Lecture 2 Plan: 1. Automatic Boolean Algebras 2. Automatic Linear Orders 3. Automatic Trees 4....
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Transcript of Lecture 2 Plan: 1. Automatic Boolean Algebras 2. Automatic Linear Orders 3. Automatic Trees 4....
![Page 1: Lecture 2 Plan: 1. Automatic Boolean Algebras 2. Automatic Linear Orders 3. Automatic Trees 4. Automatic Versions of König’s lemma 5. Intrinsic Regularity.](https://reader034.fdocuments.net/reader034/viewer/2022051622/5697bf861a28abf838c87f9e/html5/thumbnails/1.jpg)
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)
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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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.
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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).
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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.
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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.
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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).
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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
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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.
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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.
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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) .
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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}
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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)).
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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
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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.
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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.
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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:
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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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.
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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.