# Countable limit ordinal as limit of $\omega$ ordinals

I've got a most probably silly question, but I can't find the answer to it:

If $\alpha$ is a countable limit ordinal, how can we be sure that there exist ordinals $\alpha_n$ such that $\alpha = \bigcup \{\alpha_n\mid n\in\omega\}$ (rather than the usual $\alpha = \bigcup \{\beta\mid \beta<\alpha\}$)?

Cheers!

Well, first of all $\{\beta\mid\beta<\alpha\}=\alpha$ is countable, therefore it can be enumerated as $\alpha_n$'s. But I suppose that you want to ask about a strictly increasing sequence.
First enumerate $\alpha$ as $\{\beta_n\mid n<\omega\}$. Then by induction define $\alpha_n$, where $\alpha_{n+1}=\beta_k$ where $k=\min\{m\mid\alpha_n<\beta_m\}$. You can now show that this is a strictly increasing and cofinal sequence.
• Thanks for the answer. I'm new to ordinals and while I know (knew) that countable things can be enumerated as $\alpha_n$'s, I am now unsure why we can do so using only $\omega$, and not, say, $\omega^2$ indices to do so. Is that the definition of countability? Sorry for all the stupid questions. Commented May 5, 2014 at 19:19
• Well, being countable is by definition to have an injection into $\omega$, or in other words to be enumerated. Commented May 5, 2014 at 19:21