proof of fixed points of normal functions

Suppose $f$ is a $\kappa$ -normal function and consider any $\alpha<\kappa$ and define a sequence by $\alpha_{0}=\alpha$ and $\alpha_{n+1}=f(\alpha_{n})$ . Let $\alpha_{\omega}=\sup_{n<\omega}\alpha_{n}$ . Then, since $f$ is continuous,

f(\alpha_{\omega})=\sup_{n<\omega}f(\alpha_{n})=\sup_{n<\omega}\alpha_{n+1}=% \alpha_{\omega}

So $\operatorname{Fix}(f)$ is unbounded.

Suppose $N$ is a set of fixed points of $f$ with $|N|<\kappa$ . Then

f(\sup N)=\sup_{\alpha\in N}f(\alpha)=\sup_{\alpha\in N}\alpha=\sup N

so $\sup N$ is also a fixed point of $f$ , and therefore $\operatorname{Fix}(f)$ is closed.

Title	proof of fixed points of normal functions
Canonical name	ProofOfFixedPointsOfNormalFunctions
Date of creation	2013-03-22 13:29:01
Last modified on	2013-03-22 13:29:01
Owner	Henry (455)
Last modified by	Henry (455)
Numerical id	4
Author	Henry (455)
Entry type	Proof
Classification	msc 03E10