Suppose, a two-player game $$G(\omega,(X_n)_{n \in\omega},(Y_n)_{n \in\omega},A)$$ 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:$$a_0 \in X_0, a_1 \in Y_0, a_2 \in X_1, a_3 \in Y_1, a_4 \in X_2……$$ $A$, the winning set, is a subset of the set of all possible outcomes $C$, i.e. $\prod_{i \in \omega}(X_i \times Y_i)$. If the outcome $\{a_n\}_{n<\omega} \in A$, then player II wins, otherwise, player I wins.

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.

Let's fix $X_n = Y_n =Z$, for any $n \in \omega$ and $Z$ be any ordinal number from $2$ to $\omega$. As we know, the cardinality of set of strategies coincides with that of set of all possible outcomes. More precisely, $|S_{\text {I}}| = |S_{\text {II}}| = |C|= 2^{\aleph_0}$

We say two games $\alpha$ and $\beta$ are isomorphic, if there are bijective functions $f:S^{\alpha}_{\text I} \to S^{\beta}_{\text {I}}$, $g:S^{\alpha}_{\text{II}} \to S^{\beta}_{\text{II}}$ and $h:C^{\alpha} \to C^{\beta}$ such that for any pair of strategies of players I and II in game $\alpha$, $\sigma$ and $\tau$, $$f(\sigma) \star g(\tau) = h (\sigma \star \tau)$$

How to show the existence, or non-existence of such isomorphism between games with action space $Z$ of different sizes, say, $Z= 2$ and $Z = \omega$

  • $\begingroup$ Is $\sigma \circ \tau$ the result of playing $\sigma$ against $\tau$? I think this is usually written $\sigma \star \tau$ or $\sigma * \tau$. $\endgroup$ Feb 7, 2013 at 2:26
  • $\begingroup$ @TrevorWilson: You're right.Thank you for pointing that out. $\endgroup$ Feb 7, 2013 at 2:28
  • $\begingroup$ Just to clarify, your notion of isomorphism doesn't involve any payoff sets on either side? $\endgroup$ Feb 7, 2013 at 2:29
  • $\begingroup$ @TrevorWilson:No. $\endgroup$ Feb 7, 2013 at 2:31
  • $\begingroup$ @TrevorWilson: Tons of thanks for your comment. I got it wrong. $\endgroup$ Feb 8, 2013 at 23:55


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