some basic cardinal arithmetic on $\text{cf}(\aleph_{\omega_1})$

I'm reading The Joy of Sets by K. Devlin, by self-study. I've just seen a statement $\text{cf}(\aleph_{\omega_1})=\omega_1$ without proof, but I think this is slightly harder to prove than more obvious one, $\text{cf}(\aleph_{\omega})=\omega$.

Concretely, $\text{cf}(\aleph_{\omega_1})\le \omega_1$ is trivial, but.. how about the other direction? I've tried and I guess $\aleph_{\omega_1}^{\aleph_0}=\aleph_{\omega_1}$ (as cardinal exp.) implies the result desired. How can I prove the last identity, or is there a simpler proof on the cofinality?

This is a consequence of a more general fact:

Suppose $\gamma$ is a limit ordinal and I have a sequence of ordinals $\langle \alpha_{\eta}:\eta<\theta\rangle$, with each $\alpha_\eta<\omega_\gamma$. Then this induces a sequence of ordinals $\langle \beta_\eta: \eta<\theta\rangle$, each $<\gamma$, as follows: $\beta_\eta$ is the least $\delta$ such that $\alpha_\eta<\omega_\delta$.

The fact: if $\langle \alpha_\eta: \eta<\theta\rangle$ is cofinal in $\omega_\gamma$, then $\langle \beta_\eta: \eta<\theta\rangle$ is cofinal in $\gamma$.

Setting $\gamma=\omega_1$ addresses your specific question.

• Thank you, I wish I could choose all of the answers.. Aug 12, 2015 at 1:52

Suppose that $\langle\alpha_n:n\in\omega\rangle$ is an increasing sequence cofinal in $\omega_{\omega_1}$. For each $\xi<\omega_1$ there is a least $n(\xi)\in\omega$ such that $\omega_\xi\le\alpha_{n(\xi)}$. Clearly there is then an $m\in\omega$ such $X=\{\xi<\omega_1:n(\xi)=m\}$ is uncountable, and it follows easily that $\omega_{\omega_1}=\sup_{\xi\in X}\omega_\xi\le\alpha_m<\omega_{\omega_1}$, which is absurd.

• Clear and simple. Thanks! Aug 12, 2015 at 2:00
• @D.Lee: You’re welcome! Aug 12, 2015 at 2:01

The actual question has been answered, but it should still be pointed out that $\aleph_{\omega_1}^{\aleph_0}=\aleph_{\omega_1}$ is not necessarily true, or rather it is not provable without additional assumptions.

For example, it is possible that $2^{\aleph_0}=\aleph_{\omega_2}$, in which case it is trivial that $\aleph_{\omega_1}^{\aleph_0}\geq\aleph_{\omega_2}$.