show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant 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 $|f(\alpha)-f(\beta)|&lt;\epsilon$ for all $\beta&gt;\alpha$. However, I couldn't figure out how to prove this statement.
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 $f(\alpha)&lt;\alpha$ 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 $\varphi_n(\alpha)&lt;\alpha$ such that $|f(\beta)-f(\alpha)|&lt;2^{-n}$ whenever $\varphi_n(\alpha)&lt;\beta\le\alpha$. 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.)
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?