 Solution

By the natural-number recursion theorem, let ${\alpha }_0=0$ and ${\alpha }_{n+1}=F({\alpha }_n)$. These ordinals form a nondecreasing sequence by the preceding bound. If ${\alpha }_{n+1}={\alpha }_n$ for some $n$, it is already a fixed point.

Otherwise the sequence is strictly increasing. Its supremum $\lambda ={\bigcup }_{n<\omega }{\alpha }_n$ is a limit ordinal, and the sequence is cofinal in it. Continuity and monotonicity giveFor the middle equality, every $\beta <\lambda$ is below some ${\alpha }_n$, while every ${\alpha }_n<\lambda$. Therefore $F$ has a fixed point. The sequence is a set by Axiom schema of replacement, since $F$ is a definable set-theoretic class function.