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!
Best Answer
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 $$k<k^{cf(k)}=k^{\omega}=k^l.$$