4
$\begingroup$

Let $\kappa,\tau$ be two cardinals and $\{\varkappa_\alpha\}_{\alpha<\kappa}$ an indexed set of cardinals. Is it true that $$\sup_{\alpha<\kappa}(\varkappa_\alpha^\tau)=\left(\sup_{\alpha<\kappa}\varkappa_\alpha\right)^\tau ?$$

I clearly see $\leq$. However, I am unsure whether the converse inequality holds generally true. For finite $\tau$, the axiom of choice proves this statement directly noting that, for infinite $\varkappa$, $\varkappa^\tau=\varkappa$. But I'm pretty sure that the statement for finite $\tau$ can be proven without the axiom of choice. For infinite $\tau$, I also suspect that this holds in ${\sf ZF}$, but I am not so sure. Thanks in advance!

$\endgroup$

2 Answers 2

3
$\begingroup$

No, for instance under (GCH) we have:

$\sup_{n<\omega} (\aleph_n)^\omega \le \sup_{n<\omega} \aleph_{n+1} = \aleph_\omega < \aleph_{\omega+1} = (\aleph_\omega)^\omega = (\sup_{n<\omega} \aleph_n)^\omega $

$\endgroup$
3
$\begingroup$

This is typically false.

As a concrete example, take the sequence $(\beth_n)_{n<\omega}$ with $\tau=\omega$. It is easy to see that we have $(\beth_n)^{\omega}=\beth_n$ for all $n\geq 1$. Thus, the left hand side is just $\beth_\omega$, which is less than $(\beth_\omega)^{\omega}$ by König's Theorem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .