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 $|f(\alpha)-f(\beta)|<\epsilon$ for all $\beta>\alpha$. However, I couldn't figure out how to prove this statement. 
 A: 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.)
A: 
There is at least one point in $\mathbb{R}$ such that the preimage of all of its neighborhoods is uncountable. 

You can prove this by covering $\mathbb{R}$ with a sequence of closed intervals of length $1$, picking one with uncountable preimage, and then keep dividing the interval that you are looking at in half and keeping from those halves one that has uncountable preimage. 

There is only one such point in $\mathbb{R}$ for which all its neighborhoods have uncountable preimages

Assume that $a,b$ have the property that all their neighborhoods have uncountable preimages. Take $x_1\in\omega_1$ such that $|f(x_0)−a|<1/1$. Then take $x_2>x_1$ such that $|f(x_2)−b|<1/2$, and so on $x_{n+1}>x_n$ with $|f(x_{n+1})−c|<1/n$, where $c$ alternates between $a$ and $b$ depending on the parity of $n$. Note that the property of $a$ and $b$ allows to find each next term every time, since there are only countably many elements $<x_n$ and uncountably many in the preimage of every neighborhood of $a$ and of $b$.
Since $x_n$ is strictly increasing and countably many, they have a limit $\alpha=\cup_n x_n\in\omega_1$. But then $f(\alpha)$ must be arbitrarily close to both $a$ and $b$. Therefore $a=b$. 

The value $a$ is taken in a tail of $\omega_1$.

The preimage of $\{x:|x−a|>1/n\}$ must be countable, otherwise we could to the same argument above to find a point in it having all its neighborhoods with uncountable preimages. Therefore the preimage of $\{x\neq a\}$ is a countable union of countable subsets of $\omega_1$. This has an upper bound, $\beta$, the union of all those elements. 
All ordinals larger than $\beta$ must be mapped to the complement of $\{x\neq a\}$, i.e. $a$.
A: 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.
A: Here is another proof, picking up on Yai0Phah's post above in this thread in which he/she showed that for all $\epsilon>0$, there is an $\alpha$ such that $\beta\ge \alpha\Rightarrow |f(\beta)-f(\alpha)|<\epsilon.$
If we take $\epsilon_n=1/n$, we get an increasing sequence $(\alpha_n)$ with the property that $|f(\beta)-f(\alpha_n)|<1/n$ for all $\beta\ge \alpha_n$.
This shows that  that $f(x)=\sup (\alpha_n)$ for all $x\in [\sup (\alpha_n),\omega_1).$
A: 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?
