We first show how games can be used to prove perfect set theorems. Suppose A⊆ C. We define a game G∗(A) where at stageiPlayer I playsσi ∈2<ω and
Player II playsji∈ {0,1}. Together they play
x=σ0bj0bσ1bj1σ2bj2. . .∈ C. Player I wins ifx∈Aand Player II wins ifx6∈A.
Proposition 6.12 If Player I has a winning strategy inG∗(A), thenAcontains a perfect set.
Proof Letτ be a winning strategy for Player I. Define f : C →A byf(x) is the play of the game where Player I uses τ and Player II plays x(0), x(1), . . .. In other words
f(x) =τ(∅)bx(0)bτ(x0)bx1bτ(x(0), x(1))bx2bτ(x(0), x(1), x(2), x(3)). . . . Clearly if x|n = y|n, then f(x)|n = f(y)|n. Thus f is continuous. Suppose x6=y andnis least such thatx(n)6=y(n). Let
µ=τ(∅)bx(0)bτ(x(0))bx1. . .bτ(x(0), . . . x(n−1)).
Thenf(x)⊃µbx(n) andf(y)⊃µby(n). Thusf(x)6=f(y). Thusf is continu- ous and one-to-one. Hencef(C) is an uncountable closed subset ofA.
Proposition 6.13 If Player II has a winning strategy in G∗(A), then A is
countable.
Proof Let τ be a winning strategy for Player II. Consider a position p = (σ0, j0, . . . , σn, jn) where Player II has played usingτ and it is Player I’s turn
to play. Supposex∈Aandx⊃µ=σ0bj0, . . . , σnbjn. We say thatx isrejected
at pif for allσn+1, ifx ⊇µbσn+1, then x 6⊇µbσn+1bτ(σ0, . . . , σn+1). In other
words, up to stagep, it looks like it is possible that we will eventually playx, but in fact no matter what Player I does at this stage, Player II will immediately make a play which ensures that we will not eventually playx.
Claim Ifx∈A, there is a positionpsuch thatx is rejected atp.
Suppose not. Consider the following play of the game. Sincexis not rejected at the empty position. There isσ0⊂xsuch thatx⊃σ0bτ(σ0). Player I playsσ0. Letpndenote the position after Player II’snth move and letµnbe the sequence
σ0bτ(σ(0))b. . . τ(σ0, . . . , σn). We assume by induction that x⊃µn. Since x is
not rejected atpn, there isσn+1 ∈2<ω such thatx⊃µnbσn+1bτ(σ0, . . . , σn+1).
Player I plays σn+1. But then the final play of the game is Sµn = x ∈ A,
contradicting the fact thatτ is a winning strategy for Player II.
Claim There is at most onex∈Arejected at p.
Supposex is rejected at p= (σ0, τ(σ0), . . . , σn, τ(σ0, . . . , σn)). Letµ=x|k
be the portion ofxwe have decided by positionp. We claim that knowing only p we can inductively determine the remaining values of x. Suppose we have determinedx(k), . . . , x(m−1). If Player I playsx(k), . . . , x(m−1), the Player II must play 1−x(m+ 1). Thus
x(m) = 1−τ(σ0, . . . , σn,hx(k), . . . , x(m−1)i).
Thus there is a unique element ofArejected atp.
Since every element of Ais rejected at one of the countably many possible positions,Amust be countable.
Corollary 6.14 IfAis uncountable andG∗(A)is determined, thenAcontains
a perfect set.
Exercise 6.15 LetA⊆X. Prove the following without using determinacy. a) If|A| ≤ ℵ0, then Player II has a winning strategy inG∗(A).
b) IfAcontains a perfect set, then Player I has a winning strategy inG∗(A).
We have only proved this for A ⊆ C, but using the fact that any two un- countable standard Borel spaces are Borel isomorphic we see that it is true for any uncountable Polish space.
Corollary 6.16 If PD holds, the any uncountable projective set contains a per- fect subset.
There is a technique of “unfolding” games, that allows us to show that if Det(Σ1
n) holds, then every uncountableΣ1n+1set contains a perfect subset. We will illustrate this idea by giving another proof of the perfect set theorem for
Σ1
1-sets using only the determinacy of closed games. SupposeA⊆ C is Σ1
1. Let B ⊆ C × N such that A={x: ∃y (x, y)∈B}. Consider the gameG∗u(A) where at stageiPlayer I playsσi∈2<ωandy(i)∈N
and Player II responds withji ∈2. Together they play
x=σ0bj0bσ1bj1b. . . and
y= (y(0), y(1), . . .).
Player I wins if (x, y)∈A. By closed determinacy (or more correctly by 6.7), G∗
u(A) is determined.
Lemma 6.17 If Player I has a winning strategy in G∗
u(A), then A contains a
perfect subset.
Proof As in 6.12 ifτ is a winning strategy for Player I, there are continuous functionsf :C → Candg:C → N such that if Player II playsz(0), z(1), z(2), . . . and Player I uses τ, then together they play x=f(z)∈ C and y =g(z)∈ N with (x, y)∈B. As in 6.12f is one-to-one andf(C) is an uncountable closed subset ofA.
Lemma 6.18 If Player II has a winning strategy inG∗
u(A), thenAis countable. Proof Suppose x ∈A. Choose y such that (x, y)∈ B. As in 6.13 there is a positionpat which (x, y) is rejected. Letµ= (x(0), . . . , x(k−1)) be the portion ofxforced byp. If Player I now play (x(k), . . . , x(m−1)) andy(n), then
x(m) = 1−τ(σ0, y(0), . . . , σn−1, y(n−1),hx(0), . . . , x(m−1)i, y(n)i.
Indeed for each possible value ofy(n), there is at most onexrejected atp. Thus the set ofxrejected atpis countable andAis countable.
Lemmas 6.17 and 6.18 together with the determinacy of closed games gives a second proof of the Perfect Set Theorem forΣ1
1.
In§7 we will examine this game again. At that time it will be useful to note that ifxis rejected atp, thenx is recursive inτ.