Lecture 2 Plan: 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 Algebras A 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 | a  x<b}. Consider the set B L consisting of all finite unions of all intervals. Claim 1: B L is a Boolean algebra. Claim 2: Any Boolean algebra is isomorphic to B L for some linear order L.

Automatic Boolean Algebras Examples: B ω, B iω, B η are Boolean algebras. Lemma 1. The Boolean algebra B iω has an automatic presentation. Lemma 2. The Boolean algebra B η does not have an automatic presentation. Proof. Assume B η is automatic.

Automatic Boolean Algebras For each binary string σ construct an element b σ as follows. b λ =1. Assume b σ has been constructed. Find the length-lexicographically first x such that b σ ∩x≠0 and b σ ∩(1/x) ≠0. Set: b σ0 = b σ ∩x. b σ1 = b σ ∩ (1/x).

Automatic Boolean Algebras Claim 1: There is a C 1 such that |b σα |  |b σ | +C 1. Claim 2: Let X n ={b σ | |σ|=n}. There is a C 2 such that for all x in X n we have |x|  C 2 n. Claim 3: There is a C 3 such that for all y generated from X n we have |y|  C 3 n.

Automatic Boolean Algebras Thus, the set of all elements of the Boolean algebra generated by X n is a subset of {0,1}  O(n). However, the number of elements of the the Boolean algebra generated by X n is 2 to the power of 2 n. 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 ) A Boolean algebra has an automatic copy if and only if it is isomorphic to B iω 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 T n with the following properties: 1.The number of leaves is at least n 2. 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 n 2 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)=2 an+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 (L 1,  1 ). Apply the process to (L 1,  1 ) and get (L 2,  2 ).

Automatic Linear Oders Continue on. The first point (ordinal) at which (L n,  ) equals (L n+1,  n+1 ) is called the Cantor- Bendixson rank of the linear order. Example. The CB rank of  i is i. Note: if (L n,  ) = (L n+1,  n+1 ) then either (L n,  ) is singleton or (L n,  ) is isomorphic to η

Automatic Linear Orders Theorem [Ranks of Automatic Linear Orders] ( Khoussainov, Nies, Rubin, Stephan ) 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+ω 2 +ω 3 +ω 4 +…… 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 (u 1,v 1 ) with |u|=|v| and |u 1 |=|v 1 | are equivalent if: 1. v, v 1 are accepted by D. 2. D does not distinguish u and u 1 3. M does not distinguish (u,v) and (u 1,v 1 ).

Automatic Linear Orders Claim 1. The number of equivalence classes  |D| |M|. Claim 2. If (u,v) and (u 1,v 1 ) are equivalent then {uw | uw is in D & uw  v} is isomorphic to {u 1 w | u 1 w is in D & u 1 w  v 1 }

Automatic Linear Orders Consider v 1  v 2  v 3  ….. such that  v i ={x | x  v i } is isomorphic to  i. With each v i associate the set Char(v i ) consisting of all (Du, M(u,v i )), where |u|=|v i |. Consider the sequence: Char(v 1 ), Char(v 2 ), Char(v 3 ),…….. Fact 3.  i  j ( i < j & Char(v i )=Char(v i )).

Automatic Linear Orders Decompose  v j into the union of sets of the type {uw | uw is in D & uw  v j & |u|=|v j |}. Thus,  v j =X 1  X 2  …..  X k. Fact 4: One of X s is isomorphic to  v j. Do exactly the same for  v i :  v i =Y 1  Y 2  …..  Y t

Automatic Linear Orders By Claims 2 and 3 there exists Y k  X j contradicting v i < v j. 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 Trees Let 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(d n (T)),…. Definition. The first α at which d α (T)= d α+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 T x. Call x scattered if T x has countably many paths. The lemma above implies that for CB(T x ) 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 T 1 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 regularity Let 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. ({0,1}*; L, R,  pref, EqL) is Relations in ({0,1}*; L, R,  pref, EqL) is 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 M 2 ={x| x is at odd position} is intrinsically regular. Proof. We need to extract an automaton recognizing M 2 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 0 n  0 n 1  0 n  0 n  …  0 n-i 11 2i  ….  1 2n+1  1 2n+2  1 2n 0  1 2n-2 00  …1 2n-2i 0 i  ….  0 n+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, 0 n+1 > 1 2n+2. But, this is not a regular event.

Intrinsic Regularity Theorem ( Khoussainov, Rubin Stephan ). The relation M 2 ={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 M 2 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, m  n  m+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)=2 n-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))  …..  (2 n-1 -b, op(x))  (2 n-1 +b, op(x))  …  (2 n -b, op(x))  (2 n-1, op(x)), Where the addition by b is performed mod 2 n.

Intrinsic Regularity The value of the successor at (2 n-1, op(x)) is defined as follows. If op(x)+1 is not 2 n then (2 n-1, op(x))  (0, op(x)+1). Otherwise, (2 n-1, op(x))  0 n+m+1. Intuition behind the definition of the successor is the following.

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