Are the strong limit cardinals precisely those of the form $\beth_\lambda$, where $\lambda$ is a limit ordinal or $0$?

Question

I know that $\aleph_\lambda$ is a weak limit cardinal iff $\lambda$ is a limit ordinal or $0$. In the absence of GCH, can we similarly prove that $\kappa$ is a strong limit cardinal iff $\kappa=\beth_\lambda$ with $\lambda$ is a limit ordinal or $0$? I'm guessing 'no', because I imagine that ZFC has models wherein the beth numbers 'skip over' some of the strong limit cardinals.

Answer 1

I’m assuming the axiom of choice. If $\lambda$ is a limit ordinal, then by definition $\beth_\lambda=\bigcup_{\alpha<\lambda}\beth_\alpha$. Let $\kappa=\operatorname{cf}\lambda$, and let $\langle\alpha_\xi:\xi<\kappa\rangle$ be an increasing sequence cofinal in $\lambda$. Then $\langle\beth_{\alpha_\xi}:\xi<\kappa\rangle$ is cofinal in $\beth_\lambda$, so for any cardinal $\mu<\beth_\lambda$ there is a $\xi<\kappa$ such that $\mu<\beth_{\alpha_\xi}$ and hence $2^\mu\le\beth_{\alpha_\xi+1}<\beth_\lambda$. It follows that $\beth_\lambda$ is a strong limit cardinal.

Added: Conversely, suppose that $\kappa>\omega$ is a strong limit. If $\beth_\alpha<\kappa$ for some ordinal $\alpha$, then $\beth_{\alpha+1}<\kappa$. Let $\lambda=\sup\{\alpha:\beth_\alpha<\kappa\}$; then $\beth_\lambda\le\kappa$, but if $\beth_\lambda<\kappa$, then $\beth_{\lambda+1}<\kappa$, contradicting the choice of $\lambda$. Thus, $\kappa=\beth_\lambda$.