Cofinalities of ordinals $\omega_1 \omega, \omega_2 \omega_1, \omega_2 \omega_1 \omega$?

Question

I am trying to understand cofinalities so I would really appreciate it if you could confirm whether the following line of reasoning is correct. (All multiplication is ordinal multiplication)

$$ cf(\omega_1 \omega)=\omega \text{ (since } \omega_1 \omega=\sup_{n\in \omega} \omega_1 n)$$ $$ cf(\omega_2 \omega_1)=\omega_1 \text{ (since } \omega_2 \omega_1=\sup_{\alpha\in \omega_1} \omega_2 \alpha) $$ $$ cf(\omega_2 \omega_1 \omega)=\omega \text{ (since } \omega_2 \omega_1 \omega=\sup_{n\in \omega} \omega_1 \omega_2 n) $$ I am pretty sure about the first and the last one. But for the second one I would somehow have to show that there is no shorter sequence which "converges" to $\omega_2 \omega_1$. Any insights would be greatly appreciated, as I am really struggling to fully grasp the concept of cofinality.

Thanks in advance!

Answer 1

Your answers are correct. The thing about the second one is that if you had a shorter cofinal sequence, say, $\alpha_n$, then for each one we can pick the least $\omega_2\beta_n$ which is above it.

But that means that $\sup\beta_n=\omega_1$, which (at least assuming the Axiom of Choice) is impossible.

The whole idea is that cofinality is unique, so given two cofinal sequences we can interleave them (at least on a tail segment).