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

October 12, 2010

For the third talk (and a link to the second one), see here. The fourth talk took place on October 12.

We want to show the following version of Theorem 2:

Theorem. Suppose $\kappa$ is a singular strong limit cardinal of uncountable cofinality. Then the following are equivalent:

1. For each ideal $J$ on $\kappa$, player II has a winning coding strategy in $RG(J)$.
2. $2^\kappa<\kappa^{+\omega}$.

Since $2^\kappa$ has uncountable cofinality, option 2 above is equivalent to saying that the instance of ${\sf wSCH}$ corresponding to $\kappa$ holds.

Before we begin the proof, we need to single out some elementary consequences in cardinal arithmetic of the assumptions on $\kappa$. First of all, since $\kappa$ is singular strong limit, then for any cardinal $\lambda<\kappa$, we have that

$\kappa^\lambda=\left\{\begin{array}{cl}\kappa&\mbox{\ if }\lambda<{\rm cf}(\kappa),\\ 2^\kappa&\mbox{\ otherwise.}\end{array}\right.$

Also, since the cofinality of $\kappa$ is uncountable, we have Hausdoff’s result that if $n<\omega$, then $(\kappa^{+n})^{\aleph_0}=\kappa^{+n}$. I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here.

Proof. $(2.\Rightarrow 1.)$ We use Theorem 1. If option 1. fails, then there is an ideal $J$ on $\kappa$ with ${\rm cf}(\left< J\right>,{\subset})>|J|$.

Note that ${\rm cf}(\left< J\right>,{\subset})\le({\rm cf}(J,{\subset}))^{\aleph_0}$, and $\kappa\le|J|$. Moreover, if $\lambda<\kappa$, then $2^\lambda<{\rm cf}(J,{\subset})$ since, otherwise,

$({\rm cf}(J,{\subset}))^{\aleph_0}\le 2^{\lambda\aleph_0}=2^\lambda<\kappa$.

So ${\rm cf}(J,{\subset})\ge\kappa$ and then, by Hausdorff, in fact ${\rm cf}(J,{\subset})\ge \kappa^{+\omega}$, and option 2. fails.

$(1.\Rightarrow 2.)$ Suppose option 2. fails and let $\lambda=\kappa^{+\omega}$, so $\kappa<\lambda<2^\kappa$ and ${\rm cf}(\lambda)=\omega$. We use $\lambda$ to build an ideal $J$ on $\kappa$ with ${\rm cf}(\left< J\right>,{\subset})>|J|$.

For this, we use that there is a large almost disjoint family of functions from ${\rm cf}(\kappa)$ into $\kappa$. Specifically:

Lemma. If $\kappa$ is singular strong limit, there is a family ${\mathcal F}\subseteq{}^{{\rm cf}(\kappa)}\kappa$ with ${}|{\mathcal F}|=2^\kappa$ and such that for all distinct $f,g\in{\mathcal F}$, we have that ${}|\{\alpha<{\rm cf}(\kappa)\mid f(\alpha)=g(\alpha)|<{\rm cf}(\kappa)$.

In my notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions $f:{\rm cf}(\kappa)\to\kappa$, and then replace them with (appropriate codes for) the branches they determine through the tree $\kappa^{{\rm cf}(\kappa)}$. Distinct branches eventually diverge, and this translates into the corresponding functions being almost disjoint.

Pick a family ${\mathcal F}$ as in the lemma, and let ${\mathcal G}$ be a subfamily of size $\lambda$. Let $S=\bigcup{\mathcal G}\subseteq{\rm cf}(\kappa)\times\kappa$. We proceed to show that $|S|=\kappa$ and use ${\mathcal G}$ to define an ideal $J$ on $S$ as required.

First, obviously $|S|\le\kappa$. Since $\kappa<\lambda=|{\mathcal G}|$ and ${\mathcal G}\subseteq{\mathcal P}(S)$, it follows that ${}|S|\ge\kappa$, or else ${}|{\mathcal P}(S)|<\kappa$, since $\kappa$ is strong limit.

Now define

$J=\{X\subseteq S\mid\exists {\mathcal H}\subseteq{\mathcal G}\,(|{\mathcal H}|<\omega,\bigcup{\mathcal H}\supseteq X)\}.$

Clearly, $J$ is an ideal. We claim that $|J|=\lambda$. First, each singleton $\{f\}$ with $f\in{\mathcal G}$ is in $J$, so ${}|J|\ge\lambda$. Define $\Phi:[{\mathcal G}]^{<\aleph_0}\to J$ by $\Phi({\mathcal H})=\bigcup{\mathcal H})$. Since the functions in ${\mathcal G}$ are almost disjoint, it follows that $\Phi$ is 1-1. Let $G$ be the image of $\Phi$. By construction, $G$ is cofinal in $J$. But then

${}|J|\le|{\mathcal G}|2^{{\rm cf}(\kappa)}=\lambda 2^{{\rm cf}(\kappa)}=\lambda$,

where the first inequality follows from noticing that any $X\in J$ has size at most ${\rm cf}(\kappa)$. It follows that $|J|=\lambda$, as claimed.

Finally, we argue that ${\rm cf}(\left< J\right>,{\subset})>\lambda$, which completes the proof. For this, consider a cofinal ${\mathcal A}\subseteq\left< J\right>$, and a map $f:{\mathcal A}\to[{\mathcal G}]^{\le\aleph_0}$ such that for all $A\in{\mathcal A}$, we have $A\subseteq\bigcup f(A)$.

Since ${\mathcal A}$ is cofinal in $\left< J\right>$, it follows that $f[{\mathcal A}]$ is cofinal in ${}[{\mathcal G}]^{\le\aleph_0}$. But this gives the result, because

${}|{\mathcal A}|\ge{\rm cf}([{\mathcal G}]^{\le \aleph_0},{\subset})={\rm cf}([\lambda]^{\le \aleph_0},{\subset})>\lambda$,

and we are done. $\Box$

## 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.

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

September 27, 2010

For the first talk, see here. The second talk took place on September 21.

We want to prove $(2.\Rightarrow1.)$ of Theorem 1, that if ${\rm cf}(\left< J\right>,\subset)\le|J|$, then II has a winning coding strategy in $RG(J)$.

The argument makes essential use of the following:

Coding Lemma. Let $({\mathbb P},<)$ be a poset such that for all $p\in{\mathbb P}$,

${}|\{q\in{\mathbb P}\mid q>p\}|=|{\mathbb P}|.$

Suppose that ${}|H|\le|{\mathbb P}|$. Then there is a map $\Phi:{\mathbb P}\to{}^{<\omega}H$ such that

$\forall p\in{\mathbb P}\,\forall\sigma\in H\,\exists q\in{\mathbb P}\,(q>p\mbox{ and }\Phi(q)=\sigma).$

Proof. Note that ${\mathbb P}$ is infinite. We may then identify it with some infinite cardinal $\kappa$. It suffices to show that for any partial ordering $\prec$ on $\kappa$ as in the hypothesis, there is a map $\Phi:\kappa\to\kappa$ such that for any $\alpha,\beta$, there is a $\gamma$ with $\alpha\prec\gamma$ such that $\Phi(\gamma)=\beta$.

Well-order $\kappa\times\kappa$ in type $\kappa$, and call $R$ this ordering. We define $\Phi$ by transfinite recursion through $R$. Given $(\alpha,\beta)$, let $A$ be the set of its $R$-predecessors,

$A=\{(\mu,\rho)\mid(\mu,\rho) R(\alpha,\beta)\}$.

Our inductive assumption is that for any pair $(\mu,\rho)\in A$, we have chosen some $\tau$ with $\mu\prec\tau$, and defined $\Phi(\tau)=\rho$.  Let us denote by $D_A$ the domain of the partial function we have defined so far. Note that ${}|D_A|<\kappa$. Since $\{\gamma\mid\alpha\prec\gamma\}$ has size $\kappa$, it must meet $\kappa\setminus D_A$. Take $\mu$ to be least in this intersection, and set $\Phi(\mu)=\beta$, thus completing the stage $(\alpha,\beta)$ of this recursion.

At the end, the resulting map can be extended to a map $\Phi$ with domain $\kappa$ in an arbitrary way, and this function clearly is as required. $\Box$

Back to the proof of $(2.\Rightarrow1.)$. Fix a perfect information winning strategy $\Psi$ for II in $RG(J)$, and a set $H$ cofinal in $\left< J\right>$ of least possible size. Pick a $f:\left< J \right>\to H$ such that for all $A\in \left< J\right>$ we have $A\subseteq f(A)$.

Given $X\in J$, let $J(X)=\{Y\in J\mid X\subseteq Y\}$. Now we consider two cases, depending on whether for some $X$ we have ${}|J(X)|<|J|$ or not.

Suppose first that $|J(X)|=|J|$ for all $X$. Then the Coding Lemma applies with $(J,\subset)$ in the role of ${\mathbb P}$, and $H$ as chosen. Let $\Phi$ be as in the lemma.

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

1. Given $O\in\left< J\right>$, let $Y\supseteq\psi(f(O))$ be such that $\Phi(Y)=\left$, and set $F(\emptyset,O)=Y$.
2. Given $(T,O)\in J\times\left< J\right>$ with $T\ne\emptyset$, let $Y\supseteq \Psi(\Phi(T){}^\frown\left)$ be such that $\Phi(Y)=\Phi(T){}^\frown\left< f(O)\right>$, and set $F(T,O)=Y$.

Clearly, $F$ is winning: In any run of the game with II following $F$, player II’s moves cover their responses following $\Psi$, and we are done since $\Psi$ is winning.

The second case, when there is some $X\in J$ with $|J(X)|<|J|$, will be dealt with in the next talk.

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

September 25, 2010

This semester, the seminar started with a series of talks by Marion. The first talk happened on September 14.

We consider two games relative to a (proper) ideal $J\subset{\mathcal P}(S)$ for some set $S$. The ideal $J$ is not assumed to be $\sigma$-complete; we denote by $\left< J\right>$ its $\sigma$-closure, i.e., the collection of countable unions of elements of $J$. Note that $\left< J\right>$ is a $\sigma$-ideal iff $\left< J\right>$ is an ideal iff $S\notin\left< J\right>$.

The two games we concentrate on are the Random Game on $J$, $RG(J)$, and the Weakly Monotonic game on $J$, $WMG(J)$.

In both games, players I and II alternate for $\omega$ many innings, with I moving first, moving as follows:

$\begin{array}{cccccc} I&O_0\in\left< J\right>&&O_1\in\left< J_2\right>&&\cdots\\ II&&T_0\in J&&T_1\in J \end{array}$

In $RG(J)$ we do not require that the $O_i$ relate to one another in any particular manner (thus “random”), while in $WMG(J)$ we require that $O_1\subseteq O_2\subseteq\dots$ (thus “weakly”, since we allow equality to occur).

In both games, player II wins iff $\bigcup_n T_n\supseteq\bigcup_n O_n$. Obviously, II has a (perfect information) winning strategy, with $=$ rather than the weaker $\supseteq$.

However, we are interested in an apparently very restrictive kind of strategy, and so we will give some leeway to player II by allowing its moves to over-spill if needed. The strategies for II we want to consider we call coding strategies. In these strategies, II only has access to player I’s latest move, and to its own most recent move. So, if $F$ is a coding strategy, and II follows it in a run of the game, then we have that for every $n$,

$T_n=F(T_{n-1},O_n)$,

with $T_{-1}=\emptyset$.

The underlying goal is to understand under which circumstances player II has a winning coding strategy in $WMG(J)$. Obviously, this is the case if II has a winning coding strategy in $RG(J)$.

Theorem 1. For an ideal $J\subset{\mathcal P}(S)$, the following are equivalent:

1. II has a winning coding strategy in $RG(J)$.
2. ${\rm cf}(\left< J\right>,{\subset})\le|J|$.

Corollary. ${\sf GCH}$ implies that for any ideal $J\subset{\mathcal P}(S)$, II has a winning strategy in $WMG(J)$.

We can reformulate our goal as asking how much one can weaken ${\sf GCH}$ in the corollary.

Let’s denote by ${\sf wSCH}$, the weak singular cardinals hypothesis, the statement that if $\kappa$ is singular strong limit of uncountable cofinality, then for no cardinal $\lambda$ of countable cofinality, we have $\kappa<\lambda<2^\kappa$.

By work of Gitik and Mitchell, we know that the negation of ${\sf wSCH}$ is equiconsistent with the existence of a $\kappa$ of Mitchell order $o(\kappa)=\kappa^{+\omega}+\omega_1$.

Theorem 2. The following are equivalent:

1. ${\sf wSCH}$.
2. For each ideal $J$ on a singular strong limit $\kappa$ of uncountable cofinality, II has a winning strategy in $RG(J)$.

We now begin the proof of Theorem 1.

$(1.\Rightarrow2.)$ Suppose II has a winning coding strategy $F$ in $RG(J)$. We want to show that ${\rm cf}(\left< J\right>,{\subset})\le|J|$. For this, we will define a map $f:J\to\left< J\right>$ with $\subset$-cofinal range, as follows: Given $X\in J$, let $T_0=X$ and $T_{n+1}=F(T_n,\emptyset)$ for all $n$. Now set

$f(X)=\bigcup_n T_n$.

To see that $f$ is cofinal, given $O\in\left< J\right>$, let $X=F(\emptyset,O)$, so that the $T_n$ are II’s responses using $F$ in a run of the game where player I first plays $O$ and then plays $\emptyset$ in all its subsequent moves. Since $F$ is winning, we must have $f(X)\supseteq O$.