Question on constructing an auxiliary game of a closed game

Question

I'm trying to understand Lemma 20.7, page 142, Classical Descriptive Set Theory(Kechris) by working on a concrete example.

Lemma 20.7:Let $T$ be a non-empty pruned tree, and let $X \subseteq [T]$ be closed. there is a $k$-covering of $T$ that unravels $X$.

Question: Suppose, given a closed game, $G(T, X)$, with $T = 2 ^{\omega}, X = \{x \in 2^{\omega}: \text{player I doesn't play } 1 \text{from} \{0,1\}\text{before player II }\}$. What is the auxiliary game which is denoted as $G(\hat{T}, \hat{X})$in which $\hat X$ is clopen in $[\hat{T}]$? (For simplicity we only consider the case of $k = 0$ for $k$-covering.)

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Here's my attempt.At the beginning, player I should choose $(x_0, \Sigma_{\text{I}})$. If I understand correctly, there's no requirement on $x_0$ and the quasistrategy $\Sigma_{\text{I}}$ in $T_{(x_0)}$. Thus let $\Sigma_{\text{I}}= T_{2^{\omega}}$. Next, player II faces two options. One is $(x_1, u)$. Let $u = (x_2,x_3)=(1,0)$ The other is $(x_1, \Sigma_{\text{II}})$. Let the quasistrategy $\Sigma_{\text{II}}$ be $(T_X)_{(x_0,x_1)}$.

Thus we have $\hat T$ consist of all finite sequences of the form for all $l \geq 4 $: $$(x_0,x_1,1,0,x_5, \ldots, x_l)$$

or $$(x_0,x_1,x_3,x_4,x_5, \ldots, x_l) \in T_X$$

We have $\hat X = X$ and $[\hat T]\setminus \hat{X}$ is clopen, so $\hat X$ is clopen in $[\hat T]$. Is this construction correct?

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Definitions and the proof of Lemma 20.7:

$T$, as a a non-empty pruned tree, is where legal moves of both players are defined. A covering of $T$ is $(\hat {T}, \pi, \psi)$. where:

1. $\hat T$ is a a non-empty pruned tree.
2. $\pi: \hat {T} \to T$ is monotone with length($\pi(s)$) = length$(s)$, which induces a continuous function $\pi： [\hat {T}] \to [T]$.
3. $\psi$ maps the strategy space of player I(player II) in $\hat T$ into her strategy space in $T$ in such a way that $\psi(\hat\sigma | k) = \psi(\hat \sigma)| k$ for all $k$,and for all $m < n$, $\psi (\hat \sigma | m) = \psi (\hat \sigma | n)| m$.
4. If $\hat \sigma$ is one of player I(player II)'s strategy in $\hat T$, and $x \in [\psi(\hat \sigma)]$, then there exist $\hat x \in [\hat \sigma]$ such that $\pi(\hat x) = x$.

We say a covering $(\hat {T}, \pi, \psi)$ is a $k$-covering, if it is a covering such that $T|2k = \hat{T}|2k$ and $\pi|(\hat{T}|2k)$ is the identity.

We say a covering $(\hat {T}, \pi, \psi)$ unravels the payoff set $X \subseteq T$, iff $\pi^{-1}(X) = \hat X$ is clopen in $[T]$.

Lemma 20.7: Let $T$ be a non-empty pruned tree, and let $X \subseteq [T]$ be closed. For each $k \in \Bbb N$, there is a $k$-covering of $T$ that unravels $X$.

Added: This proof is incomplete, since the rest of it is devoted to the construction of $\psi$ which seems not to be that relevant to the question.

Answer 1

Dear Metta World Peace,

Very nice theorem that you are trying to understand! Nice and quite difficult actually...^^

The general idea is that you want to turn an infinite game to a finite one (i.e. clopen), and to be able to transfer strategies from the second one to the first. And you do that by complicating your alphabet.

In your example, do not forget that the quasi-strategies are not fixed! It's a move of I to choose $\Sigma_I$ amongst all the quasi-strategies he has. So of course $\Sigma_I$ could be $T_{2^{\omega}}$, but it could also be "playing only $1$'s", "playing only $0$'s until II plays three $1$'s in a row", or anything else. All those quasi-strategies have to appear in $\hat{T}$ - and the same with all the compatible quasi-strategie for II afterwards. The quasi-strategies are now in some sense part of your alphabet, and $\hat{T}$ is not a tree on $2^{\omega}$ but on $(2 \cup \omega \times Q_I \cup \omega \times 1 \times \omega^{< \omega} \cup \omega \times 2 \times Q_{II})^{\omega}$, where $Q_I$ and $Q_{II}$ are the sets of quasi-strategies respectively for I and II. As you can see, it is quite difficult, even in your "simple" example, to describe "concretely" $\hat{T}$. It is nonetheless quite easy to convince yourself that if $T$ is pruned and non-empty, then also is $\hat{T}$.

The game on $\hat{T}$ is designed to be finite. Even if the two players have to play infinitely many times, we already know who will be the winner after II auxiliary move: if II chose the first option, she will win, otherwise it is I who will.

Now $\pi$ is just the application that extracts the main run from the auxiliary game by forgetting about the auxiliary moves; note that it is continuous. $\hat{X}$ is by definition $\pi^{-1}(X)$, but then $\hat{x} \in \hat{X}$ iff the resulting sequence $\pi(\hat{x})$ is in X, that is iff for all $l \in \omega$, $(x_0, \ldots, x_l) \in T_X$. And that happens exactly when II chooses the second option for her auxiliary move! The set $\pi^{-1}(X)$ can thus be seen as a reunion of basic open sets of $[\hat{T}]$, and is hence open. Moreover it is closed since it is the preimage of a closed set by a continuous function, so that $\hat{X}$ is clopen!

Anyway, IMHO, the tricky and interesting part of the demonstration is afterwards, when you transfer the strategies! Then you really understant why you needed to define the things like that...

I got a bit carried away I'm affraid, hope it helped!

Good luck with the rest of the proof,

K.