580 -Cardinal arithmetic (3)

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.

The following easy consequence is particularly useful:

Corollary 11. (Hausdorff). (\kappa^+)^\lambda=\kappa^+\cdot\kappa^\lambda.

Proof. The result is immediate from the formula above if \lambda<\kappa^+={\rm cf}(\kappa^+). If, on the other hand, \lambda\ge\kappa^+ then certainly \kappa^+\le 2^\lambda, and it is easy to see that both sides of the equation we want to prove equal 2^\lambda. {\sf QED}

For example, \aleph_n^{\aleph_0}=\aleph_n\cdot 2^{\aleph_0} for any n\in\omega.

The following more general result is most easily established by induction on \gamma:

Homework problem 8. (Tarski). If |\gamma|\le\aleph_\beta then \aleph_{\alpha+\gamma}^{\aleph_\beta}=\aleph_{\alpha+\gamma}^{|\gamma|}\cdot\aleph_\alpha^{\aleph_\beta}.

 For example, \aleph_\omega^{\aleph_1}=\aleph_\omega^{\aleph_0}\cdot\aleph_0^{\aleph_1}=\aleph_\omega^{\aleph_0}\cdot 2^{\aleph_1}. Hence if 2^{\aleph_1}\le\aleph_\omega, then \aleph_\omega^{\aleph_0}=\aleph_\omega^{\aleph_1}.

Now we begin to explore the behavior of infinitary products, mainly as they relate to the gimel function \gimel(\kappa)=\kappa^{{\rm cf}(\kappa)}, but the results to follow are more general.

First, an easy observation.

Fact 12. Suppose \kappa_i\ge2 for all i\in I. Then (\sum_i\kappa_i)^{|I|}=(\prod_i\kappa_i)^{|I|}.

Proof. There is an obvious injection from the product set \prod_i\kappa_i into  {}^I(\bigsqcup_i\kappa_i). It follows that \sum_i\kappa_i\le\prod_i\kappa_i\le(\sum_i\kappa_i)^{|I|}, from which the result is immediate. {\sf QED}

Theorem 13. Suppose that \mu is a cardinal and (\kappa_\xi:\xi<\mu) is a strictly increasing sequence of nonzero cardinals with supremum \kappa. Then \kappa^\mu=\prod_\xi\kappa_\xi.

Note that the weaker claim that (\prod_\xi\kappa_\xi)^\mu=\kappa^\mu is an obvious consequence of Facts 1 (see lecture II.1) and 12.

Proof. Write \mu as a disjoint union of \mu sets A_\alpha for \alpha<\mu, each of size \mu. This is possible since \mu\times\mu=\mu. Since each A_\alpha has size \mu, it is necessarily cofinal in \mu. We then have \kappa^\mu\ge\prod_{i\in\mu}\kappa_i=\prod_{\alpha\in\mu}\prod_{i\in A_\alpha}\kappa_i\ge\prod_\alpha\kappa=\kappa^\mu. {\sf QED}

For example, \prod_{\alpha<\lambda}\aleph_\alpha=\aleph_\lambda^\lambda for any infinite cardinal \lambda. We will prove a more general result below, in Theorem 15, but notice for now that, in particular, \aleph_\omega^{\aleph_0}=\prod_n\aleph_n. One of the most celebrated results of Shelah’s pcf theory (established by a careful analysis of this product) is the fact that \aleph_\omega^{\aleph_0}\le 2^{\aleph_0}+\aleph_{\omega_4}. In particular, if \aleph_\omega is strong limit, so \aleph_\omega^{\aleph_0}=2^{\aleph_\omega} by the Bukovský-Hechler Theorem 9.2, then 2^{\aleph_\omega}\le\aleph_{\omega_4}. In fact, the inequality is strict, by König’s lemma, since {\rm cf}(\aleph_{\omega_4})=\omega_4<\aleph_\omega.  

The following is another consequence of Theorem 9:

Corollary 14. If there is an ordinal \beta such that for all \alpha we have that 2^{\aleph_\alpha}=\aleph_{\alpha+\beta}, then \beta<\omega.

Proof. Suppose that \beta\ge\omega and for all \alpha, 2^{\aleph_\alpha}=\aleph_{\alpha+\beta}. Let \alpha be least such that \beta<\alpha+\beta; \alpha exists since \beta<\beta+\beta. On the other hand, \alpha is a limit ordinal, since 1+\omega=\omega and \omega\le\beta, so (\lambda+n)+\beta=\lambda+(n+\beta)=\lambda+\beta for any \lambda and any n<\omega, and we would have a contradiction to the minimality of \alpha.

Let \gamma<\alpha. Then \gamma+\beta=\beta and 2^{\aleph_{\alpha+\gamma}}=\aleph_{\alpha+\beta}. By the Bukovský-Hechler theorem, 2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\beta} since \aleph_{\alpha+\alpha} is singular (it has cofinality {\rm cf}(\alpha)\le\alpha<\alpha+\alpha\le\aleph_{\alpha+\alpha}). This is a contradiction since, by assumption, 2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\alpha+\beta}>\aleph_{\alpha+\beta}. {\sf QED}

Remark. On the other hand, for any nonzero n\in\omega, Woodin proved that it is consistent with {\sf ZFC} that 2^{\aleph_\alpha}=\aleph_{\alpha+n}. The case n=1 is of course the {\sf GCH}. (For  n=2 this is a joint result of Foreman and Woodin.) However, for n>1, this situation is incompatible with the existence of strongly compact cardinals, as we will see in Section 4.   

Theorem 15. (Tarski).

  1. For any limit ordinal \beta, \prod_{\xi<\beta}\aleph_\xi=\aleph_\beta^{|\beta|}. 
  2. If \beta is zero or a limit ordinal then, for and any n\in\omega, \prod_{\xi\le\beta+n}\aleph_\xi=\aleph_{\beta+n}^{|\beta+n|}.

Remark. Theorem 15 does not follow from the proof of Theorem 13. The same argument of Theorem 13 actually shows that whenever (\kappa_\xi:\xi<\beta) is a strictly increasing sequence of nonzero cardinals and \beta is a countable limit ordinal, then \prod_\xi\kappa_\xi=\kappa^{\aleph_0}=\kappa^{|\beta|} where \kappa=\sup_\xi\kappa_\xi. However, the argument used to prove Theorem 13 does not apply if \beta=\omega_1+\omega, since there is no way of splitting \omega_1+\omega into \aleph_1 many cofinal sets.

Proof. Notice that item 2. is trivial if \beta=0, and an immediate consequence of 1. and Hausdorff’s formula, Corollary 11, if \beta is an infinite limit ordinal.

Now we prove 1.: Write \beta=|\beta|+\alpha where \alpha=0 or a limit ordinal, and proceed by induction on \alpha, noticing that the case \alpha=0 follows from Theorem 13.

Notice that any limit ordinal \alpha can be written as \alpha=\omega\cdot\delta for some nonzero ordinal \delta; this is easily established by induction. If \delta is a limit ordinal, then any sequence converging to \delta gives rise in a natural way to a sequence of limit ordinals converging to \alpha. On the other hand, if \delta is a successor, then \alpha=\lambda+\omega for some \lambda=0 or limit. In summary: Any limit ordinal is either a limit of limit ordinals, or else it has the form \lambda+\omega for \lambda=0 or limit.

We divide our induction on \alpha in two cases. Suppose first \alpha=\lambda+\omega, as above. Then \prod_{\xi<\beta}\aleph_\xi=\aleph_{|\beta|+\lambda}^{|\beta|}\prod_n\aleph_{|\beta|+\lambda+n}, by induction. Since |\beta|=|\beta|\aleph_0, we have \aleph_{|\beta|+\lambda}^{|\beta|} \prod_n\aleph_{|\beta|+\lambda+n}= \prod_n\aleph_{|\beta|+\lambda}^{|\beta|}\aleph_{|\beta|+\lambda+n}=\prod_n\aleph_{|\beta|+\lambda+n}^{|\beta|}, where the last equality follows from Hausdorff’s formula. Finally, \prod_n\aleph_{|\beta|+\lambda+n}^{|\beta|} =(\prod_n\aleph_{|\beta|+\lambda+n})^{|\beta|}=(\aleph_\beta)^{\aleph_0\cdot|\beta|}=\aleph_\beta^{|\beta|} as wanted, where the previous to last equality is by Theorem 13.

Now suppose that \alpha is a limit of limit ordinals. Choose a strictly increasing sequence (\gamma_\nu:\nu<{\rm cf}(\alpha)) of limit ordinals cofinal in \alpha. We argue according to which of the four possibilities described in Theorem 10 holds. If \aleph_\beta\le 2^{|\beta|} then \aleph_\beta^{|\beta|}=2^{|\beta|}\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}.

Notice that {\rm cf}(\alpha)={\rm cf}(\beta)={\rm cf}(\aleph_\beta) and {\rm cf}(\beta)\le|\beta|, so the second possibility does not occur.

Suppose now that we are in the third possibility, i.e., 2^{|\beta|}<\aleph_\beta and \rho\mapsto\rho^{|\beta|} is eventually constant as \rho approaches \aleph_\beta. Call \tau this eventual value. Then \aleph_\beta^{|\beta|}=\tau=\sup_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}^{|\beta|}=\sup_\nu\prod_{\xi<|\beta|+\gamma_\nu}\aleph_\xi, by induction (it is here that we use the assumption that \alpha is a limit of limit ordinals). And \sup_\nu\prod_{\xi<|\beta|+\gamma_\nu}\aleph_\xi \le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}.

Finally, if 2^{|\beta|}<\aleph_\beta and \rho\mapsto\rho^{|\beta|} is not eventually constant as \rho approaches \aleph_\beta, then \aleph_\beta^{|\beta|}=\aleph_\beta^{{\rm cf}(\alpha)} and \aleph_\beta^{{\rm cf}(\alpha)}=\prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}, by Theorem 13, but \prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu} \le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}, and we are done. {\sf QED}

Remark. Tarski conjectured that Theorem 13 also holds in this generality. This is independent of {\sf ZFC}. It follows, for example, from {\sf GCH}. Its failure is a strong refutation of the Singular Cardinals Hypothesis, {\sf SCH}. Indeed , there are counterexamples in models of Magidor where {\sf SCH} fails. Notice that the proof of Theorem 15 shows that the first time we run into trouble trying to generalize Theorem 13 has to be at some \beta of the form |\beta|+\lambda+\omega for some \lambda=0 or limit. Indeed, the inductive proof above in the case that \alpha is a limit of limit ordinals generalizes without difficulty. However, the proof in the case \alpha=\lambda+\omega for \lambda=0 or limit uses Hausdorff’s formula, and does not apply in general. In Thomas Jech, Saharon Shelah, On a conjecture of Tarski on products of cardinals, Proceedings of the American Mathematical Society 112, (1991), 1117–1124, it is shown that there is a counterexample if and only if there is one of shortest possible length, \beta=\omega_1+\omega.

4 Responses to 580 -Cardinal arithmetic (3)

  1. […] example, it is consistent that for all regular cardinals (as mentioned last lecture, the same result is consistent for all cardinals, as shown by Foreman and Woodin, although their […]

  2. […] For all by Tarski’s formula, Homework problem 8 from lecture II.3. It follows that for stationarily many But, for any countable limit ordinal we have that The […]

  3. […] Also, since the cofinality of is uncountable, we have Hausdoff’s result that if , then . I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here. […]

  4. […] For all by Tarski’s formula, Homework problem 8 from lecture II.3. It follows that for stationarily many But, for any countable limit ordinal we have that The […]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: