# 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

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 $\omega^\gamma$.