Can we find $\mu<\kappa$ with $2^{\mu^{(n)}}<\sup_{n<\omega}\mu^{(n)}$?

Question

Let $\kappa$ be inaccessible. Can we find $\mu$ with $cf(2^\mu)<\kappa$ and $2^{\mu^{(n)}}<\sup_{n<\omega}\mu^{(n)}$ for every $n<\omega$?

Answer 1

If I understand the notation correctly, if $\mu = \aleph_\alpha$, then $\mu^{(n)} = \aleph_{\alpha+n}$ and $\sup_{n<\omega}\mu^{(n)}$ is $\aleph_{\alpha+\omega}$. So you're asking for some ordinal $\alpha$ such that $2^{\aleph_{\alpha+n}} < \aleph_{\alpha+\omega}$ for all $n\in \omega$, which is equivalent to the condition that $\aleph_{\alpha+\omega}$ is a strong limit cardinal.

You also ask that $cf(2^{\aleph_\alpha})<\kappa$ for some fixed inaccessible $\kappa$. I'm not sure what the role of this condition is, since $\kappa$ doesn't appear in the rest of the statement. Why not just ask about the first inaccessible? And in any case, if $\aleph_{\alpha+\omega}$ is a strong limit, then since $\aleph_\alpha < 2^{\aleph_{\alpha}} < \aleph_{\alpha+\omega}$, we have that $2^{\aleph_{\alpha}}$ is a successor cardinal, hence regular, so $cf(2^{\aleph_{\alpha}})$ is just $2^{\aleph_{\alpha}}$. And again, if $2^{\aleph_\alpha}<\aleph_{\alpha+\omega}$, then $\aleph_\alpha<\kappa$ iff $2^{\aleph_\alpha}<\kappa$ (since if $\aleph_\alpha<\kappa$, then certainly $\aleph_{\alpha+\omega}\leq\kappa$).

So your question can be rephrased: Can we find $\mu = \aleph_{\alpha}<\kappa$ such that $\aleph_{\alpha+\omega}$ is a strong limit?

The existence of such a $\mu$ is independent from ZFC. Indeed, by Easton's theorem, it's consistent that $2^{\aleph_{\alpha+1}} = \aleph_{\alpha+\omega+1}$ for every ordinal $\alpha$ (since both these cardinals are successors, and hence regular), in which case no cardinal of the form $\aleph_{\alpha+\omega}$ is a strong limit. On the other hand, under GCH, every infinite cardinal $<\kappa$ works, since every limit cardinal is a strong limit cardinal.