# If $\omega^\alpha$ has cofinality $\omega^\beta$, and $\beta\le\gamma\le\alpha$, does it also have a cofinal subset of order type $\omega^\gamma$?

Here's a question I'm having some trouble answering:

Say we have an ordinal $\omega^\alpha$ and suppose it has cofinality $\omega^\beta$, i.e., $\omega^\beta$ is the smallest order type of a cofinal subset. (Yes, it must be an initial ordinal; it's not clear to me whether that's relevant for this particular question.) Let $\gamma$ be such that $\beta\le\gamma\le\alpha$. The question is: Must $\omega^\alpha$ have a cofinal subset of order type exactly $\omega^\gamma$?

Notes: This is obviously false if we were to replace $\omega^\alpha$ with an ordinal that might not be a power of $\omega$.

Thank you!

• Your title and your question have the order of $\alpha,\beta$ and $\gamma$ flipped. – Asaf Karagila Sep 3 '18 at 7:44
• Oops! WIll fix, thanks. – Harry Altman Sep 3 '18 at 7:47

## 1 Answer

Certainly not.

First, note that if $\omega^\beta$ is an infinite cardinal, then it must be equal to $\omega$ or to $\beta$. This is even easier to see in the case where $\omega^\beta$ is regular, which is our case anyway.

Now, say $\beta>\omega$. If $\gamma=\beta+1$, then $\omega^\gamma=\omega^\beta\cdot\omega=\beta\cdot\omega$, and this ordinal has countable cofinality. In particular, if $\delta$ is any ordinal whose cofinality is $\beta$, it does not have a cofinal subset whose cofinality is $\omega$, and in particular no cofinal subset can be of type $\omega^\gamma$.

• Oh, yeah, I guess that does it; strictly speaking this also requires that it is, in fact, possible, for an uncountable ordinal to be the cofinality of something strictly larger than itself, but that's easy enough to show. I may try re-asking a refined version of this... – Harry Altman Sep 3 '18 at 7:54