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Suppose, in $G_{\omega}(\mathbb{R})$, a player's two strategies are equivalent, if, for any strategy of his opponent, the outcome incurred are the same. It can be shown that in $G_{\omega}(\omega)$ and $G_{\omega}(2)$, the cardinalities of partition of strategy spaces derived from equivalence relation are $2^{\aleph_0}$, since cardinality of set of strategies that are constant functions(which are not equivalent)and that of strategy space are both $2^{\aleph_0}$. But in $G_{\omega}(\mathbb{R})$,the cardinalities of two corresponding sets are $2^{\aleph_0}$ and $2^{2^{\aleph_0}}$.


(Apologize for just copy and paste some content from the question I asked before)


Formally, a two-player game $$G(\omega,(X_n)_{n \in\omega},(Y_n)_{n \in\omega},)$$ where

  • $\omega$ is the number of the moves of each of the two players I and II.
  • $(X_n)_{n \in \omega}$ (or $(Y_n)_{n \in \omega} $) is a $\omega$-sequence of $X_n$(or $Y_n$), which is the action space of $n$th move of first (second) player.
  • The game is played in an alternating fashion.An outcome is a $\omega$-sequence:$$a_0 \in X_0, a_1 \in Y_0, a_2 \in X_1, a_3 \in Y_1, a_4 \in X_2……$$
  • A player moves at any stage contingent on history. So a strategy for player I take the form as s sequence of functions $\{\sigma_i\}_{i \in \omega}$: $$\sigma_i : \prod_{j<i}{X_j \times Y_j} \to X_i$$ Player II's strategy's form $\{\tau_i\}_{i \in \omega}$is defined similarly. Player I and Player II's strategy spaces are denoted as $S_{\text I}$ and $S_{\text {II}}$ respectively. Denote the binary operation $\star$ as the function that sends a pair of strategies of player I and player II to the outcome they give rise to.
  • Player I and Player II's strategy spaces are denoted as $S_{\text I}$ and $S_{\text {II}}$ respectively. Denote the binary operation $\star$ as the function that sends $\sigma$ and $\tau$, a pair of strategies of player I and player II to the outcome $a$ they give rise to$$\sigma \star \tau \mapsto a$$
  • Strategy $\sigma$ and $\sigma'$ for player I are equivalent, if for any player II's strategy $\tau$, $$\sigma \star \tau = \sigma' \star \tau$$ The equivalence class $[\sigma]$ is denoted as$\{\sigma' \in S_{\text{I}}:\forall \tau \in S_{\text{II}}(\sigma \star \tau = \sigma' \star \tau)\}$, and the corresponding partition of $S_{\text{I}}$ as $S_{\text{I}}'$. We define the equivalence relation for Player II in the same way.

What is cardinality of $S_{\text{I}}'$, when $X_n = Y_n = \mathbb{R}$?

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Here are $2^{2^{\aleph_0}}$ pairwise inequivalent strategies for player II (and it's easy to modify them to work for player I instead). Fix two distinct reals $x$ and $y$. For each set of reals $A$, let $\sigma_A$ be the strategy for player II that outputs $x$ on the first turn if Player I's first move is in $A$, and $y$ otherwise. On each subsequent turn we let $\sigma_A$ output $x$.

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  • $\begingroup$ Aleph and alpha are both he first letters of their respective alphabets, and they represent the same sounds. But they are not the same, more so mathematically. :-) $\endgroup$
    – Asaf Karagila
    Feb 15, 2013 at 4:01
  • $\begingroup$ @Asaf Oops, I always do that. Thanks for pointing it out. $\endgroup$ Feb 15, 2013 at 4:08
  • $\begingroup$ @Asaf I need a keyboard with $\aleph$ on it so I don't have to type something so similar to "alpha". $\endgroup$ Feb 15, 2013 at 4:09
  • $\begingroup$ Oh, you still have to type \aleph. LaTeX and Hebrew aren't very good friends... :-) $\endgroup$
    – Asaf Karagila
    Feb 15, 2013 at 4:32

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