# Aleph arithmetic question

We want to prove that:

$$\aleph_2^{\aleph_0} = \aleph_2\aleph_1^{\aleph_0}$$

My idea was to approach this by doing a Schroder-Bernstein style argument and proving this by showing two inequalities, but that doesn't seem to work because of the extra $\aleph_2$ on the right side. Any suggestions for this? Thanks.

• Do you know about cofinality? If not yet, what can you say about the size of the union of countably many sets, each of size $\aleph_1$? – Andrés E. Caicedo Nov 15 '13 at 16:56
• I dont yet know what cofinality is. Can you say that the size is $\aleph_1$? – barmin_kioppp Nov 15 '13 at 17:07

HINT: Prove that $\aleph_1^{\aleph_0}=2^{\aleph_0}$. Now observe that if $\aleph_2\geq2^{\aleph_0}$ equality ensues; and if the other way around then also equality ensues since in that case $\aleph_0^{\aleph_0}\leq\aleph_2^{\aleph_0}=2^{\aleph_0}$.
More generally, this is Hausdorff's formula for cardinal exponentiation applied for $\alpha=1$.
Edit: Show that generally $\kappa^\lambda=|\{A\subseteq\kappa\mid |A|=\lambda\}|$. Note that there are $\aleph_2$ ordinals of size $\aleph_1$, and every countable subset of $\aleph_2$ is in fact a countable subset of an ordinal of size $\aleph_1$.
• i managed to prove that $\aleph_1^{\aleph_0} = 2^{\aleph_0}$, but I am not quite sure why we can observe equality in the cases you pointed out -- why is that? – barmin_kioppp Nov 15 '13 at 20:43
• Both of them -- if $\aleph_2 \ge 2^{\aleph_0}$, how does that help me in proving that $\aleph_2^{\aleph_0} = \aleph_2\aleph_1^{\aleph_0}$? Sorry I don't fully understand – barmin_kioppp Nov 15 '13 at 21:40