## 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. $\Box$

## 580 -Cardinal arithmetic (3)

February 9, 2009

It is easy to solve negatively the question immediately following Homework problem 5 on lecture II.1. I asked whether if $X$ is Dedekind-finite but ${\mathcal P}(X)$ is Dedekind-infinite, then it followed that there is an infinite Dedekind-finite set $Y$ such that ${\mathcal P}(Y)\preceq X$.

To exhibit a counterexample, it is enough to know that it is consistent to have an infinite Dedekind finite set $X$ that is the countable union of finite sets (in fact, sets of size 2). Notice that $\omega$ is a surjective image of $X,$ so ${\mathcal P}(X)$ is Dedekind-infinite. Suppose that ${\mathcal P}(Y)\preceq X.$ Then certainly $Y\preceq X,$ so $Y$ is a countable union of finite sets $Y_n.$ If $Y$ is infinite then $Y_n\ne\emptyset$ for infinitely many values of $n.$ But then $\omega$ is also a surjective image of $Y$, so $\omega$ (and in fact $P(\omega)$) injects into ${\mathcal P}(Y)$ and therefore into $X,$ contradiction.

At the end of last lecture we showed Theorem 10, a general result that allows us to compute products $\kappa^\lambda$ for infinite cardinals $\kappa,\lambda,$ namely:

Let $\kappa$ and $\lambda$ be infinite cardinals. Let $\tau=\sup_{\rho<\kappa}|\rho|^\lambda.$ Then

$\displaystyle \kappa^\lambda=\left\{\begin{array}{cl} 2^\lambda & \mbox{if }\kappa\le 2^\lambda,\\ \kappa\cdot\tau & \mbox{if }\lambda<{\rm cf}(\kappa),\\ \tau & \begin{array}{l}\mbox{if }{\rm cf}(\kappa)\le\lambda,2^\lambda<\kappa,\mbox{ and }\\ \rho\mapsto|\rho|^\lambda\mbox{ is eventually constant below }\kappa,\end{array}\\ \kappa^{{\rm cf}(\kappa)} & \mbox{otherwise.}\end{array}\right.$