## 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 […]