show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant

Question

show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant.Where $S_{\mathbb{\Omega}}$ denote the first uncountable ordinal.

There is a hint that for each $\epsilon$,there is an element $\alpha$ of $S_{\mathbb{\Omega}}$ such that $|f(\alpha)-f(\beta)|<\epsilon$ for all $\beta>\alpha$. However, I couldn't figure out how to prove this statement.

Answer 1

This result is quite difficult if you don’t know the most basic form of the pressing-down lemma:

Lemma. If $f:S_\Omega\to S_\Omega$ is a pressing-down function, meaning that $f(\alpha)<\alpha$ for each $\alpha\in S_\Omega\setminus\{0\}$, then there is an unbounded $A\subseteq S_\Omega$ on which the function $f$ is constant.

You can find a proof here, immediately followed by a proof of the result that you want. Before reading those proofs, however, you might like to try using the pressing-down lemma to prove your result. Here’s a hint to get you started: if $f:S_\Omega\to\Bbb R$ is continuous, then for each $n\in\Bbb N$ and $\alpha\in S_\Omega\setminus\{0\}$ there is a $\varphi_n(\alpha)<\alpha$ such that $|f(\beta)-f(\alpha)|<2^{-n}$ whenever $\varphi_n(\alpha)<\beta\le\alpha$. For each $n\in\Bbb N$, $\varphi_n$ is a pressing-down function.

(The full form of the lemma involves the notion of a stationary set, a complication that you don’t need here; if you’re interested, you can find it here.)

Answer 2

Hint: Here the ordinal $\Omega$ is identified with the set $S_\Omega$ of ordinals $\le \Omega$, in the order topology. What does a neighbourhood of $\Omega$ look like? What does continuity of $f$ at $\Omega$ say?

Answer 3

I found another simple proof for the hint (in fact, from Munkres's Topology: A first course).$\newcommand\abs[1]{\left\lvert#1\right\rvert}$

Suppose that there's some $\epsilon_0>0$ such that for each $\alpha\in S_\Omega$, there's some $\beta>\alpha$ such that $\abs{f(\alpha)-f(\beta)}\ge\epsilon_0$.

Let $\alpha_0=\min S_\Omega$, and given $\alpha_{n-1}$, we denote the corresponding $\beta$ as $\alpha_n$, so increasing sequence $\{\alpha_n\}$ is constructed.

Since $\{\alpha_n\}$ is countable, it's upper-bounded, and since $S_\Omega$ is well-ordered, there's a minimal upper bound $\theta$.

By continuity of $f$, we choose $\beta<\theta$ such that $f((\beta,\theta])\subset(f(\theta)-\epsilon_0/2,f(\theta)+\epsilon_0/2)$. By the minimality of $\theta$, there's an $n$ such that $\beta<\alpha_n\le\theta$, and therefore $\beta<\alpha_n<\alpha_{n+1}\le\theta$, thus $f(\alpha_n),f(\alpha_{n+1})\in(f(\theta)-\epsilon_0/2,f(\theta)+\epsilon_0/2)$, contradicts the hypothesis.