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Assume that $2^{\aleph_0}=\aleph_1$ and $(\aleph_{\omega_1})^{\aleph_0}=\aleph_{\omega_1+5}$.

Prove that there exist $\alpha<\omega_1$ such that:
$\alpha$ is a limit ordinal
$(\aleph_\alpha)^{\aleph_0}\geq\aleph_{\omega_1}$
for all $\gamma<\alpha:(\aleph_\gamma)^{\aleph_0}<\aleph_{\omega_1}$

I suspect some smart usage of Tarski, Bukovsky or Hausdorff is needed here but I do not see how to use their formulas.

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  • $\begingroup$ Where did you see this claim stated? $\endgroup$ Commented Jan 25, 2018 at 19:15
  • $\begingroup$ @NoahSchweber, I found it in archival exam. $\endgroup$ Commented Jan 25, 2018 at 19:18

1 Answer 1

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Note that any $f\!:\omega\to\aleph_{\omega_1}$ is bounded, so ${}^\omega\aleph_{\omega_1}=\bigcup_{\alpha<\omega_1}{}^\omega\aleph_\alpha$. It follows that if $\aleph_\alpha^{\aleph_0}\le\aleph_{\omega_1}$ for all countable $\alpha$, then $\aleph_{\omega_1}^{\aleph_0}\le\aleph_{\omega_1}$ as well, a contradiction.

It follows that there is some countable $\alpha$ such that $\aleph_\alpha^{\aleph_0}>\aleph_{\omega_1}$. Now, Hausdorff's formula gives us that $\aleph_{\beta+1}^{\aleph_0}=\aleph_{\beta+1}\cdot \aleph_\beta^{\aleph_0}$ (by the same argument involving cofinalities as in the previous paragraph), so the least such $\alpha$ must be $0$ or a limit ordinal. But it is not zero by assumption.

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  • $\begingroup$ And just to see that I remember how things work, it's not even known to be consistent that this can even happen. Right? $\endgroup$
    – Asaf Karagila
    Commented Jan 25, 2018 at 20:17
  • $\begingroup$ What is ${}^\omega\aleph_{\omega_1}$ as opposed to ${\aleph_{\omega_1}}^\omega$? $\endgroup$ Commented Jan 25, 2018 at 20:20
  • $\begingroup$ I am using the notation ${}^AB$ for the set of functions $f\!:A\to B$ and $|B|^{|A|}$ for its cardinality. $\endgroup$ Commented Jan 25, 2018 at 20:31
  • $\begingroup$ @Asaf Hmm... I guess you are right. $\endgroup$ Commented Jan 25, 2018 at 20:40

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