# 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 ProofOfFixedPointsOfNormalFunctions 2013-03-22 13:29:01 2013-03-22 13:29:01 Henry (455) Henry (455) 4 Henry (455) Proof msc 03E10