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Home , Borel hierarchy, Computable ordinal, Pointclass

Descriptive theory

by George Barmpalias

Institute for Language, Logic and Computation

Lectures 9-13

(last updated: 30 of November 2009) Hyperarithmetic theory 1 The ∆1 is of fundamental importance in many branches of .

These are called the hyperarithmetic sets because they can be shown to be a constructive extension of the arithmetical class. . .

1 . . . in the same way that ∆1 was obtained by extending the finite Borel classes into the transfinite.

This is just one of the many striking analogies between the 1 1 classes ∆1, ∆1 and the theories behind them.

1 We get the theorems about ∆1 by ‘effectivizing’ (i.e., 1 constructivising) the arguments about ∆1. This often requires a fair amount of recursion/.

Remember that the difference between the boldface and the lightface theory is that the operations in the lightface theory have to be computable (effective unions, intersections) while in the boldface theory this restriction is not there.

1 Up to the Borel level ∆1 operations (unions, intersections) are countable. 1 1 But in the projective classes Σ1, Σ2,... and the analytical 1 1 classes Σ1, Σ2,... the operations become uncountable.

The set of Borel codes can be defined as the smallest class of reals which contains the codes for closed sets and is closed under the operations of ‘union’ and ‘complement’ (as these were defined when we defined the notion of ‘α codes A’).

By transfinite induction we can show that a set is Borel iff there is a Borel code α such that the the relation ‘α codes Borel set A’ holds (as this was defined formally).

Also you can stratify the Borel codes α according to the length of the tree Tα. If α is a Borel code, then Tα is well founded and has countable length.

Hence we have a hierarchy of Borel codes of length ℵ1.

Using the current methods it can be shown that the levels in the hierarchy of Borel codes correspond to the transfinite hierarchy of Borel classes. 1 We showed that if B has Borel code α then it is ∆1 in parameter α.

How about restricting ourselves to the computable Borel codes α?

1 In that case B is ∆1 without parameter.

1 Is the converse true? Does every ∆1 set have a recursive Borel code?

We will answer this question in the positive, by ‘effectivizing’ the methods we have seen. Read what follows against the theory used above to show that 1 ∆1 is the Borel class.

Computability-theoretic details are replaced by intuitive descriptions (as computability theory was not required for this course).

Definition Let f : N → N be a . A real δ is a code for f iff for all k we have δ(k) = 0 exactly when k = hs, ti such that f (Ns) ⊆ Nt (equivalently, Ns ⊆ Nt ). Continuous functions

The behavior or a continuous function between two Polish spaces is completely determined by its behavior on the basic open sets of the spaces.

If we know the code of f , i.e. all s, t such that f (Ns) ⊆ Nt , then we can ‘compute’ f (x), given x.

This is because if f (x) 6= y then there are neighborhoods Ns, Nt of x, y respectively such that f (Ns) 6⊆ Nt .

So given x, we ‘compute’ f (x) inductively as follows. Continuous functions

If we have ‘computed’ an initial segment σ of f (x), to determine the next digit we ‘look’ for a segment τ ⊂ x and some string 0 σ ⊃ σ such that f (Nτ ) ⊆ Nσ0 is recorded somewhere in the code of f .

Since f : X → Y is continuous, smaller open areas in X are mapped to smaller open areas in Y , so σ0, τ will be found.

Hence the next digit of f (x) is determined. Continuous functions

One thing to remember is that, if f is continuous then f (x) = y iff:

for all Nt with y ∈ Nt there is Ns with x ∈ Ns such that f (Ns) ⊆ Nt .

Continuous functions are a generalization of computable functions.

Computable functions are continuous. A continuous function is computably coded if it has a code which is computable.

A computably coded Borel set is a Borel set which has a computable Borel code.

Theorem If B is a computably coded Borel set and f is a computably coded continuous function, then f −1(B) is computably coded Proof:

Suppose that the code β of B is computable.

Then the items Tβ, Fβ associated with it (the tree and the assignment that we defined earlier) are also computable, since they are computable in β.

Recall that f −1(D ∪ C) = f −1(D) ∪ f −1(C) (same for countable unions) and N − f −1(D) = f −1(N − D).

Using a program for f and the computable tree Tβ we can define a program generating a computable code for f −1(B). This is done by recursion on the well-founded tree Tβ and generates computable codes for the inverse images of the sets ω represented by each labelled node Fβ(s), s ∈ ω of the tree.

This happens in the same way we computably unravel a code of a Borel set and produce the codes of the sets it is made of (through unions and negations).

To generate the code of the inverse image of the set −1 corresponding to the root of Tβ (that is, f (B)) we ask the same questions about the nodes in the next lower level of the tree.

And so on. . . Since Tβ is well founded, the program will halt and will give the right answer.

Of course the program must give an immediate answer for the sets corresponding on the leaves of the tree (nodes without successors).

Note that the leaves are closed sets with computable codes, so 0 Π1 sets (set of infinite paths through a computable tree).

0 It is not hard to show that the inverse image of a Π1 set (through 0 a computably coded continuous function) is Π1 (exercise). The height of a well founded tree is a countable ordinal (exercise).

For each countable ordinal σ, the set WFσ is countable (exercise).

Therefore it can be coded into a real, a point in the .

Theorem Suppose that T is a well-founded tree. There is a function ϕ computable in T such that for all s the number ϕ(s) is a code for a program printing a code for WF . ||T s|| Proof:

Consider fs(P) = P  s for each tree P of N .

Then fs is a computably coded continuous function.

If Q is a well-founded tree then

P ∈ WF||Q|| ⇐⇒ ||P|| ≤ ||Q|| ⇐⇒ ∀k ∃` ||P  sk || ≤ ||Q  s`||

where sn is the one-bit sequence with digit n. Hence WF = ∩ ∪ f −1(WF ). ||Q|| k ` sk ||Qs`|| So for each sequence s,

WF = ∩ ∪ f −1(WF ). ||T s|| k ` sk ||T s∗`||

This gives a recipe how to make a code for WF using ||T s|| codes corresponding to the predecessors of s in T , i.e. codes for WF , k, ` ∈ . ||T s∗`|| N Then as in the previous theorem a program can be defined by recursion on the the tree T , giving a code for WF for each T s node s on T .

The output on a node of the tree depends on the output of the program on the nodes below it.

For a node s which is a leaf, the tree T  s is the empty tree.

Therefore a code for the set of empty trees can be immediately calculated.

The program always halts since the tree is well founded. Corollary

If T is a well-founded tree, then WF||T || is Borel with a code computable in T .

An ordinal is computable if it is the height of a computable tree.

Therefore computable ordinals are countable.

CK Let ω1 be the least non-.

CK Then ω1 is countable, as the supremum of a countable collection of countable ordinals. Definition Let WFG be the set of Gödel numbers of well-founded computable trees. Let WFGσ be the set of Gödel numbers of well-founded computable trees of height < σ.

Note that WFG = WFG CK . ω1

CK We have shown that for each σ < ω1 the set WFσ is a recursively coded Borel set.

Theorem 1 1 The set WFG is Π1 but not Σ1. Proof

This is similar to the analogous boldface version.

1 1 A usual method to show that something is Π1 but not Σ1 is to 1 show that it is Π1-complete.

1 This means that knowing this set makes us compute all Π1 sets.

So ‘complete’ means universal, and we constructed such sets when we proved the hierarchy theorems. 1 1 1 1 If a Π1-complete set is Σ1, then all Π1 are Σ1, which is a contradiction.

1 So let A be Π1. Then n ∈ A ⇐⇒ ∀β∃mP(β, m, n) for a computable predicate P. By the normal form theorem (see Assignment 4 for an exercise asking for a complete proof), there is a computable sequence of trees Tn such that

n ∈ A ⇐⇒ Tn ∈ WF ⇐⇒ hTni ∈ WFG.

Since the lightface classes are closed under computable 1 1 1 substitution, if WFG was Σ1 then all Π1 would be Σ1, which we know it is not true. Theorem 1 Every ∆1 set of reals is a computably coded continuous inverse CK image of some WFσ for σ < ω1 . Proof. 1 Follows by effectivizing the corresponding argument for ∆1.

CK The Boundedness Theorem now gives a bound < ω1 by the same argument, using WFG instead of WF (and the theorem above).

Every theorem used in the boldface argument is replaced with the lightface analog, which is one of the theorems described above. Corollary 1 Every ∆1 set of reals is a computably coded Borel set. In the lightface theory instead of Borel codes we have computable Borel codes.

1 These can be called hyperarithmetic codes. Every set in ∆1 has one.

These codes α correspond to trees Tα as before, and the heights of the trees which are computable ordinals.

0 0 The classes Σσ, Πσ for computable ordinals σ are defined by transfinite induction, as the transfinite Borel classes were defined, but here we only consider effective operations (e.g. unions). If a set has a computable Borel code α then it is (at least in the) 0 0 α level of the hierarchy, i.e. in Σσ or Πσ.

This is the hyperarithmetic hierarchy and corresponds to infinitary languages about arithmetic.

0 0 1 By induction the hyperarithmetic hierarchy Σσ, Πσ exhausts ∆1.

A predicate is hyperarithmetic if it is in the hyperarithmetic 1 hierarchy. Equivalently, if it is ∆1. Theorem 1 The class ∆1 is the smallest class which contains the computable sets and is closed under effective unions and complements. Proof. 1 Exercise. First show that ∆1 has the mentioned properties.

Then by induction on the computable ordinals show that every 1 class M with these closure properties contains ∆1.

Instead of the Σ, Π classes of the hyperarithmetic hierarchy you 1 can use the fact that ∆1 contains the sets which have computable Borel codes.

Show that for every computable ordinal σ the sets which have Borel code α with ||Tα|| = σ belong to M. Theorem 1 0 Every ∆1 set is the image of a Π1 set through a 1–1 computably coded continuous function from N to N . Proof. Exercise. Follow the corresponding boldface proof and replace the notions with their effective counterparts. Note on the

The projection functions, e.g. f (x, y) = y, are continuous.

1 Verify that Σ1 consists of the images of closed sets under a projection.

1 1 Σ2 consists of the images of Π1 sets under a projection.

and so on. . .

That is why it is called the projective hierarchy. Analytic sets

A is a complete separable metric space.

Theorem For every Polish space M there is a continuous surjection of the Baire space onto M. Proof

Fix a countable dense D = {r0, r1,... } of M and to α each α ∈ N assign the sequence (xn ) = (xn) by induction.

Let x0 = rα(0) and ( r , if d(x , r ) < 2−n x = α(n+1) n α(n+1) n+1 −n xn, if d(xn, rα(n+1)) ≥ 2 .

−n α We have d(xn, xn+1) < 2 so (xn ) is Cauchy and it has a limit:

π(α) = lim xα. n n Proof

α β Then π is continuous since if α  n = β  n then x  n = x  n, so

α β −n+1 −n+1 −n+2 d(π(α), π(β)) ≤ d(π(α), xn )+d(xn , π(β)) ≤ 2 +2 = 2 .

Finally, given x ∈ M let

−n−1 α(n) = least k such that d(x, rk ) < 2

It is easy to check that π(α) = limn rα(n) = x. Definition The analytic sets A are the continuous images of Borel sets. The co-analytic sets CA are the complements of the analytic sets. The class PCA consists of the continuous images of CA.

We already gave a syntactic characterization of the class of Borel sets. Theorem 1 The class of analytic sets of some space X is Σ1.

Proof: First let A be analytic and let f be continuous such that f (B) = A for a Borel set B.

Since f is continuous, the relation f (x) = y is equivalent to:

for all Nt with y ∈ Nt there is Ns with x ∈ Ns such that f (Ns) ⊆ Nt . . . . which is equivalent to

∀t ∃s [y ∈ Nt ⇒ (x ∈ Ns ∧ f (Ns) ⊆ Nt )]

0 which is Π2 in the code of the function. Hence it is Borel.

1 Now y ∈ A iff ∃x [x ∈ B ∧ f (x) = y] which is Σ1. 1 For the other direction, let A be in Σ1. Then A is the projection of some closed set C ⊂ X × N .

Then C is a metric space with the metric it inherits from X × N .

It is a Polish space. Therefore there is a continuous surjection f : N → C.

So P is the image of N under the composition of a projection with f .

Hence P is the continuous image of the Borel set N . 1 Separation for Σ1

Definition A Γ has the separation property if, given any disjoint sets A, B ∈ Γ there is a set C ∈ Γ ∪ ¬Γ (where ¬Γ consists of the complements of the sets in Γ) such that A ⊆ C and B ∩ C = ∅. Theorem The class of analytic sets have the separation property. For a proof, we need the following basic fact. Lemma If A = ∪i∈IAi and B = ∪j∈J Bj (where I, J are arbitrary) and for each i ∈ I, j ∈ J the set Cij separates Ai , Bj (i.e. Ai ⊂ Cij and Bj ∩ Cij = ∅) then the set C = ∪i∈I ∩j∈J Cij separates A from B. Proof of lemma:

To show that B ∩ C = ∅ it suffices to show that B ∩ (∩j∈J Cij ) = ∅ for each i ∈ I.

Indeed, if b ∈ B then b ∈ Bj0 for some j0 ∈ J so b ∈/ Cij0 so b ∈/ ∩j∈J Cij .

Next, to show that A ⊆ C, let a ∈ A. Then a ∈ Ai0 for some i0 ∈ I.

Then a ∈ Ci0j for all j ∈ J by hypothesis. Hence a ∈ ∩j∈J Ci0j.

So a ∈ C. Back to the theorem. Theorem Given two disjoint analytic sets A, B in a Polish space, there is a Borel set C in the space such that A ⊆ C and B ∪ C = ∅.

Proof: Let f : N → A and g : N → B be continuous functions onto A, B respectively.

Let Aσ = f (Nσ) and Bσ = g(Nσ) for each string σ of numbers.

Notice that Aσ ⊂ A and Bσ ⊂ B.

We have Aσ = ∪i∈ωf (Nσ∗i ) and Bτ = ∪j∈ωg(Nτ∗j ). By previous lemma, if for some σ, τ the sets Aσ, Bτ are Borel inseparable, then there are i, j ∈ N such that Aσ∗i , Bτ∗j are Borel inseparable.

Suppose that A, B are Borel inseparable.

Inductively, starting with the empty sequence, we can define

(σi ), (τi ) such that |σi | = |τi | = i and for all i ∈ N the sets Aσi ,

Bτi are Borel inseparable.

Define x = ∪i σi and y = ∪i τi . f (x) ∈ f ([x n]) = A ⊆ A and g(x) ∈ g([x n]) = B ⊆ B.  xn  xn

Since A ∩ B = ∅ we have f (x) 6= g(y).

So we can find a neighborhood U around f (x) and a neighborhood V around g(y) which are disjoint. By continuity of f , g there is n ∈ N such that f ([x  n]) ⊆ U and g([y  n]) ⊆ V .

Hence A = A is separated from B = B by the Borel (in σn xn τn yn fact open!) set U.

This is a contradiction. So A, B are Borel separable. Corollary

If X is a Polish space and (An) is a pairwise disjoint sequence of analytic sets then there are pairwise disjoint Borel sets (Bn) such that An ⊆ Bn. Proof. Exercise in assignment 5. Theorem If Γ is one of the Σ/Π and has the reduction property, then ¬Γ has the separation property. Proof. Given disjoint P, Q ∈ Γ relation on a Polish space X, let P1 = X − P and Q1 = X − Q.

∗ ∗ ∗ ∗ Choose P1 , Q1 to reduce P1, Q1, so P1 ⊆ P1, Q1 ⊆ Q1, ∗ ∗ ∗ ∗ P1 ∪ Q1 = P1 ∪ Q1 and P1 ∩ Q1 = ∅.

∗ ∗ ∗ Notice that X = P1 ∪ Q1 and take R = Q1.

Then P ⊆ R and R ∩ Q = ∅. Theorem If Γ is one of the Σ/Π pointclasses and has the reduction property, then it does not have the separation property. Proof. Exercise in assignment 5. Corollary 1 The class Π1 does not have the separation property. 1 Theorem ( theorem for Σ1) 1 Every uncountable Σ1 set in a Polish space contains a perfect set.

We need the following.

Lemma If a set A in a Polish space X is uncountable, then there are disjoint open sets U, V such that both A ∩ U and A ∩ V are uncountable. Proof

For each n ∈ N let (Un,i )i be an open cover of the space such −n that Un,i is an open ball of diameter 2 .

Then for each n there exists some xn such that A ∩ Un,xn is uncountable.

Consider the intersection of the closures ∩nUn,xn .

Since the diameter of these closed sets tends to 0, the intersection contains at most one element. So there is a least n such that (∩ U ) − U contains 0 n

Indeed, otherwise at each step of this countable intersection, only countably many points of A would leave, thus uncountably many elements of A would remain in the total intersection.

Formally, U0,x0 = ∩nUn,xn ∪ [∪k (∩n

Hence there is some y 6= x such that A ∩ (U − U ) is n0 n0,y n0,xn0 uncountable.

Let U = U and V = U − U and observe that U, V n0,xn0 n0,y n0,xn0 are open, disjoint and each contain uncountably many elements of A. Corollary If f : N → X and V is an open subset of N such that f (V ) is uncountable, then there exist W0, W1 disjoint open of N such that f (Wi ) are uncountable. Proof: By previous lemma find U0, U1 open disjoint sets such that f (V ) ∩ Ui are both uncountable.

−1 If we let Wi = V ∩ f (Ui ) then you can easily verify that f (Wi ) = f (V ) ∩ Ui .

Hence f (Wi ) are uncountable and disjoint. Corollary If f : N → X and V is an open subset of N such that f (V ) is uncountable, then there exist W0, W1 disjoint basic open subsets of N such that f (Wi ) are uncountable. Proof: Notice that f (∪i Ii ) = ∪i f (Ii ).

So if f (∪i Ii ) is uncountable, there must be some i ∈ N such that f (Ii ) is uncountable.

Now the result follows from the previous corollary. Proof of theorem:

Let f be a continuous function such that f (N ) = A.

Using the previous corollary we can inductively construct a <ω ω function τσ : 2 → ω such that the following hold:

I t∅ = ∅

I if ρ ⊂ η iff τρ ⊂ τη

I f (Nτσ ) is uncountable for all σ.

I f (Nτσ∗0 ) ∩ f (Nτσ∗1 ) = ∅. This is a perfect tree where the nodes are uncountable sets in the Baire space, ordered by inclusion.

Let g : 2ω → N , g(x) = ∪ τ . n xn

This is continuous and 1–1, hence by the construction f ◦ g : 2ω → A is 1–1 and continuous.

Since 2ω is compact, f ◦ g(2ω) is closed.

Since f ◦ g(2ω) is also uncountable, it contains a perfect set. Definition Say that a class of sets Γ has the reduction property if whenever A, B ∈ Γ there are A0 ⊆ A and B0 ⊆ B such that A0, B0 ∈ Γ,A0 ∩ B0 = ∅ and A0 ∪ B0 = A ∪ B.

Theorem 1 Π1 has the reduction property. Proof

1 Suppose that A, B ∈ Π1.

Let f , g : N → Tr (where Tr is the set of codes for trees) be continuous functions such that A = f −1(WF) and B = g−1(WF).

1 These exist by the normal form theorem for Π1 sets.

Let A0 = {x ∈ A | ||f (x)|| ≤ ||g(x)||} and B0 = {x ∈ B | ||g(x)|| < ||f (x)||}. 1 These sets are Σ1 by known theorem and continuous substitution.

Obviously A0 ⊆ A, B0 ⊆ B and A0 ∩ B0 = ∅ by definition.

To show that A0 ∪ B0 = A ∪ B, let x ∈ A ∪ B.

If x ∈ A ∩ B then either ||f (x)|| ≤ ||g(x)|| or ||f (x)|| > ||g(x)|| (both f (x), g(x) are well=founded trees).

Then x is in A0 or B0 according to which case holds, respectively. If x ∈ A − B then f (x) is well-founded and g(x) is not, hence ||f (x)|| ≤ ||g(x)|| and x ∈ A0.

If x ∈ B − A then g(x) is well-founded and f (x) is not, hence ||g(x)|| < ||f (x)|| and x ∈ B0. Wadge reducibility

Definition Given Polish spaces X, Y and A ⊆ X,B ⊆ Y we say that X is Wadge reducible to Y (denoting X ≤w Y ) if there is a continuous function f : X → Y such that x ∈ X ⇐⇒ f (y) ∈ Y for all x ∈ X. Definition Given a class Γ in a Polish space we say that X is Γ-complete if X ∈ Γ and Y ≤w X for all Y ∈ Γ.

Example: In the Baire space, the set of all sequences that 0 have limit 0 form a Σ2-complete set. A function f : X → Y is Borel measurable (or just Borel) if −1 f (Ns) is Borel for all s ∈ N (where Ns are the basic open sets of Y ).

Theorem Let f : X → Y be a function between two Polish spaces A, B. The following are equivalent:

I f is Borel

I The graph of f is Borel

I The graph of f is analytic. Proof:

−1 Notice that f (x) = y iff x ∈ f (Ns) for all basic open neighborhoods around y.

Indeed, one direction is trivial and if f (x) 6= y then there would be an open neighborhood Ns such that y ∈ Ns and f (x) 6∈ Ns.

Hence if f is Borel then the graph of f is Borel.

Now suppose that the graph of f is analytic. Then

−1 x ∈ f (Ns) iff ∃y [f (x) = y ∧ y ∈ Ns] iff ∀y [f (x) = y ⇒ y ∈ Ns].

1 −1 This gives a ∆1 definition of the relation x ∈ f (Ns), so it is Borel. Theorem Let X be a Polish space. The class of Borel functions f : X → R is the smallest class of functions from X to R which contains all the continuous functions and is closed under taking pointwise limits of sequences of functions (i.e. if fn : X → R belong in the class and f (x) = limn fn(x) for each x , then f is in the class too).

Theorem Let Y be a Polish space and f : N → Y be continuous. If A ⊆ X is closed and f  A is injective, then f (A) is Borel. Proof:

Let Bs = f (A ∩ Ns) for all finite sequences s of numbers.

Let B∅ = f (A) and Bs = ∪nBs∗n.

Notice that Bs is analytic.

Also, (Bs) is a Lusin scheme i.e. Bs∗i ∩ Bs∗j = ∅ for i 6= j and Bs∗i ⊆ Bs. 0 By separation for analytic sets we can find Borel sets Bs such 0 0 0 that B∅ = Y , Bs ⊆ Bs and (Bs) is a Lusin scheme.

∗ Now define the following Lusin scheme (Bs ) by induction:

∗ 0 B∅ = B∅ B∗ = B0 ∩ B (n0) (n0) (n0) B∗ = B0 ∩ B∗ ∩ B . (n0,...nk ) (n0,...,nk ) (n0,...,nk−1) (n0,...,nk ) By induction on k we have B ⊆ B∗ ⊆ B . (n0,...,nk ) (n0,...,nk ) (n0,...,nk )

∗ ∗ Also (Bs ) is also a Lusin scheme and Bs are Borel.

∗ If we show that f (A) = ∩k ∪s Bs then clearly f is Borel.

To prove this, let x ∈ f (A) and a ∈ A such that f (a) = x.

∗ Then x ∈ ∩k Ba k and thus x ∈ ∩k B .  ak ∗ For the other direction, let x ∈ ∩k ∪s Bs .

∗ Since Bs is a Lusin scheme there is a unique a ∈ N such that ∗ x ∈ ∩k B . ak

Then x ∈ ∩ B , so in particular B 6= ∅. k ak ak

So A ∩ N 6= ∅ for all k, which means that a ∈ A since A is ak closed. So f (a) ∈ ∩ B . We claim that f (a) = x. k ak

Indeed, otherwise let f (a) = y 6= x and U be an open nbhd of y such that x 6∈ U.

Since f is continuous , there is an open nbhd N of a such ak0 that f (N ) ⊆ U. ak0

Then x 6∈ f (N ) ⊃ B which is a contradiction. ak0 ak0 Games

Let A be a nonempty set and let X ⊆ AN (as set of sequences of points in A).

Consider the following two-player infinite game.

Player I plays α0 ∈ A, then player II plays some a1 ∈ A, then player I plays a2 ∈ A and so on.

Player I wins iff (an) ∈ X. We say that X is the payoff set.

We denote this game by G(A, X) or G(X) (if A is understood). Strategies

A strategy for player I is a map ϕ : AN → AN such that |ϕ(s)| = |s| + 1 and s ⊆ t ⇒ ϕ(s) ⊆ ϕ(t).

Intuitively, if ϕ(∅) = (a0), ϕ((a1)) = (a0, a2), ϕ((a1, a3)) = (a0, a2, a4),... then I plays a0, a2, a4 ... when II plays a1, a3,... .

Equivalently, a strategy for I can be viewed as a map < ϕ : A N → A with I playing a0 = ϕ(∅), a2 = ϕ((a1)), a4 = ϕ((a1, a3)) when II plays a1, a3,... . Strategies as trees

We can also view a strategy for I as a tree T ⊆ A

I T is nonempty and pruned (every node has an extension on the tree)

I if (a0, a1,..., a2j ) ∈ T then for all a2j+1 we have (a0,..., a2j , a2j+1) ∈ T

I if (a0, a1,..., a2j−1) ∈ T then for a unique a2j we have (a0, a1,..., a2j−1, a2j ) ∈ T

Interpretation: Player I starts with the unique a0 in T , then (a0, a1) ∈ T so there is a unique a2 with (a0, a1, a2) ∈ T which is I’s next move, etc. Strategy for player II is defined analogously. Winning strategies and

A strategy of I is winning in G(A, X) if for every run of the game (a0, a1,... ) in which I follows this strategy, (an) ∈ X. Winning strategy for II is defined similarly.

We say that G(A, X) (or just X) is determined if one of the two players has a winning strategy.

Using the axiom of choice, not all games are determined.

For example, show that a set X that does not have the perfect property, nor its complement has it (like the one we constructed, Lusin’s example) is not determined. We expect that ‘definable’ games should be determined. Games with rules

Consider games where players cannot play arbitrary points from A, but have to obey certain rules.

So along with A we are given a nonempty pruned tree T ⊆ A

I and II take turns playing a0, a1,... so that (a0,..., an) ∈ T for each n.

I wins if (an) ∈ X.

Clearly the previous games are special instances of games with rules. Conversely, such a game with rules on T is equivalent to G(A, X 0) where

X 0 = {x ∈ AN |

(x ∈ [T ] ∧ x ∈ X) ∨ [(∃n, x  n 6∈ T ) ∧ (least such n is even)]

}

So it is the same notion of game. Determinacy of closed games

Theorem (Gale-Stewart) Let T be a nonempty pruned tree on A. Let X ⊆ [T ] be closed or open in [T ]. Then G(T , X) is determined.

Proof: Assume first that X is closed. Assume also that II has no winning strategy.

We will find a winning strategy for I.

Given a position p = (a0, a1,..., a2n−1) ∈ T with I to play next, we say that it is not losing for I if II has no winning strategy from then on, i.e. no winning strategy in the game G(Tp, Xp) where Tp = {s | p ∗ s ∈ T } and Xp = {x | p ∗ x ∈ X}. So ∅ is not losing for I.

Crucial fact: If p is not losing for I there is a2n with (a2n) ∈ Tp such that for any a2n+1 with (a2n, a2n+1) ∈ Tp position p ∗ (a2n, a2n+1) is also not losing for I.

We use this fact to produce a strategy for I as follows.

I starts by choosing a0 with (a0) ∈ T such that for all a1 with (a0, a1) ∈ T the position (a0, a1) is not losing for I.

Player II then plays a1 with (a0, a1) ∈ T . Player I responds with a2 such that (a0, a1, a2) ∈ T and for all a3 with (a0, a1, a2, a3) ∈ T position (a0, a1, a2, a3) is not losing for I, etc.

To show that this strategy is winning for I, let (a0, a1,... ) be a run of the game in which player I plays this strategy.

Then (a0, a1,..., a2n−1) is not losing for I, for all n.

If (an) 6∈ X, since ¬X is open in [T ], there is k such that

N(a0,...,a2k−1) ∩ [T ] ⊆ ¬X. But then (a0,..., a2k−1) is losing for I, as II has a trivial winning strategy from then on (i.e. she plays arbitrarily). This is a contradiction.

For the case when X is open, player II has a winning strategy, assuming that I does not. Quasi-determinacy of closed games without choice.

Definition A quasi-strategy for I in T is a pruned nonempty subtree Σ ⊆ T such that if (a0,..., a2j ) ∈ Σ and (a0,..., a2j , a2j+1) ∈ T then (a0,..., a2j , a2j+1) ∈ Σ.

Since Σ is pruned, if (a0,..., a2j−1) ∈ Σ then there is some a2j such that (a0,..., a2j−1, a2j ) ∈ Σ, but this may not be unique. So the difference with the notion of strategy is that we don’t uniquely choose the next move of I upon a move of II.

We define quasi-strategies for II in the same way. If X ⊆ [T ] is given, we say that a quasistrategy Σ for I is winning in G(T , X) if [Σ] ⊆ X (similarly for II).

Notice that if Σ ⊆ T is a winning quasi-strategy for I in G(T , X) then there is a winning strategy σ ⊆ Σ for I in G(T , X) using the Axiom of Choice.

A game G(T , X) is called quasi-determined if at least one of the players has a winning quasi strategy.

The argument in the Gale-Stewart result can be modified in order to show that every such closed game is quasi-determined. Game quantifiers

If X ⊆ AN then the statement ‘I has a winning strategy in G(A, X)’ can be abbreviated as:

∃a0∀a1∃a2∀a3 ··· (an) ∈ X.

Similarly, ∀a0∃a1∀a2∃a3 · · · ¬(an) ∈ X abbreviates the statement that II has a winning strategy.

Hence the determinacy of G(A, X) can be expressed as

¬∃a0∀a1 ··· (an) ∈ X ⇐⇒ ∀a0∃a1 · · · ¬(an) ∈ X. Game quantifiers

So determinacy can be thought of as an infinitary analog of the basic rule in logic that

¬∃a0∀a1 ··· Qan−1X(a0,..., an−1)

is equivalent to ˆ ∀a0∃a1 ··· Qan−1X(a0,..., an−1)

Notice that this rule asserts the determinacy of the finite game where I plays a0, a2,... and II plays in the odd stages a1, a3,... and I wins iff X(a0,..., an−1). Game quantifiers

We wish to show that this infinitary rule is true for Borel sets.

We know that it fails in general, using the axiom of choice.

For any nonempty set A the game quantifier GA is defined by

GAyP(x, y) ⇐⇒ ∃a0∀a1∃a2∀a3 ··· P(x, (an))

where P ⊆ X × AN. Game quantifiers

ˆ The dual game quantifier GA is defined by ˆ GAyP(x, y) ⇐⇒ ∀a0∃a1∀a2∃a3 ··· P(x, (an)).

So if all games G(A, Px ) are determined, then ˆ ¬GAyP(x, y) ⇐⇒ GAy¬P(x, y) Borel determinacy

Theorem (Martin) Let T be a nonempty pruned tree on A and let X ⊆ [T ] be Borel. Then G(T , X) is determined.

The idea for the proof is to associate to G(T , X) an auxiliary game G(T ∗, X ∗) such that

I it is known to be determined (for example closed or open game)

I a winning strategy for any of the players in it gives a winning strategy for the corresponding player in G(T , X). Definition (Covering of a game) Let T be a nonempty pruned tree on A. A covering of T is a triple (Tˆ, π, ϕ) where (i) Tˆ is nonempty pruned tree (on some A).ˆ (ii) π : Tˆ → T is monotone with length(π(s)) = length(s). Thus π induces a continuous function from [Tˆ] to [T ] also denoted by π :[Tˆ] → [T ]. (iii) ϕ maps strategies for I (resp. II) in Tˆ to stragegies for player I (resp. II) in T in such a way that ϕ(ˆσ) restricted to positions of length ≤ n depends only on σˆ restricted to positions of length ≤ n, for all n. (iv) If σˆ is a strategy for I (resp. II) in Tˆ and x ∈ [T ] follows ϕ(ˆσ) (i.e. x ∈ [ϕ(ˆσ)]), then there is xˆ ∈ [Tˆ] played according to σˆ such that π(xˆ) = x. Covering of a game

(ii) π maps Tˆ → T ina canonical way. (iii) ϕ maps strategies in Tˆ to strategies in T in a canonical way. ϕ is really defined on partial strategies σˆ  n in a monotone way (m ≤ n ⇒ ϕ(ˆσ  m) = ϕ(ˆσ  n)  m). It is extended to strategies naturally, by following what it does in the initial segments of strategies. (iv) If σˆ is a strategy (for I or II) in Tˆ and x ∈ T is a run of the original game that follows ϕ(ˆσ) then there is a run xˆ of the simulated game which corresponds to it via π follows σˆ.

I The game G(T , X) is ‘simulated’ by the auxiliary game G(Tˆ, Xˆ ), where Xˆ = π−1(X). Lemma (Winning strategies) The map ϕ preserves winning. That is, if σˆ is a winning strategy (for I or II) in G(Tˆ, Xˆ ) then ϕ(ˆσ) is winning (for I or II resp.) in G(T , X). Proof. Otherwise there is x ∈ [ϕ(ˆσ)] with x 6∈ X (resp. x ∈ X). But by (iv) we can find xˆ ∈ [ˆσ] with π(xˆ) = x. Then xˆ ∈ Xˆ because σˆ is a winning strategy (resp. xˆ 6∈ Xˆ ) so x ∈ X (resp. x 6∈ X) which is a contradiction. For technical reasons, in order to carry on an induction in the proof, we extend the notion of covering.

Definition (k-coverings) For k ∈ N we say that (Tˆ, π, ϕ) is a k-covering if it is a covering ˆ ˆ such that T  2k = T  2k and π  (T  2k) is the identity.

Intuitively this means thatin the uxiliary game G(Tˆ, Xˆ ) the first k moves of each player are identical to those of G(T , X). Lemma If (Tˆ, π, ϕ) is a k-covering then for any strategy σˆ in Tˆ we have ϕ(ˆσ)  2k =σ ˆ  2k. Proof. By (iv) if (Tˆ, π, ϕ) is a k-covering then for any strategy σˆ in Tˆ we have ϕ(ˆσ)  2k ⊆ σˆ  2k..

ˆ Since T  2k = T  2k and both ϕ(ˆσ)  2k and σˆ  2k are both strategies for the same player in T wi must have ϕ(ˆσ)  2k =σ ˆ  2k Definition (Unraveling) We say that covering (Tˆ, π, ϕ) unravels X ⊂ [T ] if π−1(X) = Xˆ is clopen (in [Tˆ]). Now clearly if unravels G(T , X), by Gale/Stewart and the preceding fact G(T , X) is determined.

So the theorem is implied by the following. Theorem (Martin) If T is a nonempty pruned tree on A and X ⊆ [T ] is Borel, then for each k ∈ N there is a k-covering of T which unravels X.

We are only interested in k = 1, but we in order to prove this we need an induction that proves the above for all k. There are two main lemmas for the proof of this theorem. Lemma (Base case) Let T be a nonempty pruned tree and let X ⊆ [T ] be closed. For each k ∈ N there is a k-covering of T that unravels X. Lemma (Induction/ Existence of inverse limits) Let k ∈ N and (Ti+1, πi+1, ϕi+1) be a (k + i)-covering of Ti , i = 0, 1,... . Then there is a pruned tree T∞ and π∞,i , ϕ∞,i such that (T∞, π∞,i , ϕ∞,i ) is a (k + i)-covering of Ti and

πi+1 ◦ π∞,i+1 = π∞,i and ϕi+1 ◦ ϕ∞,i+1 = ϕ∞, i. Proof of theorem (using the lemmas)

By induction on the levels of the .

We prove by inductin on 1 ≤ ξ < ω1 that for all T , k ∈ N and 0 X ⊆ [T ] in Σξ ([T ]). there is a k-covering of T that unravels X.

By symmetry in the definition, if a k-covering unravels X it also unravels[T ] − X, so by the first lemma the statement is true for ξ = 1.

Assume now that it holds for all η < ξ.

0 Then for each T , each Y ∈ Πη with Y ⊆ [T ], η < ξ and each k there is a k-covering that unravels [T ] − Y , thus also Y itself. 0 Let X ∈ Ση with X ⊆ [T ] and k ∈ N.

Then X = ∪ X for X ∈ Π0 with X ⊆ [T ] and ξ < ξ. i∈N i i ξi i i

Let (T1, π1, ϕ1)be a k-covering of T0 = T . that unravels X0.

Then π−1(X ) (subset of [T ]) is also in Π0 for i ≥ 1 since π is 1 i 1 ξi 1 continuous.

By recursion define (Ti+1, πi+1, ϕi+1) to be a (k + i)-covering of −1 −1 −1 Ti that unravels πi ◦ πi−1 ◦ · · · π1 (Xi ). Now let (T∞, π∞,i , ϕ∞,i ) be as in the ‘induction lemma’.

Then (T∞, π∞,0, ϕ∞,0) unravels every Xi .

−1 −1 So π∞,0(X) = ∪i π∞,0(Xi ) is open in T∞.

Now use the ‘base lemma’ and let (Tˆ, π, ϕ) be a k-covering of −1 T∞ that unravels π∞ .

ˆ Then (T , π∞,0◦π, ϕ∞,0 ◦ φ) be a k-covering.