Infinite combinatorial games 
*

*Hercules vs. Hydra: Recall the story where every time Hercules cuts of a head, two more heads grow instead. Now suppose the following: The hydra starts off with one head, but every time Hercules cuts off a head, $\aleph_0$ grow anew. Show that Hercules will be able to defeat the Hydra before $\omega_1$ steps. That is, we will reach a step $\lambda<\omega_1$ where the Hydra doesn't have any heads.

*Two players play in the following game: In each turn $n \in \mathbb{N}$, the player whose turn it is chooses a set $A_n$ that is contained in the previous set,i.e $A_n \subseteq A_{n-1}$ such that $A_n$ are stationary sets $A_n \subseteq \omega_1$. Player 1 starts. After $\aleph_0$ such turns, Player 2 wins if the intersection $\cap_nA_n$ contains at most 1 ordinal (we can say Player 1 wins otherwise). Show that Player 2 has a winning strategy, that is a way to choose the $2n+1$ set to ensure he wins.
 A: OK, let's see. 
For the first question, enumerate the heads as they are added: The original head is head $0$. We can enumerate the heads added in stage $\alpha$ by ordinals in the ordinal interval $(\omega\alpha,\omega(\alpha+1)]$. If we suppose for no $\alpha<\omega_1$ the Hydra has no heads at the beginning of stage $\alpha$ then, for a club $C$ of countable ordinals $\alpha$, the heads enumerated so far at the beginning of stage $\alpha$ are indexed precisely by the ordinals less than $\alpha$. 
This allows us to define a regressive function on $C$, that to each $\alpha\in C$ assigns $\beta<\alpha$ iff $\beta$ is the index of the head removed at stage $\alpha$. By Fodor's lemma, the same index is used more than once (in fact, stationarily often). But this is impossible, and we have a contradiction. (This is a nice problem. I forget where it originated.)

The second question reminds me of arguments of Mycielski and Solovay from the early 70s. I give here an argument that applies not just to subsets of $\omega_1$, but also to subsets of any regular $\kappa\le\mathfrak c$, the cardinality of the continuum. I suspect this restriction is a distraction, but I'll use it below. 
Accordingly, suppose $\kappa$ is regular, $\kappa\le\mathfrak c$, and we play the game you describe, except that the $A_n$ are stationary in $\kappa$. Since $\kappa\le\mathfrak c$, we may fix an injection of $\kappa$ into the interval $[0,1]$. In what follows, we identify subsets of $\kappa$ with subsets of $[0,1]$ via this injection. Also, to each bounded set $A\subseteq[0,1]$ with infimum $m=m(A)$ and supremum $M=m(A)$, we associate a strictly increasing sequence $m=a_0=a_0(A)<a_1<\dots$ converging to $M$, with $a_1=(m+M)/2$, $a_2=(a_1+M)/2$, etc. (So $a_1$ is the midpoint of the interval $[m,M]$, $a_2$ is the midpoint of $[a_1,M]$, etc.)
Now we describe how player II must respond to the moves of player I. Say that at stage $n$, player I has played $A=A_{2n}$. There must be an interval $[a_k(A),a_{k+1}(A))$ whose intersection with $A$ is (via the injection fixed above) a stationary subset of $\kappa$. The reason is that the stationary set $A\setminus\{M(A)\}$ is the countable union of these sets, and the countable intersection of clubs is club, so one of these sets must be stationary as well.
More precisely, if $i$ is the injection, then $A\setminus\{i^{-1}(M)\}=\bigcup_k i^{-1}([a_k,a_{k+1})\cap i[A])$.
Let $A_{2n+1}$ be this stationary set $i^{-1}([a_k,a_{k+1})\cap i[A_{2n}])$ (where, if you wish, we let $k$ be least so that this is stationary).
The point is this: Call the size of $A_n$ the number $M(A_n)-m(A_n)$. Then the size of $A_{2n+2}$ is at most half the size of $A_{2n}$. Since the sets $i[A_n]$ shrink to size $0$, their intersection is empty or a singleton. 
