Can the function $S(\alpha)$ have a fixed point ordinal?

Question

I feel like there is no fixed point of the successor function, as for any ordinal $\alpha$, the ordinal $\alpha \cup \{\alpha\}$ must be greater. Though maybe I’m wrong and some ordinals with inaccessible cardinality are so large that the successor function means literally nothing, so I thought I’d ask.

Answer 1

Not every function can, or should, admit fixed points. Much like not every function from $\Bbb{R\to R}$ is continuous, or even continuous at a single point.

One condition which guarantees the existence of a fixed point is aptly named "normality". Recall that a function $f\colon\lambda\to\lambda$ is normal if $f(\alpha)\geq\alpha$ and for limit ordinals $f(\alpha)=\sup\{f(\beta)\mid\beta<\alpha\}$.

In that case, either $\lambda$ is of countable cofinality, or $f$ admits a fixed point (or both, of course). In fact, in case that $\lambda$'s cofinality is uncountable, one could argue that "most points are fixed points".

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Of course, the successor function does not satisfy the second condition, which is known as continuity, since if $\alpha$ is a limit ordinal, $\sup\{S(\beta)\mid\beta<\alpha\}=\alpha<S(\alpha)$. Which is exactly why continuity is necessary.

I leave you to think why also the requirement that $f(\alpha)\geq\alpha$ is necessary.