Characterization of projective sets by finer topologies

Matthias Enders

5. Juli 2001

1 Characterization of Borel and analytic sets by finer topologies

We work in Polish spaces

Definition 1.1. A topological space (X, τ) is called a Polish space if (X, τ) is a separable, completely metrizable space, i.e., there exists a complete metric on X and τ is the topology of the metric space.

The following result is classical:

Theorem 1.2. Let X be a Polish space. A ⊆ X is Borel iff there exists a finer topology t on A which is Polish.

For a construction of the finer topology see [Kech95, 13.1]. For the converse note that the image of an Borel set under a continuous injection is again Borel ([Kech95, 15.1]), so using the identity mapping (A, t) ⟶ (X, τ) proves the result.

Professor Howard Becker found results for Σ_{n}^{1} sets in the spirit of the above theorem in [Be1] and [Be2]. He used for this a characterization of a Polish space by a strong Choquet space. We introduce this notion:

Let X be a nonempty topological space. The Choquet game G_{X} of X is defined as follows: Players I and II take turns in playing nonempty open subsets of X

I  U_{0}  U_{1}  …
II  V_{0}  V_{1} …,

such that U_{0} ⊇ V_{0} ⊇ U_{1} ⊇ V_{1} ⊇ … We say II wins this run of the game if ⋂_{n} V_{n} = ⋂_{n} U_{n} ≠ ∅. Otherwise I wins.

The strong Choquet game G_{X}^{s} is defined similiar, but in addition to the Choquet game player I is required to play a point x_{n} ∈ U_{n} on every turn and then player II must play V_{n} ⊆ U_{n} with x_{n} ∈ V_{n}. So we have

I  U_{0}, x_{0}  U_{1}, x_{1}  … II  V_{0}  V_{1} …,

U_{0} ⊇ V_{0} ⊇ …, x_{n} ∈ U_{n}, x_{n} ∈ V_{n} for n ∈ ω. A (winning) strategy for one of the players is defined in the usual way.

Definition 1.3. A nonempty topological space is called a (strong) Choquet space if player II has a winning strategy in the (strong) Choquet game G_{X} (G_{X}^{s}).

Now the characterization of the Polish space looks as follows:

Theorem 1.4 (Choquet). A nonempty, second countable topological space is Polish iff it is T1, regular and strong Choquet.

It is an easy exercise to prove that a Polish space has this properties. The converse is somewhat more difficult and makes use of the axiom of choice.

This characterization gives now a result for Σ_{1}^{1} sets instead of Borel sets as in the above theorem. We only have to drop the regularity property.

Theorem 1.5. Let X be a Polish space. A subset A ⊆ X is Σ_{1}^{1} iff there exists a topology t on A such that

• t ⊇ T |A

• t is second countable

• t is strong Choquet.

The topology we construct on A is the Gandy-Harrington topology (or its relativized version). For a definition and the proof that this topology is strong Choquet see [HKL90]. The converse is proved in [Be1].

2 A finer topology on a Σ_{n}^{1} set

We are now interested in results similar to that of chapter 1 for higher classes of the projective hierarchy. So we have to consider further ways of weakening the topological conditions in our space. A finer topology of a Hausdorff space will always remain Hausdorff and we already dropped the regularity property. One could ask what happens if we weaken the strong Choquet property to the Choquet property. The next Proposition shows that this leads nowhere.

Proposition 2.1. Let (X, T ) be a Polish space and A an arbitrary subset of X. Then there exists a topology t on A such that t is finer than T and t is regular, second countable and Choquet.

Proof: Let B be a basis for (A, T |A). Let C be the closure of B under complements and finite intersections. Pick a point x_{C} in each nonempty C ∈ C. Now let D = C ∪ {{x_{C}} | C ∈ C} be the countable basis for the topology t. Since the basis consists of clopen sets t is regular. The isolated points are dense in t, so player II wins the Choquet game in his first move by playing one of the x_{C}’s.

By this proposition the only topological condition that remains to be considered is the second countable condition. The key fact for the proof of the strong Choquet property in the Gandy-Harrington topology was that Σ_{1}^{1} sets are ω-Suslin. If we would have a Suslin representation for Σ_{n}^{1} sets we could imitate this proof. We work now under ZF + DC + projective determinacy where we have indeed such an result.

Theorem 2.2 (PD). For all n ≥ 0 the following holds:

(i) Every Σ_{2n+2}^{1}-set is δ_{2n+1}^{1}-Suslin. (ii) Every Σ_{2n+2}^{1}-set A is κ_{2n+1}(A)-Suslin for a cardinal κ_{2n+1}(A) < δ_{2n+1}^{1}.

Proof: First notice that if a set A is κ-Suslin then p[A] is κ-Suslin. To see this let T be a tree on ω^{k+1} × κ such that A ⊆ N^{k} is given by A = p[T]. Fix a bijection

f : ω × κ ⟶ κ

This leads to a bijection

f^{∗} : (ω × κ)^{<ω} ⟶ k^{<ω}

We define a tree T′ on ω^{k} × κ by

(s_{1}, …, s_{k}, η) ∈ T′ :⇔ (s_{1}, …, s_{k}, f^{∗−1}(η)) ∈ T

(1) p[T′] = p[A] Proof:

(x_{1}, …, x_{k}) ∈ p[T′] ⇔ ∃u ∈ κ^{ω} (x_{1}, …, x_{k}, u) ∈ [T′]
⇔ ∃u ∈ κ^{ω} ∀n(x_{1}|n, , …, x_{k}|n, u|n) ∈ T′
⇔ ∃u ∈ κ^{ω} ∀n(x_{1}|n, …, x_{y}|n, f^{∗−1}(u|n)) ∈ T
⇔ ∃u ∈ κ^{ω}∃x_{k+1} ∈ N ∀n(x_{1}|n, …, x_{k+1}|n, u|n) ∈ T
⇔ ∃u ∈ κ^{ω}∃x_{k+1} ∈ N (x_{1}, …, x_{k+1}, u) ∈ T
⇔ ∃x_{k+1} ∈ N (x_{1}, …, x_{k+1}) ∈ p[T] = A
⇔ (x_{1}, …, x_{k}) ∈ p[A]

q.e.d.(1)

So for (i) it is enough to prove that each Π_{2n+1}^{1}-set is δ_{2n+1}^{1}-Suslin. But by Second Periodicity Theorem from Moschovakis we know that each Π_{2n+1}^{1} set has a Π_{2n+1}^{1}-scale. Since a Π_{2n+1}^{1}-scale on a Π_{2n+1}^{1}-set is a δ_{2n+1}^{1}- scale we are done since we know that if a set admits a λ-scale this set is λ-Suslin For (ii) Let A be a Σ_{2n+1}^{1} set and let B ∈ Π_{2n}^{1} such that A = p[B]. Since B ∈ Δ_{2n+1}^{1} there exists a Π_{2n+}^{1}-scale (φ_{i})_{i∈ω} on B. Each φ_{i} is a Δ_{2n+1}^{1}- norm on B, so |φ_{i}| < δ_{2n+1}^{1}. The length of the scale is sup_{i<ω} |φ_{i}| and since cf(δ_{2n+1}^{1}) > ω the sequence |φ_{i}| is bounded below δ_{2n+1}^{1}. Hence there is a cardinal κ_{2n+1}(A) < δ_{2n+1}^{1} such that |φ_{i}| ≤ κ_{2n+1}(A) for all i ∈ ω. Thus (φ_{i})_{i} is a κ_{2n+1}(A)-scale and the associated tree tells us that B is κ_{2n+1}(A)-Suslin. By the above remark we are done.

Now we can give the main result of this section:

Theorem 2.3 (PD). Let (X, T ) be a Polish space. For n ≥ 1 let A ∈ Σ_{n}^{1}(X). Then there exists a finer topology t on A such that t is strong Choquet and has a basis of length a cardinal less than δ_{n}^{1} (less as an ordinal, not necessarily less in cardinality).

Proof: Let n ≥ 1. WLOG we can assume that A is a subset of N. Let κ be a cardinal less than δ_{n}^{1}. Let T be a tree on ω × κ such that A = p[T]. (Note that using a bijection between an ordinal λ and its cardinality κ each λ-Suslin set is κ-Suslin.) For s ∈ T the subtree T_{s} consists of all sequences compatible with s. Let A = {p[T_{s}] | s ∈ T }. Then A is of cardinality κ. Let B be the closure of A and all the T -basic open sets of A under finite intersections. The cardinality of B is κ. B is the basis for the topology t. It remains to show that this topology t is strong Choquet. We do this by describing a winning strategy for II. Assume player I starts by playing (x_{0}, U_{0}). Then choose a basic open set of the form p[T_{r_{0}^{0}}] ∩ p[T_{r_{0}^{1}}] ∩ … ∩ p[T_{r_{0}^{m_{0}}} ] ∩ N_{u_{0}}, u_{0} ∈ ω^{<ω}, such that this set is a subset of U_{0} and contains the point x_{0}. We want to make sure our set is not just a T -basic open set, so intersect the basic set with p[T] if necessary. Since x_{0} ∈ p[T_{r_{0}^{i}}], 0 ≤ i ≤ m_{0}, there exists an η_{0}^{i} ∈ κ^{ω} such that (x_{0}, η_{0}^{i}) ∈ T_{r_{0}^{i}}. Set s_{0} = x_{0}|1, t_{0}^{0,i} = η_{0}^{i}|1. Then (x_{0}, η_{0}^{i}) ∈ (T_{r_{0}^{i}})_{(s_{0},t_{0}^{0,i})} (Of course this operation really only applies here if r_{0}^{i} is the empty sequence). Let II play

V_{0} = p[(T_{r_{0}^{0}})_{(s_{0},t_{0}^{0,0})}] ∩ … ∩ p[(T_{r_{0}^{m_{0}}} )_{(s_{0},t_{0}^{0,m_{0}})}] ∩ N_{u_{0}}

Let player I’s answer be (x_{1}, U_{1}) with x_{1} ∈ U_{1} ⊆ V_{0}. Since x_{1} ∈ V_{0} there exists for 0 ≤ i ≤ m_{0} an \overline{η}_{0}^{i} ∈ κ^{ω} such that (x_{1}, \overline{η}_{0}^{i}) ∈ (T_{r_{0}^{i}})_{(s_{0},t_{0}^{0,i})}. Set s_{1} = x_{1}|2, t_{1}^{0,i} = \overline{η}_{0}^{i}|2. Then s_{0} ⊆ s_{1}, t_{0}^{0,i} ⊆ t_{1}^{0,i} and x_{1} ∈

p[(T_{r_{0}^{i}})_{(s_{1},t_{0}^{0,i})}]. Choose now again a basic set p[T_{r_{1}^{0}}] ∩ p[T_{r_{1}^{1}}] ∩ … ∩ p[T_{r_{1}^{m_{1}}} ] ∩ N_{u_{1}} such that this is a subset of U_{1} and contains x_{1}. Let η_{1}^{i} be in κ^{ω} for 0 ≤ i ≤ m_{1} such that (x_{1}, η_{1}^{i}) ∈ T_{r_{1}^{i}}. Set t_{0}^{1,i} = η_{1}^{i}|1. Then x_{1} ∈ p[(T_{r_{1}^{i}})_{(s_{0},t_{0}^{1,m_{1}})}]. In particular x_{1} ∈ p[(T_{r_{1}^{0}})_{(s_{0},t_{0}^{1,0})}] ∩ … ∩ p[(T_{r_{1}^{m_{1}}} )_{(s_{0},t_{0}^{1,m_{1}})}] ∩ N_{u_{1}}. Now

V_{1} = p[(T_{r_{0}^{0}})_{(s_{1},t_{1}^{0,1})}] ∩ … ∩ p[(T_{r_{0}^{m_{0}}} )_{(s_{0},t_{0}^{0,m_{0}})}] ∩ N_{u_{0}} ∩ p[(T_{r_{1}^{0}})_{(s_{0},t_{0}^{1,0})}] ∩ …
∩p[(T_{r_{1}^{m_{1}}} )_{(s_{0},t_{0}^{1,m_{1}})}] ∩ N_{u_{1}}

is a legal move for player II. Proceeding this way, when I plays (x_{0}, U_{0}), (x_{1}, U_{1}), … II produces V_{0}, V_{1}, … with U_{0} ⊇ V_{0} ⊇ U_{1} ⊇ V_{1} ⊇ …, x_{n} ∈ V_{n} and moreover one defines for each n basic sets A_{n} with x_{n} ∈ A_{n} = p[T_{r_{n}^{0}}] ∩ … ∩ p[T_{r_{n}^{mn}}] ∩ N_{u_{n}} ⊆ U_{n} and sequences s_{0} ⊆ s_{1} ⊆ s_{2} …, t_{o}^{n,0} ⊆ t_{1}^{n,0} ⊆ t_{2}^{n,0} …, …, t_{0}^{n,m_{n}} ⊆ t_{1}^{n,m_{n}} ⊆ t_{2}^{n,m_{n}} ⊆ … with (s_{k}, t_{k}^{n,i}) ∈ T_{r_{n}^{i}}, 0 ≤ i ≤ m_{n}, such that for each k the finite sequences s_{k}, t_{k}^{n,i} have length k + 1 and

V_{k} = p[(T_{r_{0}^{0}})_{(s_{k},t_{k}^{0,0})}] ∩ … ∩ p[(T_{r_{0}^{m_{0}}} )] ∩ N_{u_{0}}
∩ p[(T_{r_{1}^{0}})_{(s_{k−1},t_{k−1}^{1,0})}] ∩ … ∩ p[(T_{r_{1}^{m_{1}}} )_{(s_{k−1},t_{k−1}^{1,m_{1}})}] ∩ N_{u_{1}}
∩ …
∩ p[(T_{r_{k}^{0}})_{(s_{0},t_{0}^{k,0})}] ∩ … ∩ p[(T_{r_{k}^{m_{k}}})_{(s_{0},t_{0}^{k,m_{k}})}] ∩ N_{u_{k}}

Let x = ⋃_{k} s_{k}. Claim: x ∈ ⋂_{n} A_{n} = ⋂_{n} V_{n} Proof: Consider A_{n} = p[T_{r_{n}^{0}}] ∩ … ∩ p[T_{r_{n}^{mn}}] ∩ N_{u_{n}}. Let 0 ≤ i ≤ m_{n}. Let η_{n}^{i} = ⋃_{k} t_{k}^{n,i}. We have (s_{k}, t_{k}^{n,i}) ∈ T_{r_{n}^{i}} for all k. Therefore (x, η_{n}^{i}) ∈ [T_{r_{n}^{i}}] and thus x ∈ p[T_{r_{n}^{i}}]. It remains to show that x ∈ N_{u_{n}}. For this let S be the full tree on ω^{ω}. For a sequence s ∈ ω^{<ω} we have [S_{s}] = N_{s} our basic set in the Baire space. It suffices to show now that s_{k} ∈ S_{u_{n}} for every k. Note that s_{n} = x_{n}|n+1. Since x_{n} ∈ N_{u_{n}} we have s_{n} and u_{n} compatible, therefore s_{k} ∈ S_{u_{n}} for k ≤ n. Assume towards a contradiction that for k > n the sequence s_{k} is not in S_{u_{n}}. This implies x_{k} ∉ N_{u_{n}}. In particular, x_{k} ∉ V_{n}, but x_{k} ∈ V_{k} ⊆ V_{n}, a contradiction.

3 Characterization of projective sets by finer topologies

In this chapter we want to give a sketch of the proof for the converse of Theorem 2.3. Unfortunately we are not able to prove this under the same assumptions as Theorem 2.3. Instead of PD we need the much stronger axiom AD_{ℝ}. We put down the theorem:

Theorem 3.1 (ZF+DC+AD_{ℝ}). Let (X, T ) be a Polish space, A ⊆ X and n ∈ ω. If there exists a topology t on A such that

• t is finer that T

• t is strong Choquet

• t has a basis of length less than δ_{n}^{1}

then A is a Σ_{n}^{1} subset of (X, T ).

3.1 Sketch of the proof

Fix now an n ∈ ω. Note first that under AD the projective ordinals are successor cardinals, see for this [Kech78]. So in the spirit of our theorem we are interested in the predecessors of the δ_{n}^{1}. Let κ be the predecessor of δ_{n}^{1}. Second we claim that there exists a Δ_{n}^{1}-scale (φ_{i})_{i∈ω} on a Δ_{n}^{1} subset W ⊆ N such that φ_{0} is a surjection on κ and the length of all other norms of the scale is less or equal κ. We code elements from κ by φ_{0}. Let now X, T , A, t be given. Let B = {B_{i} | i ∈ ω} be a basis for (X, T ).

Let d be a complete metric compatible with T . Let C = {C_{ξ} | ξ < κ} be a basis for (A, t). Let σ be a winning strategy for player II in the strong Choquet game G_{(A,t)}^{s} that plays elements of C with diameter less than \frac{1}{i} in the i-th move. Let W ⊆ N be a Δ_{n}^{1} set, φ_{i} : W ⟶ κ a Δ_{n}^{1} scale on W such that im(φ_{0}) = κ.

Definition 3.1.1. A countable subset S ⊆ κ is called ξ-honest, if ξ ∈ S and there exists an w ∈ W such that φ_{0}(w) = ξ and φ_{i}(w) ∈ S for all i ≥ 1. S is called honest, if it is ξ-honest for all ξ ∈ S.

We need now the following Lemma. A proof can be found in [KW80].

Lemma 3.1.2. There exists a Lipschitz function F : κ^{ω} ⟶ ω^{ω} such that

(a) ∀f ∈ κ^{ω} ∀i (F (f))_{i} ∈ W

(b) ∀f ∈ κ^{ω} : {f(0), f(1), … } is honest ⇒ ∀i ∈ ω φ_{0}((F (f))_{i}) = f(i)

Fix such an F.

A game G

We define now a game G in the following way:

I  α_{0}, ξ_{0}, x_{0}  α_{1}, ξ_{1}, x_{1}  …
II  β_{0}, η_{0}  β_{1}, η_{1} …

where α_{i}, ξ_{i}, β_{i}, η_{i} ∈ κ and x_{i} ∈ A for i ∈ ω. The players must obey the following rule R: The players must play so that the finite initial segments of (∗)

I  (C_{ξ_{0}}, x_{0})  (C_{ξ_{1}}, x_{1})  …
II  C_{η_{0}}  C_{η_{1}} …

are legal moves in the strong Choquet game for (A, t) with diam(C_{η_{i}}) < \frac{1}{i}. The first player to fail loses. The payoff set is the following: Let us assume no player violates the rule. Let f = (α_{0}, ξ_{0}, β_{0}, η_{0}, α_{1}, ξ_{1}, β_{1}, η_{1}, …) ∈ κ^{ω}. Set η̂_{i}(f) = φ_{0}((F (f))_{4i+3}). Let P = {f ∈ κ^{ω} | ∃x ∈ A∀m ∈ ωx ∈ Cη̂_{m}(f)}. II wins the round of the game ⇔ f ∈ P. Remark: The definition of F implies that if f is honest, then

η̂_{i}(f) = φ_{0}((F (f))_{4i+3} = f(4i + 3) = η_{i}

Hence if f is honest, then f ∈ P ⇔ II wins the round of the strong Choquet game (∗)

Lemma 3.1.3. II has a winning quasi-strategy τ for G with the following property: Suppose s = ((α_{0}, ξ_{0}, x_{0}), (β_{0}, η_{0}), …, α_{m−1}, ξ_{m−1}, x_{m−1})) and s′ = ((α_{0}, ξ_{0}, x′_{0}), (β_{0}, η_{0}), …, α_{m−1}, ξ_{m−1}, x′_{m−1})) are two positions in G which are legal and consistent with τ. Let (β_{m−1}, η_{m−1}) ∈ κ × κ. If x_{m−1}, x′_{m−1} ∈ C_{η_{m−1}} and s^{∧}(β_{m−1}, η_{m−1}) is consistent with τ, then s′^{∧}(β_{m−1}, η_{m−1}) is also consistent with τ.

The tree T_{0} Let τ be a winning strategy for II as in the above Lemma 3.1.3. τ is essentially a tree on (κ×κ×A×κ×κ). Let us call this tree T_{0} and we assume all positions in T_{0} legal, that is if I loses by violating the rule (∗) we remove this finite branch from the tree.

The tree T_{1} Define a tree T_{1} on κ^{4} by

((α_{0}, ξ_{0}, β_{0}, η_{0}), …, (α_{m−1}, ξ_{m−1}, β_{m−1}, η_{m−1})) ∈ T_{1}
⇔ ∃x_{0}, … x_{m−1} ∈ A such that
((α_{0}, ξ_{0}, x_{0}, β_{0}, η_{0}), …, (α_{m−1}, ξ_{m−1}, x_{m−1}, β_{m−1}, η_{m−1})) ∈ T_{0}

Lemma 3.1.4. Let f = (α_{0}, ξ_{0}, β_{0}, η_{0}, α_{1}, ξ_{1}, β_{1}, η_{1}, …) be an infinite branch in T_{1}. If f is honest, then ⋂_{i} C_{η_{i}} contains a point x_{f} ∈ A.

The tree T Let finally T be the following tree on ω × κ^{4}: ((i_{0}, α_{0}, ξ_{0}, β_{0}, η_{0}), …, (i_{m−1}, α_{m−1}, ξ_{m−1}, β_{m−1}, η_{m−1})) ∈ T ⇔

(i) For all k diam(B_{i_{k}}) < \frac{1}{k}

(ii) For all k \overline{B_{i_{k+1}}} ⊆ B_{i_{k}}

(iii) ((α_{0}, ξ_{0}, β_{0}, η_{0}), …, (α_{m−1}, ξ_{m−1}, β_{m−1}, η_{m−1})) ∈ T_{1}

(iv) For all k B_{i_{k}} ∩ C_{η_{k}} ≠ ∅

The set A′ Define A′ ⊆ X by

x ∈ A′ ⇔ ∃y ∈ ω^{ω}∃\overline{α}, \overline{ξ}, \overline{β}, \overline{η} ∈ κ^{ω}

[(y, \overline{α}, \overline{ξ}, \overline{β}, \overline{η}) ∈ T and x ∈ ⋂_{m} B_{y(m)}

and \{\overline{α}(m), \overline{ξ}(m), \overline{β}(m), \overline{η}(m) | m ∈ ω\} is honest ]

We claim that this set A′ is Σ_{n}^{1}. To see this we use the powerful Coding Lemma from Moschovakis that assures that T is Δ_{n}^{1}-in-the-codes.

Now we finish the proof by showing that A = A′.

A′ ⊆ A Let x ∈ A′. Let y, \overline{α}, \overline{ξ}, \overline{β}, \overline{η} witness that x ∈ A′.

Let f = (α_{0}, ξ_{0}, β_{0}, η_{0}, α_{1}, ξ_{1}, β_{0}, η_{0}, …). Then (y, f) ∈ [T] and f is honest. By definition of T we have f ∈ [T_{1}]. So by Lemma 3.1.4 there exists an x_{f} ∈ A such that x_{f} ∈ ⋂_{m} C_{η_{m}}. Since x is the only point in ⋂_{m} B_{y(m)} (by (i) of the definition of T) it suffices to show that x_{f} ∈ ⋂_{m} B_{y(m)} because then x = x_{f} ∈ A. Claim: x_{f} ∈ ⋂_{m} B_{y(m)} Proof: Assume not. So there is an m ∈ ω with x_{f} ∉ B_{y(m)}. Therefore d(B_{y(m)}, x_{f}) > 0, let us say d(B_{y(m)}, x_{f}) = ε > 0. But now there exists an k > m with diam(C_{η_{k}}) < \frac{ε}{4} (because of the rule R in the definition of G) and x_{f} ∈ C_{η_{k}}. Also diam(B_{y(k)}) < \frac{ε}{4} by (i) of the definition of T and B_{y(k)} ⊆ B_{y(m)}. By (iv) of the definition there is an z ∈ B_{y(k)} ∩ C_{η_{k}}. Since z, x_{f} ∈ C_{η_{k}} we have d(z, x_{f}) < \frac{ε}{4}. But also z ∈ B_{y(k)} ⊆ B_{y(m)} and hence d(B_{y(m)}, x_{f}) = inf\{d(z′, x_{f}) | z′ ∈ B_{y(m)}\} ≤ d(z, x_{f}) < \frac{ε}{4}. This contradicts d(B_{y(m)}, x_{f}) > ε.

A ⊆ A′ Let x ∈ A. Let h : ω^{ω} ↠ A be a coding of A by the reals. Let Z be a countable subset of ω^{ω} such that

1. there is an \overline{x} ∈ Z with h(\overline{x}) = x

2. the set of ordinals less than κ with codes in Z is honest

3. ∃ν < κ such that x ∈ C_{ν} and ν has a code in Z

4. every position in G consistent with τ with all moves from elements coded from Z has an extension consistent with τ and with all moves from elements coded from Z

To prove the existence of such a set we define by recursion countable sets Z_{i} for i ∈ ω (using AC_{ω} in every other step of the construction) and the take Z to be the union of all Z_{i}. To make an easy thing not look to complicated (by jumping back and forth between the “coded game” and G) note that if there is a countable set of ordinals less than κ one can get by AC_{ω} a countable subset of W coding these ordinals. Simultaneously one can get for a countable subset of A a countable set of reals coding the elements of the subset throug h. Let \overline{x} ∈ ω^{ω} such that h(\overline{x}) = x ∈ A and let \overline{y} ∈ W such that φ_{0}(\overline{y}) = ν and x ∈ C_{ν}. Set Z_{0} = \{\overline{x}, \overline{y}\}. Let now Z_{i} countable be given for an i ∈ ω. We want to define Z_{i+1}. Let T_{0} ↾Z_{i} be the tree on κ × κ × A × κ × κ restricted to elements coded by Z_{i}. Consider in T_{0} ↾ Z_{i} the countable set of all finite sequences s ∈ T_{0} ↾ Z_{i} which have no proper extension. Let s be such a finite sequence and let s^{∧} = {(α, ξ, x, β, η) ∈ κ × κ × A × κ × κ | s^{∧}(α, ξ, x, β, η) ∈ T_{0}}. By AC_{ω} we find a countable set Z̃_{i+1} such that for all such s there is a proper extension of s from s^{∧} in Z̃_{i+1}. Again by AC_{ω} and the above remark there is a countable set Z′_{i+1} of reals coding these elements. Choose further codes for the ordinals φ_{k}(\overline{z}) for \overline{z} ∈ Z_{i} ∩ W, k ∈ ω and let M_{i} be the set of these codes. Then let Z_{i+1} = Z_{i} ∪ M_{i} ∪ Z′_{i+1}. Z_{i+1} is countable. Set Z = ⋃_{i∈ω} Z_{i}. By definition of Z_{0} 1. and 3. are satisfied. If \overline{w} ∈ Z_{i} for some i ∈ ω then for φ_{k}(\overline{w}) there is a code in M_{i} ⊆ Z_{i+1} for all k ∈ ω. Hence the set of ordinals less than κ with codes in Z is honest. If s is a position in G consistent with τ and all moves are in Z then there is an i ∈ ω such that s ∈ T_{0} ↾ Z_{i} and there is an extension so s ∈ T_{0} ↾Z_{i+1} ⊆ T_{0} ↾Z. So this extension is also consistent with τ. Using such an Z there is a round of the game G such that

(i) all moves are in Z (that is, coded by Z)

(ii) the round is consistent with II’s winning quasi-strategy τ for G

(iii) x_{m} = x for all m (so player I always plays the same element x ∈ A)

(iv) ξ_{0} = ν, ξ_{m+1} = η_{m+1}

(v) α_{0}, α_{1}, … is an enumeration of the ordinals less than κ with codes in Z

Since such a round g = (α_{0}, ξ_{0}, x, β_{0}, η_{0}, α_{1}, ξ_{1}, x, β_{1}, η_{1}, …) of G is consistent with τ, we know that g is an infinite branch in T_{0}. By definition of T_{1} the sequence f = (α_{0}, ξ_{0}, β_{0}, η_{0}, α_{1}, ξ_{1}, β_{1}, η_{1}, …) is an infinite branch through T_{1}. Using this f we want to get an infinite branch in T. Property (iii) in the definition of T is allready satisfied. Now let i_{0}, i_{1}, … be such that x ∈ ⋂_{m} B_{i_{m}} and (i) and (ii) of the definition of T holds. Since II wins the round g of G all C_{η_{m}} are legal moves of II and therefore are moves in the strong Choquet game. So x ∈ C_{η_{m}} for all m. This implies B_{i_{m}} ∩ C_{η_{m}} ≠ ∅ for all m and therefore (iv) in the definition of T holds. So (i_{0}, α_{0}, ξ_{0}, β_{0}, η_{0}, i_{1}, α_{1}, ξ_{1}, β_{1}, η_{1}, …) ∈ [T]. To show now that x ∈ A′ it remains to show {α_{m}, ξ_{m}, β_{m}, η_{m} | m ∈ ω} is honest. But all this elements were chosen in Z and the α_{m} are all ordinals less than κ coded by Z and this set is honest by the construction of Z.

References

[Be1] Howard Becker, Finer topologies on pointsets in Polish spaces, unpublished notes, 1991.

[Be2] Howard Becker, Playing around with finer topologies, unpublished notes, 1992.

[HKL90] Leo A. Harrington, Alexander S. Kechris, Alain Louveau, A Glimm-Effros dichotomy for Borel equivalence relations, Journal AMS 3, 1990, p. 903-928

[Kech78] Alexander S. Kechris, AD and projective ordinals, IN: Kechris, Moschovakis (eds.): Cabal Seminar 1976-1977, Lecture notes in mathematics 689, Springer Verlag 1978.

[Kech95] Alexander S. Kechris, Classical descriptice set theory, Graduate texts in mathematics, Vol. 156, Springer 1995.

[KW80] Alexander S. Kechris, W. Hugh Woodin, Generic codes for uncountable ordinals, partition properties and elementary embeddings, unpublished notes, 1980.