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 $X \subseteq [\omega]^\omega$.
- Player I start by choosing an infinite subset $A_0 \subseteq \omega$.
- Player II responds by choosing $x_0 \in A_0$, and then some $B_0 \subseteq A_0$ with $x_0 < \min(B_0)$.
- Player I responds by choosing some $A_1 \subseteq B_0$.
- Player II responds by choosing $x_1 \in A_1$, and then some $B_1 \subseteq A_1$ with $x_1 < \min(B_1)$.
- 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 $x_0,x_1,\dots,x_n,B_n$.
If Player I has a winning strategy, does Player I necessarily have a uniform strategy?
I'm also interested in how much the axiom of choice plays a part in the above statement.
Best Answer
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. $\Box$