Why does every countable limit ordinal have cofinality $\omega$?

Question

According to Wikipedia, if $\alpha$ is a countable limit ordinal, then $\mathrm{cf}(\alpha)=\omega$. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal must be $\geq\omega$. We can now look at what the ordinal ends with. If it has a single $\omega$ at the end, then that is a cofinal subset and we are done. It may, however, not end in $\omega$. It could, for example, end in $\omega$ $\omega$s or $\omega^2$. Then we can pick one element of each $\omega$ constituting the $\omega^2$, and the set of these elements is cofinal in $\alpha$, and its type is $\omega.$ If $\alpha$ ends in $\omega^3$, we can take one element of each $\omega^2$ constituting the $\omega^2$, and get a cofinal subset of type $\omega.$

It seems to me that this reasoning should work. It should probably be done inductively. How should I do it?

Answer 1

There are two approaches here:

1. $\omega\leq\operatorname{cf}(\alpha)\leq|\alpha|\leq\alpha$, for limit ordinals $\alpha$. If you show this inequality then you are done, since countable ordinals have cardinality $\omega$.

2. You can actually construct a cofinal $\omega$-sequence. Let $f\colon\omega\to\alpha$ be a bijection. Let $\alpha_0=f(0)$, and let $\alpha_{n+1}=f(k)$ such that $k$ is the least for which $f(k)>\alpha_n$. Because ordinals are well-ordered there are no decreasing sequences, and therefore there exists such $k$. It is not hard to show that the sequence $\alpha_n$ is cofinal.

Answer 2

Hint:Let $\gamma$ be a countable ordinal. Enumarate $\gamma=\{\alpha_n:n<\omega\}$, construct $\beta_n$ as $\beta_n=\alpha_{m}$, where $m$ is the least $m<\omega$ with $\alpha_m>\beta_{n-1}$, then what is $\lim_{n\rightarrow \omega}\beta_n$?