## Set theory seminar – Marion Scheepers: Coding strategies (III)

September 28, 2010

For the second talk (and a link to the first one), see here. The third talk took place on September 28.

In the second case, we fix an $X\in J$ with ${}|J(X)|<|J|$. We can clearly assume that $S$ is infinite, and it easily follows that ${}|{\mathcal P}(X)|=|J|$. This is because any $Y\in J$ can be coded by the pair $(Y\cap X,X\cup(Y\setminus X))$, and there are only ${}|J(X)|$ many possible values for the second coordinate.

In particular, $X$ is infinite, and we can fix a partition $X=\bigsqcup_n X _n$ of $X$ into countably many pieces, each of size ${}|X|$. Recall that we are assuming that $\text{cf}(\left< J\right>,{\subset})\le|J|$ and have fixed a set $H$ cofinal in $\left< J\right>$ of smallest possible size. We have also fixed a perfect information winning strategy $\Psi$ for II, and an $f:\left< J\right>\to H$ with $A\subseteq f(A)$ for all $A$.

For each $n$, fix a surjection $f:{\mathcal P}(X_n)\setminus\{\emptyset,X_n\}\to{}^{<\omega}H$.

We define $F:J\times\left< J\right>\to J$ as follows:

1. Given $O\in \left< J\right>$, let

$A=\Psi(\left)\setminus X$,

and

$B\in{\mathcal P}(X_0)\setminus\{\emptyset, X_0\}$ such that $f_0(B)=\left< f(O)\right>$,

and set $F(\emptyset,O)=A\cup B$.

2. Suppose now that $(T,O)\in J\times\left< J\right>$, that $T\ne\emptyset$ , and that there is an $n$ such that $\bigcup_{j, $T\cap X_n\ne\emptyset,X_n$, and $T\cap\bigcup_{k>n}X_k=\emptyset$. Let

$B\in{\mathcal P}(X_{n+1})\setminus\{\emptyset,X_{n+1}\}$ be such that $f_{n+1}(B)=f_n(T\cap X_n){}^\frown\left< f(O)\right>$,

and

$A=\Psi(f_{n+1}(B))\setminus X$

and set $F(T,O)=A\cup\bigcup_{j\le n}X_j\cup B$.

3. Define $F(T,O)=\emptyset$ in other cases.

A straightforward induction shows that $F$ is winning. The point is that in a run of the game where player II follows $F$:

• Player II’s moves code the part that lies outside of $X$ of player II’s moves in  a run ${\mathcal A}$ of the game following $\Psi$ where I plays sets covering the sets in the original run. For this, note that at any inning there is a unique index $n$ such that player II’s move covers $\bigcup_{j, is disjoint from $\bigcup_{j>n}X_j$, and meets $X_n$ in a set that is neither empty nor all of $X_n$, and this $n$ codes the inning of the game, and the piece of player II’s move in $X_n$ codes the history of the run ${\mathcal A}$ played so far.
• $X$ is eventually covered completely, so in particular the parts inside $X$ of player II’s responses in the run ${\mathcal A}$ are covered as well.

This completes the proof of Theorem 1. $\Box$

By way of illustration, consider the case where $J$ is the ideal of finite sets of some set $S$. Then whether II has a winning coding strategy turns into the question of when it is that $\text{cf}({\mathcal P}_{\aleph_1}(S))\le|S|$. This certainly holds if ${}|S|={\mathfrak c}$ or if ${}|S|<\aleph_\omega$. However, it fails if ${}|S|=\aleph_\omega$.

This example illustrates how player II really obtains an additional advantage when playing in $WMG(J)$ rather than just in $RG(J)$. To see that this is the case, consider the same $J$ as above with ${}|S|=\aleph_\omega$. This is an instance of the countable-finite game. We claim that II has a winning coding strategy in this case. To see this, consider a partition of $S$ into countably many sets $S_n$ with ${}|S_n|=\aleph_n$. For each $n$, pick a winning coding strategy $\sigma_n$ for the countable-finite game on $S_n$, and define a strategy in $WMG(J)$ so that for each $n$ it simulates a run of the game $WMG(J\cap{\mathcal P}(S_n)$ with II following $\sigma_n$, as follows: In inning $n$, II plays on $S_i$ for $i\le n$; player I’s moves in the “$i$-th board” are the intersection with $S_i$ of I’s moves in $WMG(J)$, and I’s first move occurred at inning $i$. (II can keep track of $n$ in several ways, for example, noticing that, following the proof of Theorem 1 produces coding strategies that never play the empty set.)

Note that this strategy is not winning in $RG(J)$, the difference being that there is no guarantee that (for any $i$) the first $i$ moves of I in the $(i+1)$-st board are going to be covered by II’s responses. On the other hand, the strategy is winning in $WMG(J)$, since, no matter how late one starts to play on the $i$-th board, player I’s first move covers I’s prior moves there (and so, II having a winning coding strategy for the game that starts with this move, will also cover those prior moves).

The first place where this argument cannot be continues is when $|S|=\aleph_{\omega_1}$. However, ${\sf GCH}$ suffices to see that player I has a winning strategy in $RG(J)$ in this case, and so we can continue. This illustrates the corollary stated in the first talk, that ${\sf GCH}$ suffices to guarantee that II always has a winning coding strategy in $WMG(J)$.

The natural question is therefore how much one can weaken the ${\sf GCH}$ assumption, and trying to address it leads to Theorem 2, which will be the subject of the next (and last) talk.