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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.

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  • $\begingroup$ 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 $\aleph_1$? $\endgroup$ – Andrés E. Caicedo Nov 15 '13 at 16:56
  • $\begingroup$ I dont yet know what cofinality is. Can you say that the size is $\aleph_1$? $\endgroup$ – barmin_kioppp Nov 15 '13 at 17:07
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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 $\aleph_1$.

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  • $\begingroup$ 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? $\endgroup$ – barmin_kioppp Nov 15 '13 at 20:43
  • $\begingroup$ Which of them have you found problematic? $\endgroup$ – Asaf Karagila Nov 15 '13 at 21:07
  • $\begingroup$ 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 $\endgroup$ – barmin_kioppp Nov 15 '13 at 21:40
  • $\begingroup$ Well, the second one should be fairly obvious. In the meantime I've come up with a better hint for the first one, let me give it some more thought on how to formulate it well. $\endgroup$ – Asaf Karagila Nov 15 '13 at 21:43
  • $\begingroup$ still doesn't entirely make sense to me, but thanks for trying -- I apreciate it $\endgroup$ – barmin_kioppp Nov 15 '13 at 22:46

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