk.

Fact 4: One of Xs is isomorphic to vj.

Do exactly the same for vi:

vi =Y1 Y2 …..Yt

Automatic Linear Orders

By Claims 2 and 3 there exists Yk Xj contradicting

vi < vj. The theorem is proved.

Corollaries:1. The rank of automatic linear order can be computed.

2. It is decidable if automatic lo is well-order.

3. The isomorphism problem for automatic ordinals is decidable.

Automatic TreesLet T=(T, ) be an automatic & infinite tree. We assume that T is finitely branching.d(T) is the sub-tree consisting of all x in T such that there is a split above x and the split is on two inf paths. Consider the sequence: T, d(T), d(d(T)),…, d(dn(T)),….

Definition. The first α at which dα(T)= dα+1(T) is the Cantor-Bendixson rank of T.

Automatic Trees

Lemma If T has countably many infinite

paths and is finitely branching then CB(T) is

finite.

Proof. For each u in T consider S(u) the set of

all immediate successors of u. Order S(u) via

llex. Now we can define the Kleene-Brower

order on T as follows:

Automatic Trees

x kb y if x is above y or y is “right of” x.

Thus, we have the linear order (T, kb )

which is automatic.

Claim:

1. (T, kb ) is scattered.

2. CB(T) does not exceed CB (T, kb )+1.

This ends the proof of the lemma.

Automatic Trees

Theorem I [Ranks of automatic trees] (Khoussainov, Rubin, Stephan, 2003)If T is automatic finitely branching tree then CB(T) is finite.

Proof. For each x in T consider the tree Tx.

Call x scattered if Tx has countably many paths. The lemma above implies that for

CB(Tx) is bounded by a fixed n. This implies That CB(T) must be at most n.

Automatic Trees

Theorem II [ranks of automatic trees]

(Khoussainov, Rubin, Stephan, 2003)

If T is automatic tree then CB(T) is finite.

The proof is based on constructing a finitely

branching automatic tree T1 whose rank

bounds the rank of T.

Konig’s Lemma (automatic versions)

Claim. Given an automatic tree it is decidable

if it has an infinite path.

This is proved by constructing the Kleene-

Brower order.

Assume that T has an infinite path and is

finitely branching.

Konig’s Lemma (Automatic versions)

Here is a FO+ ω definition of an infinite

path. Good(x) if any y below or equal to x is

the <llex-first immediate successor of its parent

such that there are infinitely many z above y.

Observation. If {x | x is on an infinite path} is

regular then we can remove the assumption

that T is finitely branching.

Konig’s Lemma (automatic versions)

Pruning Lemma For any automatic tree

T the set {x | x is on infinite path} is

regular.

Theorem I (Khoussainov, Rubin, Stephan, 2003)

Every automatic infinite tree with an infinite

path has a regular infinite path.

Konig’s Lemma (Automatic versions)

Theorem II (Khoussainov, Rubin, Stephan, 2003)

Let T be an automatic tree with countably many

infinite paths. Then each path is regular.

Proof. Use the fact that the CB(T) is finite and

use Prunning Lemma.

Intrinsic regularityLet A be an automatic structure. Let R be a

relation in it.

Definition. R is intrinsically regular if R is

regular in all automatic presentations of A.

Example 1. All relations definable in FO+

logic are intrinsically regular.

Intrinsic Regularity

Example 2.

Theorem (Semenov) A relation R in (, +) is

intrinsically regular iff R is definable.

Example 3.

Relations in ({0,1}*; L, R, ({0,1}*; L, R, prefpref, EqL) is , EqL) is

intrinsically regular iff it is definable.intrinsically regular iff it is definable.

Intrinsic regularity

Consider (, ). A unary relation in this

structure is definable iff it is finite or co-finite.

Proposition. The set M2={x| x is at odd position} is

intrinsically regular.

Proof. We need to extract an automaton

recognizing M2 from any given automatic

presentation of (, ). So, fix a presentation.

Intrinsic Regularity

Decidability Theorem III

(Khoussainov, Rubin, Stephan)

1. All relations definable in FO++(n,m) are

intrinsically regular.

2. The FO++(n,m) theory of any automatic

structure is decidable.

Intrinsic Regularity

Consider (, S). A unary relation in this

structure is definable iff it is finite.

Proposition (Khoussainov, Rubin Stephan). The

relation is not intrinsically regular in (, S).

Proof. We need to build an automatic copy of

(, S) in which is not regular.

Intrinsic Regularity

0n 0n1 0n-1111 0n-211111 …

0n-i112i …. 12n+1 12n+2 12n0 12n-200 …12n-2i0i …. 0n+1

We thus have the following automatic

structure. The domain is 0*1*, and S(x)=y iff

x y. This is isomorphic to (, S). Clearly,

0n+1 > 12n+2 . But, this is not a regular event.

Intrinsic Regularity

Theorem (Khoussainov, Rubin Stephan). The

relation M2={x| x is at odd position} is not

intrinsically regular in (, S).

Proof (sketch and idea). We need to build an

automatic copy of (, S) in which M2 is not

regular. The alphabet is {0,1}. For each string

x over this alphabet define ep(x) and op(x).

Intrinsic Regularity

So, for x=01011 : ep(x)=0011, op(x)=11

For a string x, set n=|ep(x)| and m=|op(x)|.

So, mnm+1.

Think of x as the pair (ep(x), op(x)) with

op(x) being a parameter.

A start point is any x of the type (0, op(x)).

A mid point is any x with ep(x)=2n-1.

Intrinsic Regularity

Thus, we have x of length n+m. Assume that x is a start point. Set b=2op(x) +1. Here is now how we define the successor:

(0,op(x))(b, op(x)) (2b, op(x)) ….. (2n-1-b, op(x)) (2n-1+b, op(x)) … (2n-b, op(x)) (2n-1, op(x)),

Where the addition by b is performed mod 2n.

Intrinsic Regularity

The value of the successor at (2n-1, op(x)) is

defined as follows. If op(x)+1 is not 2n then

(2n-1, op(x)) (0, op(x)+1). Otherwise,

(2n-1, op(x)) 0n+m+1.

Intuition behind the definition of the successor

is the following.

Intrinsic Regularity

Consider the sequence

(0,op(x)), (b, op(x)),…..,(2n-1-b, op(x)), [L-side]

(2n-1,op(x)) (mid point)

(2n-1+b, op(x)),…,(2n-b, op(x)) [R-side]

This is a “normal” sequence in which oddness

or evenness of positions is a regular event.

Intrinsic Regularity

The successor disrupts this regularity by putting the mid point at the end of the sequence

(0,op(x)), (b, op(x)),…..,(2n-1-b, op(x)), [L-side](2n-1+b, op(x)) (mid point)

(2n-1+2b, op(x)),…,(2n-b, op(x)) , (2n-1, op(x)), [R-side]

Thus, in order to know if (v,op(x)) is in odd or even position one needs to know (v,op(x)) is on L-side or R-side of the sequence. This is not a regular event.