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Let $f: \omega_1 \to \mathbb{R}$ be a contionuous function. Prove that $f$ is eventually constant.

I was trying to prove it by contradiction but I did not have any idea.

Grzegorz Drzyzga
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  • What is $\omega_1$? – Math1000 May 06 '19 at 22:37
  • $\omega_1$ is a first uncountable oridinal with order topology – Grzegorz Drzyzga May 06 '19 at 22:40
  • Hint to get started: can you show that there exists $\alpha_0 < \omega_1$ such that whenever $\alpha_0 \le \beta < \omega_1$ and $\alpha_0 \le \gamma < \omega_1$, then $|f(\beta) - f(\gamma)| < 1$? – Daniel Schepler May 06 '19 at 22:50
  • This question does not duplicate the cited question. The cited question uses the fact that the function $f$ is increasing. This question is more general because it assumes only that $f$ is continuous. – Robert Shore May 06 '19 at 23:37
  • 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 the one with uncountable preimage, and then keep dividing it in half and keeping one of the half that has uncountable preimage. – logarithm May 07 '19 at 00:10
  • Now assume that two points $a,b$ have the property that all its neighborhoods have uncountable preimages. Take $x_0\in\omega_1$ such that $|f(x_0)-a|<1/2$ then take $x_2>x_1$ such that $|f(x_2)-b|<1/4$, and so on $x_{n+1}>x_n$ with $|f(x_{n+1})-c|<1/2^n$, 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, since there are only countably many elements $<x_n$. – logarithm May 07 '19 at 00:10
  • Since $x_n$ is strictly increasing and countably many they have a limit $\alpha$. But then $f(\alpha)$ must be arbitrarily close to $a$ and $b$. Therefore $a=b$. Now, 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 neighborhood 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. All ordinals larger than that bound must be mapped to $a$. – logarithm May 07 '19 at 00:10
  • @AsafKaragila Maybe show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant is a better duplicating target? – YuiTo Cheng May 07 '19 at 02:01
  • @YuiToCheng: Ah, yes, I was thinking of that one, actually. But found the other one first. – Asaf Karagila May 07 '19 at 05:54

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