I'm reading through Moschovaki's proof that all $\Sigma_2^0$ games are determined, and the second part of the proof is confusing me. I follow up to the point where they prove $u\in W^{\xi}\implies $ I wins $A(u)$. (Still stuck on the converse direction, but we'll get there.)
Firstly, I'm having trouble understanding what this $H^{\xi,i}$ set is. I'm reading the definition, but not understanding why it's useful. This is used in the proof that $u\in W^{\xi}\implies $ I wins $A(u)$, so we'll just mention it when it comes up.
What exactly are the two cases that they're splitting it into? So $k$ steps have been made, and are we claiming in Case I that Player I's strategy doesn't hit any of the $T^i$'s? And why does the induction hypothesis guarantee I wins?
Case 2 is where Player I hits one of the $T^i$'s? Why is this important..? I'm still not seeing why this means Player I wins.
Another problem seems to be coming up, too. It could be the case that we're moving from tree to tree in this. (on the line before Case 1). Why doesn't it cause an issue if we're moving from tree, say $T^i$ to $T^j$ to $T^k$?
Next, I'm not really seeing why $\eta\leq\xi\implies H^{\eta,i}\subseteq H^{\xi,i}$. Why should the $H^{\xi,i}$'s be growing? Is the reason why they can't grow forever because all of these are countable? And... they kind of lose me past here. So they have a point where the induction closes off, and they call this ordinal $\kappa$ (is this $\omega_1 ^{CK}$)? They start by Player II playing the closed game $H^{\kappa,0}$, and argue that some $k_0$ exists where the final play is not in $F^{0}$, and then iterating this process.
Is someone able to answer some of these questions, and give me a better idea of how the second half of this proof goes? I'm sure some of my questions are answered in the text, but I've been staring at it for so long that it's really not clicking. Thanks.
EDIT: I'm first going to give an alternate proof, which is a little simpler, but which is not so constructive, and doesn't give as much information. It is analogous to the short proof of open determinacy, so recall this: Let $A\subseteq\omega^\omega$ be an open set. Suppose that player 1 (who wants to get into $A$) has no winning strategy. Then player 2 has the following winning strategy: given a partial play $p$ with player 2 to move, just play $n<\omega$ such that player 1 does not have a winning strategy from $p\frown(n)$. This works: suppose $q$ has even length, so player 1 is to move, but does not have a winning strategy from $q$. Then it must be that for every possible move $m<\omega$ for player 1, there is a response $n<\omega$ for player 2 such that player 1 does not have a winning strategy from $q\frown (m,n)$. (Otherwise play a counterexample $m$, and then for each $n<\omega$, let $\sigma_n$ be a winning strategy for player 1 from $q\frown (m,n)$. By putting $m$ together with $n\mapsto\sigma_n$ together, we get a winning strategy for player 1 from position $q$, contradiction.) Therefore whatever $m$ player 1 plays, player 2 can just respond with a witness $n$, so that player 1 still has no winning strategy from $q\frown (m,n)$.
Now generalize this: Let$F=\bigcup_{i<\omega}F_i$ where each $F_i$ is closed. Suppose player 1 has no winning strategy for $G_F$ (player 1 wants to get into $F$); we want to see there is one for player 2 (to get into the complement of $F$). Let $A$ be the complement, so $A=\bigcap_{n<\omega}A_i$ where each $A_i$ is open. I prefer to deal with the $A_i$'s than the $F_i$'s, but will keep player 1 and 2 as before, hence as in Moschovakis.
For $i<\omega$ let $C_i\subseteq{^{<\omega}\omega}$ be the set of finite sequences $s$ such that $N_s\subseteq A_i$ (where $N_s$ is as in Moschovakis, i.e. $N_s=\{x\in{^\omega}\omega\bigm|s\subseteq x\}$). So player 1 has no winning strategy for $G_{A^c}$ (aiming to get into $A^c=F$). We find one for player 2 to get into $A$. Let $B$ be the set of bad partial plays, which are those from which player 1 has a winning strategy (to get into $A^c$, starting from that position). So the assumption is that $\emptyset\notin B$. Let $C'_i=C_i\backslash B$. Let $A'_i$ be the corresponding open set, i.e. $A'_i=\bigcup_{s\in C'_i}N_s$. Since $A'_i$ is open, one of the players has a winning strategy for it. I claim it is player 2, to get into $A'_i$. For if player 1 had a strategy $\sigma$ (to get into $A'_i$-complement) then have player 1 play the game for $A^c$ by starting with $\sigma$, until some partial play $p$ gets into $C_i$, if ever, at which point note that we must have $p\in B$. At that point, switch to using a strategy to get into $A^c$. If not such $p$ is reached then note that the real played is not in $A_i$, so in $A^c$. This is winning for player 1 in $G_{A^c}$, contradiction. So player 2 has a winning strategy $\sigma_0$ to get into $A'_0$. Have player 2 begin the game $A$ by using $\sigma_0$. Use this until reaching some partial play $p_0$ in $C'_0$ (note such a play must be reached). At this point $p_0$, the eventual real will definitely be in $A_0$, and player 1 still has no winning strategy from $p_0$ since $p_0\notin B$. Now arguing as above but with the prefix $p_0$, player 2 can find a winning strategy $\sigma_1$ to get into $A'_1$ starting from $p_0$. So have player 2 continue after $p_0$ using $sigma_1$, until reaching $p_1\in C'_1$, and then switch to $\sigma_2$ for $A'_2$, etc. This produces a real in all $A'_i$, hence in $A_i$, hence in $A$, so player 2 has won the run.
The preceding proof was by contradiction, and just starting with the set $B$ then as given. In the proof in Moschovakis, the set $B$ is built up by recursion instead. Now to return to the proof given in Moschovakis and your questions on this...
The overall strategy is to compute positions (finite partial plays) from which player 1 is next to play and has a winning strategy, and $W^\xi$ is a such a set of positions. The positions in $W^0$ have a very simple strategy: just choose $i$ witnessing the definition and play to get into $F_i$. Now $H^{1,i}$ is a closed set enlarging $F_i$. It consists of those reals either in $F_i$ or having passed through some finite stage in $W_0$ and then remained in $W_0$ after that (at the remaining even stages). So the point is that if player 1 has a strategy $\sigma$ to get into $H^{1,i}$ then they have a strategy to win overall: initially play according to $\sigma$. If no finite stage $p$ is reached after starting play which is in $W_0$ then note that the end real is in $F_i$, so player 1 has won the run of the game. If there is such a finite stage $p$ reached then at the first such, choose $i'<\omega$ witnessing that $p\in W_0$ and play from that point on to get into $F_{i'}$ (now ignoring $\sigma$). (These were the 2 cases in the proof you asked about, i.e. either such a $p$ is reached (case 1) or not (case 2). Note that I'm assuming $u=\emptyset$ for simplicity.) The set $H^{1,i}$ might be a little confusing because e.g. above we didn't actually use the fact that reals in that set are either in $F_i$ or all even finite segments are in $W_0$, but just that there was some such segment. But we want $H^{1,i}$ to be closed, and also we want to handle arbitrary prefixes $u$ (whereas I assumed $u=\emptyset$ above).
Moving from tree to tree is no problem because each time you do, you drop the rank $\xi$ of the set $W_\xi$ you're starting from, so that can only happen finitely often. The eventual strategy for player 1, if $\emptyset\in W_\kappa$, is to choose $i$ witnessing that $\emptyset\in W_\kappa$ and play according to a strategy $\sigma$ that witnesses this, until reaching a point $p$ at which $p\in W_\alpha$ for some $\alpha<\kappa$, at which point choose a new $i'$ witnessing this, and strategy $\sigma'$, and continue from $p$ using $\sigma'$, until reaching $p'\in W_\beta$ for some $\beta<\alpha$, then switch to some $i''$ and $\sigma''$, etc. We can only drop in rank finitely often. So e.g. if we don't drop any more times after $\beta$, then note that the eventual real played is in $F_{i''}$, so player 1 wins the run.
The fact that $\eta\leq\xi$ implies $H^{\eta,i}\subseteq H^{\xi,i}$ is just because the underlying set involves a union of more sets, i.e. in particular $\bigcup_{\alpha<\eta}W^\alpha$ versus $\bigcup_{\alpha<\xi}W^\alpha$. And they can't keep growing forever because they are just sets of reals. (Actually they depend on the sets $W^\xi$, which are subsets of a countable set, and increasing, so $\kappa<\omega_1$.)
Now if $\emptyset\notin W_\kappa=W_{\kappa+1}$ then we construct a strategy for player 2, trying to avoid all $F_i$s. These are closed, so $F_0$ is avoided iff some $p$ is reached which guarantees that any real extending it is not in $F_0$. So player 2 has to at least reach such a $p$, in order to have any chance of winning. But he/she does not want to go for such a $p$ from which player 1 has a winning strategy. So the plan is to reach some such $p$, of even length, which itself is not in $W_\kappa$. (If $p\in W_\kappa$ then player 1 can win from there.) This corresponds to some open subset of the complement of $F_0$, i.e the reals which pass thru some such $p$. So either player 2 has a winning strategy for it, or player 1 for its complement (by determinacy of open sets). One shows that if player 1 has the winning strategy then $\emptyset\in W^{\kappa+1}$, contradiction. So player 2 has such a strategy $\sigma_0$. Have player 2 play according to $\sigma_0$ until reaching $p$ as desired. So $p$ has even length, is not in $W_\kappa$, and no real extending it is in $F_0$. So regarding $F_0$, player 2 is now safe. Next they deal with $F_1$, in a similar manner, then $F_2$, etc, always avoiding partial plays in $W_\kappa$. Since this gets into all $F_i$s, player 2 wins the run.
Of course I'm just repeating parts of the proof in the book, in somewhat different words, but hopefully it helps...