Does $\sf GCH$ imply that every uncountable cardinal is of the form $2^\kappa$?

Question

I think that this is a popular fallacy that GCH implies that every uncountable cardinal is of the form $2^\kappa$ for some $\kappa$. I think it does imply that for successor cardinals only. It cannot be true for all limits, right?

Answer 1

In fact, one can prove in ZFC that some infinite cardinals are not of form $2^\kappa$, regardless of whether the GCH holds or not. For example, $\aleph_\omega$ cannot be $2^\kappa$ for any $\kappa$, since it has cofinality $\omega$, and so $(\aleph_\omega)^\omega>\aleph_\omega$ by König's theorem, whereas $(2^\kappa)^\omega=2^{\kappa\cdot\omega}=2^\kappa$ for any infinite $\kappa$.

Answer 2

Yes, you are right. But even more than that. Since $2^\kappa=\kappa^+$ for every infinite cardinal, it follows that there is no $\kappa$ such that $2^\kappa$ is a limit cardinal.

To see why, note that if $2^\kappa=\lambda$ then $\kappa<\lambda$, if $\lambda$ is a limit cardinal then $2^\kappa=\kappa^+<\lambda$ as well.

(If you look closely at this argument, you'll see that it shows that if $\delta$ is a limit ordinal, then $\beth_\delta$ is not $2^\kappa$ for any $\kappa$. And under $\sf GCH$ the definitions of $\aleph$ and $\beth$ coincide.)

If you want to talk about "most", then in some sense this show that most cardinals are not $2^\kappa$, since the limit cardinals make a closed and unbounded class in the ordinals (and in the cardinals). Therefore we can regard them as "full measure", in some sense.