# Transfinite fixed points of a function

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

$F(0) = \aleph_0$

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

$F(\lambda) = \sup\{F(\alpha): \alpha \lt \lambda\}$ 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.

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

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.

## 1 Answer

Fixed points are points such that $F(\alpha)=\alpha$. It is certainly not true that $F(\omega_\omega)=\omega_\omega$, and it depends on the continuum function (which can be changed quite wildly between models of $\sf ZFC$) whether or not $2^{\aleph_n}=\aleph_k$ for $k<\omega$ at all.

The easy part, that a fixed point is always a cardinal, just show that $F(\alpha)$ is always a cardinal.

To show that there is a fixed point at all, you need to construct a sequence which has gaps that are increasing further and further. What happens $\lambda_{n+1}=F(\aleph_{\lambda_n})$? What is the limit of this sequence?

(As a side remark, this function is often denoted by $\beth$, and the numbers are called $\beth$ numbers.)