I'm reading Jech's Set Theory and I'm struggling with a remark he makes when speaking about cardinal exponentiation without GCH.
Theorem 5.16 says that without GCH the following hold:
(i) If $\kappa < \lambda$ then $2^\kappa \leq 2^\lambda$.
(ii) $ \mathrm{cf} \; 2^\kappa > \kappa.$
(iii) If $\kappa$ is a limit cardinal then $2^\kappa = (2^{<\kappa})^{\mathrm{cf} \; \kappa}$.
After the statement of the theorem Jech remarks:
For regular cardinals, the only conditions Theorem 5.16 places on the continuum function are $2^\kappa > \kappa$ and $2^\kappa \leq 2^\lambda$ if $\kappa < \lambda$.
Now, I'm having trouble understanding the first condition that $2^\kappa > \kappa$. My suspicion is that it follows from (ii) via the statement in the title of the question, i.e.
If $\kappa$ is a regular cardinal, so is $2^\kappa$.
With GCH we would have $2^\kappa = \kappa^+$ and as successor cardinal $\kappa^+$ is always regular. But without GCH I have no idea how to prove this.
Any help is appreciated!
Best Answer
No, that's not necessarily the case. König's theorem says that the cofinality of $2^\kappa$ is at least $\kappa^+$, but nothing about it being regular.
Indeed, $2^{\aleph_0}$ could be $\aleph_{\omega_1}$, and Easton's theorem implies that it is consistent with $\sf ZFC$ that if $\kappa$ is regular, then $2^\kappa=\aleph_{\kappa^+}$, which is always singular.