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In the proof that there is a payoff set $X$ such that the Gale-Stewart game is not determined(see here, Proposition 3.1.). enter image description here

I don't know why $X$, the set of all outcomes generated by a fixed strategy of one player, constitutes a perfect set. I can see it's true, if the fixed strategy is a constant function. The trouble for me is why it's true in general.

Martin Sleziak
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Metta World Peace
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1 Answers1

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Suppose that Player I uses the strategy $\sigma:\bigcup_{n\in\omega}{}^{2n}\{0,1\}\to\{0,1\}$. A game $\alpha=\langle \alpha_k:k\in\omega\rangle$ is possible iff $\alpha_{2k}=\sigma(\alpha\upharpoonright 2k)$ for each $k\in\omega$. Let $A$ be the set of possible games.

Clearly $A\ne\varnothing$. Suppose that $\alpha=\langle \alpha_k:k\in\omega\rangle\in A$. For $n\in\omega$ let $$B_n(\alpha)=\left\{\beta\in{}^\omega\{0,1\}:\beta\upharpoonright 2n=\alpha\upharpoonright 2n\right\}\;,$$ and let $$V_n(\alpha)=\{\beta\in B_n(\alpha):\beta_{2n+1}\ne\alpha_{2n+1}\}\;.$$

Then $V_n(\alpha)\cap A\ne\varnothing$, so choose $\beta^n\in V_n(\alpha)\cap A$. Clearly $\beta^n\in B_n(\alpha)\setminus\{\alpha\}$, and $\{B_n(\alpha):n\in\omega\}$ is a nbhd base at $\alpha$, so $\left\langle\beta^n:n\in\omega\right\rangle$ is a sequence in $A\setminus\{\alpha\}$ converging to $\alpha$, and $\alpha$ is not an isolated point of $A$. This shows that $A$ is dense-in-itself.

Finally, suppose that $\left\langle\alpha^n:n\in\omega\right\rangle$ is a convergent sequence in $A$ with limit $\alpha\in{}^\omega\{0,1\}$. Then for each $n\in\omega$ there is an $m\in\omega$ such that $\alpha^k\upharpoonright n=\alpha\upharpoonright n$ for each $k\ge m$; it follows that $\alpha_{2k}=\sigma(\alpha\upharpoonright 2k)$ for each $k\in\omega$ and hence that $\alpha\in A$, so $A$ is closed.

The argument for the case in which Player II’s strategy is fixed is entirely similar.

Brian M. Scott
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  • Perhaps mentioning that the set $A$ is just a copy of Cantor's set may further clarify the matter. – Andrés E. Caicedo Apr 05 '13 at 00:20
  • Thank you as always,sir. Hope you don't mind a quesion about notation. What is the difference between ${}^\omega{0,1}$ and ${0,1}^{\omega}$? – Metta World Peace Apr 05 '13 at 03:14
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    @MettaWorldPeace: You’re welcome. As for the notation, like some others with a set-theoretic background, I tend to use ${}^AB$ for the set of functions from $A$ to $B$. It didn’t really matter here, but had I written $2$ instead of ${0,1}$, it would have avoided a potential ambiguity: $2^\omega$ is the cardinal number, ${}^\omega2$ the set of functions. – Brian M. Scott Apr 05 '13 at 03:54