How does cf(cf $\alpha$) = cf $\alpha$?

Question

Assume $\alpha > 0$ is a limit ordinal, and cf $\alpha=$ the least ordinal $\beta$ such that there is an increasing $\beta$-sequence $\langle \alpha_\xi \, \colon\, \, \xi < \beta\rangle$ that is cofinal in $\alpha,$ i.e., with lim$_{\xi \to\beta}\, \alpha_\xi = \alpha.$

The proof goes like this: If $\langle \alpha_\xi \,\colon\, \, \xi < \beta\rangle$ is cofinal in $\alpha$ and $\langle \xi(\nu) \, \colon\, \, \nu < \gamma\rangle$ is cofinal in $\beta,$ then $\langle \alpha_{\xi(\nu)} \colon\, \, \nu < \gamma\rangle$ is cofinal in $\alpha.$

If we're supposing that cf $\alpha = \beta,$ then cf $\beta = \gamma \not= \beta$. I don't really think I understand anything that's going on at all, and any help will be appreciated.

Answer 1

It should be clear that $\operatorname{cf}(\alpha)\leq \alpha$ for all $\alpha$, so specifically $\operatorname{cf}(\operatorname{cf}(\alpha))\leq \operatorname{cf}(\alpha)$.

Let's assume $\gamma = \operatorname{cf}(\operatorname{cf}(\alpha)) < \operatorname{cf}(\alpha) = \beta$. Then what you have above shows that you can construct a cofinal sequence of length $\gamma$ in $\alpha$, i.e. $\operatorname{cf}(\alpha)\leq\gamma$ which contradicts our assumption.

Answer 2

I'll flesh out the details for you.

Write $\beta = \mathrm{cf}(\alpha)$ and $\gamma = \mathrm{cf}(\beta)$. Certainly $\gamma \le \beta$ since $\langle \zeta : \zeta < \beta \rangle$ is a cofinal sequence in $\beta$ of length $\beta$, so it suffices to prove $\gamma \ge \beta$.

Let $\langle \alpha_{\xi} : \xi < \beta \rangle$ be a cofinal sequence in $\alpha$; by definition of cofinality, $\beta$ is the least ordinal for which such a sequence exists.

Let $\langle \xi(\nu) : \nu < \gamma \rangle$ be a cofinal sequence in $\beta$. Then $\langle \alpha_{\xi(\nu)} : \nu < \gamma \rangle$ is a cofinal sequence in $\alpha$ of length $\gamma$, so by minimality of $\beta$ we must have $\gamma \ge \beta$.

So $\gamma = \beta$.