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!
Best Answer
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 $