Consider the following game, played by two players, 
called Q and A, in a time frame t = 1, 2, ....
At every time point i, Q mentions some $Q\_i \subset \mathbb{R}$,
after which A mentions $A\_i$ such that either $A\_i = Q\_i$ or
$A\_i = Q\_i^c$.


Define $C(i) = \bigcap\_{k \lt i} A\_i$ and
$C(\infty) = \bigcap\_{k \in \mathbb{N}} A\_i$.


Player Q is declared winner if either:


1) $C(i)$ has only one element for some $i < \infty$


or


2) $C(\infty)$ is empty.


(Player A wins in all other cases.)


Question: 
Which player (if any) has a winning strategy?


Edit:
Some additional explanatory remarks:


This question occurred to me while thinking about Brouwer style choice sequences in connection with Borel sets.


Before posting, I was not 100% sure yet, but pretty confident that player A has a winning strategy when Q is restricted to giving Borel sets. This has been confirmed, in the meantime, in one of the answers below. (More generally A has a winning strategy when Q is restricted to sets having the Baire property.)


The general case is still open, though.


Some easy observations that might help to clarify the puzzle:


a) Player Q obviously has a strategy to make sure that $C(\infty)$ has at most one element. But that's not enough to win the game. 


b) Whenever $C(i)$ is countable for some (finite) $i$, player Q can win the game, by (from time $i$ onwards) eliminating the elements of that countable set, one by one.


c) Player A has an easy strategy to make sure that $C(i)$ is uncountable for all (finite) $i$. But that's not enough to describe a winning strategy for player A. For example, when $Q\_i = A\_i = (0, 1/i)$ for every $i$, then $C(i)$ is uncountable for every (finite) $i$, but $C(\infty)$ is empty, and therefore A loses the game.



Game involving 'asking questions about a real'
Neither player has a winning strategy, see below.


However, when the game is restricted so that Q can only play sets with the [Baire property](http://en.wikipedia.org/wiki/Property_of_Baire), then A has a winning strategy. Note that sets with the Baire property form a $\sigma$-algebra which includes all analytic sets, and assuming the axiom of projective determinacy, all projective sets.


In the restricted game, we can write $A\_i=U\_i\vartriangle\bigcup\_jM\_{p(i,j)}$, where $U\_i$ is open, $M\_k$ are nowhere dense, and $p\colon\omega\times\omega\to\omega$ is a bijective pairing function such that $p(i,j)\ge i,j$. The winning strategy for A is then to maintain a chain of nonempty bounded open intervals $I\_n$ so that


* $I\_0\supseteq\overline{I\_1}\supseteq I\_1\supseteq\overline{I\_2}\supseteq I\_2\supseteq\cdots$,
* $I\_i\subseteq U\_i$,
* $I\_i\cap M\_i=\varnothing$.


This can be arranged as follows. Let $Q\_i=V\vartriangle M$ be given, where $V$ is open and $M$ is meager. If $I\_{i-1}\cap V\ne\varnothing$, A chooses $A\_i=Q\_i$, otherwise $A\_i=Q\_i^c$ (note that in the latter case, we will have $I\_{i-1}\subseteq U\_i$). This determines $U\_i$ and $M\_{p(i,j)}$, and it remains to find $I\_i$. Now, $I\_{i-1}\cap U\_i$ is a nonempty open set. Moreover, $M\_i$ is nowhere dense, hence the open set $I\_{i-1}\cap U\_i\smallsetminus\overline{M\_i}$ is still nonempty, and therefore it contains an interval $I\_i$. By shortening it if necessary, we can make sure that $\overline{I\_i}\subseteq I\_{i-1}$.


When the game finishes using this strategy, then $\bigcap\_iI\_i=\bigcap\_i\overline{I\_i}$ is nonempty by compactness. Any its element is included in every $U\_i$, and avoids all $M\_{p(i,j)}$, hence it lies in every $A\_i$. On the other hand, each $A\_i$ contains an interval minus a meager set, hence it cannot itself be meager, and in particular, it has more than one element.


EDIT: The argument above does not require full axiom of choice, it goes through in ZF + DC. Shelah has shown the relative consistency of ZF + DC + “every set of reals has the Baire property”, hence it is consistent with ZF + DC that A has a nondeterministic winning strategy in the unrestricted game.




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A also has a winning strategy if Q is restricted to Lebesgue measurable sets. The strategy is to maintain a chain $P\_0\supseteq P\_1\supseteq P\_2\supseteq\cdots$ of perfect sets with positive measure such that $P\_i\subseteq A\_i$. Given $Q\_i$, we choose $A\_i$ so that $P\_{i-1}\cap A\_i$ has positive measure. Using inner regularity, there exists a perfect set $P\_i$ of positive measure included in $P\_{i-1}\cap A\_i$. This strategy guarantees that each $\bigcap\_{i<n}A\_i$ is uncountable, and $\bigcap\_{i\in\omega}A\_i\supseteq\bigcap\_{i\in\omega}P\_i$ is nonempty by compactness.




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In ZFC, neither player has a winning strategy in the unrestricted game. For Q, this is shown in Joel David Hamkins' answer.


**Theorem.** A does not have a winning strategy.


For simplicity, I will consider the game played with the product space $2^\omega$ instead of $\mathbb R$. For $t\in2^{<\omega}$, let $B\_t=\{f\in2^\omega\mid f\supseteq t\}$ be a basic clopen set, let $D\_n=\{f\in2^\omega\mid f(n)=1\}$, and let $C$ be the Boolean algebra consisting of sets of the form $X\vartriangle Y$, where $X$ is clopen and $Y$ is finite. Given a sequence of sets $Q=\langle Q\_0,\dots,Q\_n\rangle$ as moves of the first player and a strategy $\sigma$ of A, I will write $\sigma(Q)=A\_n\in\{Q\_n,Q\_n^c\}$ for A's move provided by the strategy, and $Q^\sigma=\bigcap\_{i\le n}A\_n$. The latter notation can also be used for infinite sequences $Q$. If $Q,R$ are sequences, then $Q\smallfrown R$ is their concatenation, $|Q|$ is the length of $Q$, and $Q\subseteq R$ means that $Q$ is an initial segment of $R$.


**Lemma.** If A has a winning strategy in the game played on a subset $G\subseteq2^\omega$, then $G$ contains a perfect subset.


Proof: let's consider finite sequences $Q$ consisting of elements of $C$. First, assume that there exists such $Q$ so that for every finite $R,S\supseteq Q$, $R^\sigma\cap S^\sigma\ne\varnothing$. Then it is easy to see that there exists an ultrafilter $F\subseteq C$ such that $\sigma(R)\in F$ for every $R\supseteq Q$. Let $R$ be the infinite sequence $\langle D\_n\mid n\in\omega\rangle$. Then $|(Q\smallfrown R)^\sigma|\le1$. On the other hand, since $\sigma$ is a winning strategy, $(Q\smallfrown R)^\sigma$ is nonempty (and intersects $G$), hence it equals $\{\alpha\}$ for some $\alpha\in2^\omega$. Then either $(Q\smallfrown\{\alpha\})^\sigma=\{\alpha\}$ or $(Q\smallfrown\{\alpha\}\smallfrown R)^\sigma=\varnothing$, contradicting $\sigma$'s being a winning strategy.


Thus, for each $Q$ there exist $R,S\supseteq Q$ such that $R^\sigma\cap S^\sigma=\varnothing$. By induction, we can construct $\{Q\_t\mid t\in2^{<\omega}\}$ such that


* $t\subseteq s\Rightarrow Q\_t\subseteq Q\_s$,
* $Q\_{t\smallfrown0}^\sigma\cap Q\_{t\smallfrown1}^\sigma=\varnothing$.


Moreover, $Q\_t^\sigma\in C$, hence we can write it as $X\_t\vartriangle Y\_t$ with $X\_t$ clopen and $Y\_t$ finite. By extending $Q\_t$ (at the point where it is being constructed) with $D\_i$, $i\le|t|$, we can make sure that 


* $Q\_t\subseteq B\_s$ for some $s\in2^{<\omega}$, $|s|\ge|t|$.


By extending $Q\_t$ with $Y\_t$, we can make sure that $Q\_t^\sigma$ actually equals $X\_t\smallsetminus Y\_t$. Then $X\_{t\smallfrown0}\cap X\_{t\smallfrown1}$ is a finite clopen set, hence it is empty, thus:


* $\overline{Q\_{t\smallfrown0}^\sigma}\cap\overline{Q\_{t\smallfrown1}^\sigma}=\varnothing$.


For every $f\in2^\omega$, let $Q$ be the infinite sequence $\bigcup\_{t\subseteq f}Q\_t$. Since $\sigma$ is a winning strategy, $Q^\sigma\cap G$ is nonempty, hence it contains an element $\phi(f)$. The properties of $Q\_t$ ensure that $\phi\colon2^\omega\to2^\omega$ is injective and continuous, hence its range is a perfect subset of $G$, finishing the proof of the Lemma.


In order to prove the theorem, observe that there are $\mathfrak c=2^\omega$ perfect subsets of $2^\omega$, so we can enumerate them as $\{P\_\alpha\mid\alpha<\mathfrak c\}$. (This is the place which breaks in ZF + DC, we need $2^\omega$ to be well ordered.) We construct disjoint sequences $\{a\_\alpha\mid\alpha<\mathfrak c\},\{b\_\alpha\mid\alpha<\mathfrak c\}\subseteq2^\omega$ by induction as follows: each $P\_\alpha$ has cardinality $\mathfrak c$, whereas $S=\{a\_\beta,b\_\beta\mid\beta<\alpha\}$ has smaller cardinality, hence we can choose disctinct $a\_\alpha,b\_\alpha\in P\_\alpha\smallsetminus S$. Let $X=\{a\_\alpha\mid\alpha<\mathfrak c\}$. By the construction, neither $X$ nor $X^c$ contains a perfect subset.


Assume that $\sigma$ is a winning strategy for A in the game played on $2^\omega$. Take $Q\_0=X$, and let $A\_0=\sigma(Q\_0)$. Then $\tau(R):=\sigma(Q\_0\smallfrown R)$ defines a winning strategy for A in the game played on $A\_0$, hence by the Lemma, $A\_0$ has a perfect subset, a contradiction.


