cf for countable limit ordinal

Question

My question is related to the cofinality for ordinal number. We know $cf(c)>\omega$ , $cf(\alpha)=1$ for any successor ordinal. I am trying to show the $cf(\alpha)=\omega$ for any limit ordinal $\alpha<\omega_1.$ my guess this can be done by constructed a cofinal map. Any help will appreciated.

Answer 1

Asking about countable $\alpha<\omega$ doesn't make much sense; surely the actual question is to show that any countable limit ordinal has cofinality $\omega$.

For the revised version, it's clear that the cofinality cannot be finite so we need only show it's no larger than $\omega$. Since $\alpha$ is countable we can write $$\alpha=\{a_j:j\in\omega\}.$$Since $\alpha$ is a limit ordinal, if $F\subset\alpha$ is finite there exists $x\in\alpha$ with $x>y$ for all $y\in F$.

So we can define $f:\omega\to\alpha$ by saying that $f(n)$ is the smallest element of the set $\{x\in\alpha: x>a_j,0\le j\le n\}$. It's clear that $f$ is cofinal.