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.

We are ready to address the Theorem.

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

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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.

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