Exponentiation of singular cardinals

Question

A cardinal $\kappa$ is regular if and only if $\kappa=cf(\kappa)$. So if $\kappa$ is regular, $\lambda<\kappa$ and $f:\lambda\to\kappa$, then $f$ is bounded in $\kappa$, i.e. there is $\mu<\kappa$ such that $f\in\mu^\lambda$.

This leads to the observation that $$\kappa^\lambda=\bigcup_{\mu<\kappa}\mu^\lambda.$$

I would like to find a counterexample to this equality in the case $\kappa$ is a singular cardinal.

My attempt consists of considering $\kappa=\aleph_\omega$, $\lambda=cf(k)=\aleph_0$ and using the fact that $\aleph_\omega^{\aleph_0}>\aleph_\omega$, to conclude computing $\bigcup_{\mu<\kappa}\mu^\lambda=\bigcup_{n\in\omega}\aleph_n^{\aleph_0}$, which I would like to be $\aleph_\omega$. Does this work? If yes, how one can prove this last fact?

Thanks in advance!

Answer 1

Let $k=\beth (\omega)=\cup_{n\in \omega}\beth (n)$ where $\beth(0)=\omega$ and $\beth(n+1)=2^{\beth (n)}.$ Let $l=\omega.$

If $m<k$ then for some $n\in \omega$ we have $m\le \beth(n)$ so $$m^l\le (\beth(n))^l\le (2^{\beth (n)})^l=2^{\beth(n)\cdot l}=2^{\beth(n)\cdot \omega}= 2^{\beth (n)}=\beth (n+1)\le k.$$ So we have $$\bigcup_{m<k}\,m^l\le \left|\bigcup_{m<k}(\{m\}\times k)\right|=k\cdot k=k.$$

By Konig's Theorem (a.k.a. Konig's Lemma ) we have $$k<k^{cf(k)}=k^{\omega}=k^l.$$

Answer 2

That's the right idea, but it's not that simple: keep in mind that we could have e.g. $2^{\aleph_0}=\aleph_{\omega^2+17}$, so $(\aleph_\omega)^{\aleph_0}>\aleph_\omega$ does not obviously tell us that $(\aleph_\omega)^{\aleph_0}>(\aleph_n)^{\aleph_0}$ for all $n<\omega$.

Of course, if we assume GCH then we get $(\aleph_n)^{\aleph_0}=\aleph_n$ for all $n<\omega$. This suggests that we replace the $\aleph$-hierarchy, which may or may not play well in this context, with something else - and then look at the $\omega$th term of that hierarchy.