I think my question applies to most games, but for the sake of concreteness, I shall consider one specific game in this question. We consider the game posed by Ilias Kastanas in his paper *On the Ramsey property for sets of reals*: We first fix some $X \subseteq [\omega]^\omega$.


1. Player I start by choosing an infinite subset $A\_0 \subseteq \omega$.
2. Player II responds by choosing $x\_0 \in A\_0$, and then some $B\_0 \subseteq A\_0$ with $x\_0 < \min(B\_0)$.
3. Player I responds by choosing some $A\_1 \subseteq B\_0$.
4. Player II responds by choosing $x\_1 \in A\_1$, and then some $B\_1 \subseteq A\_1$ with $x\_1 < \min(B\_1)$.
5. Etc.


Player I wins iff $\{x\_0,x\_1,\dots\} \in X$.


Assume that Player I has a winning strategy $\sigma$. If $(A\_0,x\_0,B\_0,\dots,A\_n,x\_n,B\_n)$ is a partial play, then clearly the only things that determine what Player I should respond with are the finite sequence $(x\_0,\dots,x\_n)$ and $B\_n$, Player II's last played set. However, it appears to me that we cannot assume that $\sigma$ satisfies this property. Let's say that $\sigma$ is *uniform* if $\sigma(A\_0,x\_0,B\_0,\dots,A\_n,x\_n,B\_n)$ only depends on $x\_0,x\_1,\dots,x\_n,B\_n$.



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> If Player I has a winning strategy, does Player I necessarily have a uniform strategy?
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I'm also interested in how much the axiom of choice plays a part in the above statement.



Uniform strategy on Kastanas' game
This is a great question — definitely enjoyed.


Assuming the axiom of choice, then the answer is yes.


**Theorem.** Assume there is a well ordering of the real numbers. If player I has a winning strategy, then there is a uniform winning strategy.


**Proof.** Suppose that $\sigma$ is a winning strategy for player I. Fix a well ordering of the collection of sets $B\subseteq\omega$.


Now, suppose that we are playing the game, faced with the actual position $(A\_0, x\_0, B\_0, \ldots,A\_n,x\_n,B\_n)$. But let us look now only at the allowed information for a uniform strategy $(x\_0,\ldots,x\_n,B\_n)$. Consider the least with respect to the well ordering subset $B\_n'\subseteq B\_n$, such that there is an imaginary play of $B\_i'$ that accords with $\sigma$ and the data $(x\_0,\ldots,x\_n,B\_n')$, such that furthermore each $B\_i'\subseteq A\_i'$ is chosen as least at that stage, subject to being compatible with the data up to $n$. Let $\sigma$ respond to this imaginary play to give us a set $A\_{n+1}$, which we now play as the result for this new strategy.


Since the play depended only on $(x\_0,\ldots,x\_n,B\_n)$, I have described a uniform strategy. Let me prove that it is winning. What I claim is that every sequence $x\_0,x\_1,x\_2,\ldots$ arising from a play according to this uniform strategy can be realized as arising in a play according to $\sigma$. The key point is that as $n$ increases, the imaginary sets $B\_i'$ will eventually stabilize for each $i$. To see this, consider what happens at stage $n+1$, after we've played $A\_{n+1}$. The actual play will respond with some $x\_{n+1}\in A\_{n+1}$ and $B\_{n+1}\subseteq A\_{n+1}$, and we will forget about the actual $B\_n$ and find some minimal $B\_{n+1}''\subset B\_{n+1}$ and new imaginary sets $B\_i''$ that produce a play according with $\sigma$ and the data $(x\_0,\ldots,x\_{n+1},B\_{n+1}'')$, and is minimal at each stage $i$. But the point now is that the requirement on $B\_i''$ is a bit lighter than that on the imaginary set $B\_i'$ chosen at stage $n$, provided that $x\_{n+1}$ is in $B\_n$, which it is. Earlier, we needed $B\_i'$ to contain $B\_n'$, but now we only need it to contain $B\_{n+1}''$, which is strictly smaller. So the set $B\_i''$ will be either equal to $B\_i'$ or preceed it in the well order. But it can only go down in the well order finitely often, and so eventually, for any given $i$, the imaginary versions of $B\_i'''$ will stabilize. In other words, if we have an infinite play according to the uniform strategy, we can find choices of $B\_i'''$ that give rise to the whole play. And since this imaginary play accords with $\sigma$, the set $\{x\_0,x\_1,\ldots\}$ will be in $X$, as desired. $\Box$


