# Assuming $2^{\aleph_0}=\aleph_1$,$(\aleph_{\omega_1})^{\aleph_0}=\aleph_{\omega_1+5}$ prove existence of $\alpha\in LIM$ fulfilling some conditions

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.

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

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.
• What is ${}^\omega\aleph_{\omega_1}$ as opposed to ${\aleph_{\omega_1}}^\omega$? Jan 25, 2018 at 20:20
• I am using the notation ${}^AB$ for the set of functions $f\!:A\to B$ and $|B|^{|A|}$ for its cardinality. Jan 25, 2018 at 20:31