Consider $\mathrm{AD}\_X$, determinacy for games where players pick moves from $X$. We know that it is consistent for $X = \omega$ or $\mathbb{R}$ (under large cardinal assumptions), but inconsistent for $X = \omega\_1$.


Since this implies determinacy is inconsistent for any set with $\omega\_1 \leq X$, this answers the question of consistency for most cardinalities that we might consider (e.g. $\omega\_1 \cup \mathbb{R}$, $\mathcal{P}(\mathbb{R})$, all uncountable ordinals).
Of course, in the absence of choice other cardinalities might exist. This leads me to the following two questions:



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> **Q1:** Is the existence of infinite Dedekind finite sets consistent with $\mathrm{AD}$?
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> **Q2:** If so, is 'there exists an infinite Dedekind finite set $X$ such that $\mathrm{AD}\_X$ holds' consistent? (Let's call this statement $\mathrm{AD}\_\mathrm{DF}$.)
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(Note that assuming the consistency of $\mathrm{AD\_\mathbb{R}}+\mathrm{DC}$, we have that $\mathrm{AD}\_\mathrm{DF}$ can't be a consequence of either $\mathrm{AD}$ or $\mathrm{AD\_\mathbb{R}}$.)


Games on infinite Dedekind finite sets seem very strange. For example, consider the game where the first player to play a move that's already been played loses. By Dedekind finiteness somebody has to win this game (and in finite time!), but both players always have not-losing moves at every point. I initially thought this game would be nondetermined, but someone pointed out the following to me:


Let $X$ be infinite Dedekind finite. Then $2 \times X$ is as well. Player 2 has a winning strategy for this game on $2 \times X$: given that an odd number of moves have been played, if nobody has lost yet then there must be some $x \in X$ such that one but not both of $(0, x)$ and $(1, x)$ have been played. Thus we play the other one.


(This means that Player 1 has a winning strategy for this game on $2 \times X \setminus (0, x)$, by pretending they are Player 2 in the game on $2 \times X$ and that the first move was $(0, x)$.)



Is determinacy on an infinite Dedekind finite set consistent?
The answer is no, you cannot have determinacy for all games on an infinite Dedekind-finite set. Indeed, one cannot even have clopen determinacy for games on such a set. So the answer to question 2 is negative.


**Theorem.** If $X$ is an infinite Dedekind-finite set, then there is a non-determined clopen game on $X$. 


In particular, $\text{AD}\_X$ fails, and hence also $\text{AD}\_{\text{DF}}$ is false.


**Proof.** Suppose that $X$ is an infinite Dedekind-finite set. Fix some particular element $a\in X$, and consider the following game. Player I plays distinct elements of $X$, while player II responds with the element $a$, until Player I plays $a$, at which time player II must play an element not yet played. If he can do so, he wins, and otherwise player I wins. 


This game must terminate in finitely many moves, because player I cannot play distinct elements forever, as $X$ has no countably infinite subset. So this is a clopen game. 


It is clear that player I cannot have a winning strategy, since that strategy would call for her to play some elements of $X$ and eventually the element $a$, during which time all the while player II does something very easy, just playing the element $a$, until player I plays $a$, at which time player II can defeat the strategy by simply playing some new element. 


But I claim that also player II cannot have a winning strategy. If there were a winning strategy for player II, then we could use that strategy to generate a countable subset of $X$, since it provides a systematic way to extend any particular finite sequence of distinct elements of $X$. Just have player I play that finite sequence, and then $a$, and player II will respond with a fresh element of $X$. Since $X$ has no countably infinite subsets, there can be no such strategy. 


So the game is not determined.
**QED**


This game is related to the classical fact that clopen determinacy implies the axiom of choice, since in the comparatively trivial game, where player I plays a nonempty set from a fixed collection, and player II playing an element of that set (and winning upon doing so), it is clear that the first player cannot have a winning strategy, but a winning strategy for player II provides a choice function.


