elementary set theory - Fixed points and cardinal exponentiation

Let the function $F: On \rightarrow On$ be defined by the following recursion:

$F(0) = \aleph_0$

$F(\alpha+1) = 2^{F(\alpha)}$ (cardinal exponentiation)

$F(\lambda) = \sup\{F(\alpha):\alpha \in \lambda\}$ for $\lambda$ a limit ordinal

Prove that there is a fixed point for $F$, i.e. an ordinal $\kappa$ with $F(\kappa) = \kappa$.

Are such fixed points always cardinals?

Thoughts:
So I can see that such a fixed point is going to have to be for a limit ordinal, since the function is strictly increasing for successor ordinals.

$F(\lambda) = \sup\{\aleph_{\alpha}: \alpha \in \lambda\}$

I feel as if $\aleph_{\omega}$ might be a fixed point and suspect that any fixed points have to be cardinals, but I don’t have a justification for either.

I’m not sure how to go about proving a fixed point exists and whether it has to always be a cardinal.

Answer

What you have defined is in fact the $\beth$ function.

It is a continuous function, namely, it is increasing and the limit is evaluated as the limit of previous results. Therefore there is a proper class of fixed points. The fact that cardinal exponentiation returns a cardinal, and the supremum of cardinals is a cardinal ensures that such fixed points will necessarily be cardinals.

$\aleph_\omega$ is never a fixed point, though. Because even if $\aleph_\omega$ is a strong limit (namely, $2^{\aleph_n}<\aleph_\omega$ for all $n<\omega$), it is not the case that $\aleph_\omega=F(\omega_\omega)$. The first fixed point, with---and certainly without---assuming some relatively tame continuum function is going to be unfathomably larger than $\aleph_\omega$. How large is it going to be? Well, just think about it. It will be an ordinal which satisfies $\alpha=F(\alpha)$. It will have $\alpha$ cardinals which are strictly smaller than itself. $\aleph_\omega$ has only $\omega$ (well, $\omega+\omega$ if you count the finite cardinals).

There is no way to describe it "nicely" from below in any reasonable way.