Vtyjkyuk

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 $betalegammalealpha$. 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!

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 $omega^gamma=omega^betacdotomega=betacdotomega$, 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 $omega^gamma$.