Proving the equality of two cardinals. I'm having problems proving that $\aleph_{\omega}^{\aleph_{1}}=\aleph_{\omega}^{\aleph_{0}}2^{\aleph_{1}}$, with $\omega$ the ordinal of natural numbers. I don't know how to start. Thanks:)
 A: I assume you can prove the $\geq$ half and the problem is the $\leq$ half. The idea here is that any function $f:\aleph_1\to\aleph_\omega$ is determined by specifying, for all $n\in\omega$, the set $A_n=\{\alpha\in\aleph_1:f(\alpha)\in\aleph_n\}$ and the restriction $f_n=f\upharpoonright A_n$. So to count the $f$'s, we should count the number of possibilities for $A_n$ and $f_n$.
For any fixed $n$, there are $2^{\aleph_1}$ possibilities for $A_n$; the problem is to count (in a useful way) the possibilities for $f_n$. Clearly, the number of possibilities for $f_n$ is at most $\aleph_n^{\aleph_1}$. I claim that this cardinal is bounded by $2^{\aleph_1}\cdot\aleph_n$. If I can prove this claim, then the desired result follows, because there are, for each $n$, at most $2^{\aleph_1}\cdot\aleph_n$ pairs $(A_n,f_n)$, and so the number of possible $f$'s is at most $\prod_{n\in\omega}(2^{\aleph_1}\cdot\aleph_n)\leq 2^{\aleph_1}\cdot\aleph_\omega^{\aleph_0}$.
So it remains to prove the claim that $\aleph_n^{\aleph_1}\leq 2^{\aleph_1}\cdot\aleph_n$, and I'll do this by induction on $n$. The cases $n=0$ and $n=1$ are easy, so I'll do the induction step. Consider functions $g:\aleph_1\to\aleph_{n+1}$, where $n+1\geq2$. Any such $g$ has range bounded by some $\alpha<\aleph_{n+1}$ (because $\aleph_{n+1}$ is a regular cardinal), so the set of all such $g$'s, whose cardinality is $\aleph_{n+1}^{\aleph_1}$, is the union over all $\alpha<\aleph_{n+1}$ of the sets $\{g:g\text{ maps }\aleph_1\text{ into }\alpha\}$.  So it's the union of $\aleph_{n+1}$ sets, each of cardinality at most $\aleph_n^{\aleph_1}$. Apply the induction hypothesis to bound $\aleph_n^{\aleph_1}$ and you're done.
