Consider the following infinite game: two players, I and II, are alternating and choosing a descending sequence of subsets of $\mathbb R$ of cardinality $\frak c$, so I chooses a set $A\_1\subseteq\mathbb R$, II chooses a set $A\_2\subseteq A\_1$, I chooses $A\_3\subseteq A\_2$ etc., all having the size continuum. Let $A=\cap\_{i=1}^\infty A\_i$. I wins if $A$ is nonempty, otherwise II wins.



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> Who has a winning strategy in this game?
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This game was posed to me as a puzzle from a person I don't have any means of contact with anymore (so I can't ask them about this). However, I don't know whether he himself knew the answer or not (as far as I know it might be independent of ZFC or unknown) which is why I'm posting this on MO and not Math.SE (the same person has posed as a puzzle a combinatorial problem which I happened to know is open).


All I know is that it's consistent with ZF that II has the winning strategy: if $\mathbb R$ had countable cofinality, i.e. was a countable union of sets $S\_i,i\in\mathbb N$ of cardinality $<\frak c$, then II could play on his turns $A\_{2k}=A\_{2k-1}\setminus(S\_1\cup\dots\cup S\_n)$, which would have size $\frak c$. This doesn't go through in ZFC, since there $\frak c$ has uncountable cofinality.


One can also consider the problem with $\mathbb R$ replaced by some other set, for example an ordinal. For some sufficiently well-behaved sets (not Dedekind-finite, for example) the above reasoning shows that if a cardinal has countable cofinality, I has a winning strategy. I am especially interested in the case of $\omega\_1$, which is equivalent to the problem at hand if we assume CH.


I am interested in this problem both in context of ZF and ZFC, and I am willing to assume CH to resolve the original problem if it is of help there.



Choosing subsets of $\mathbb R$ of cardinality $\frak c$, who wins?
In ZFC, the player aiming for the empty set has a winning strategy in the game played on any infinite set, including the reals. Using the axiom of choice, we can well-order the set and thereby pretend that we are playing sets of ordinals. Now, whenever it is the empty-set player's turn, she should look at the order type of the ordinals remaining, and play so as to omit the first element of every successive block of $\omega$ many elements in the enumeration of the set. In this way, no element can survive into the final intersection. If an element is the $(k+1)^{th}$ in it's $\omega$-block, then it is the $(\omega\cdot\gamma+k+1)^{th}$ element of $A\_n$ for some ordinal $\gamma$, and after the empty-set player plays, it will become the $(\omega\cdot\gamma+k)^{th}$ element of the new set, strictly smaller. When the other player plays, if it was the $\alpha^{th}$ element, it becomes the $\beta^{th}$ element for some $\beta\leq\alpha$. So as play progresses, the index of the element in the enumeration will steadily descend. Since it cannot descend forever, the element must eventually fall out of the set and so the final intersection is empty, as desired. 


In ZF, [the argument I gave at the other question](https://mathoverflow.net/a/137367/1946) shows that the non-empty player cannot have a winning strategy in the game on the reals. 


