Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As a continuation to this question. Let $X$ be the ordinal space $[0,\Omega)$, with the order topology, where $\Omega$ is the first uncountable ordinal.

Let $f:X \rightarrow \mathbb R$ be a continuous real valued function defined on $X$.

Can we say that $f$ is constant on some segment $(\alpha,\Omega)$, where $\alpha$ is a countable ordinal?

Thank you!

share|improve this question

1 Answer 1

Yes, every continuous $f \colon X \to \mathbb{R}$ is eventually constant.

There is a sequence $(\alpha_n)$ in $X$ such that for all $n$

$$\sup_{\beta > \alpha_n} \lvert f(\beta) - f(\alpha_n)\rvert \leqslant 2^{-n}.$$

For otherwise, there would be a $k\in \mathbb{N}$ such that for every $\alpha \in X$ there is a $\beta > \alpha$ with $\lvert f(\beta) - f(\alpha)\rvert > 2^{-k}$. Then we could construct a sequence $(\gamma_n)$ with $\gamma_n < \gamma_{n+1}$ and $\lvert f(\gamma_{n+1}) - f(\gamma_n)\rvert > 2^{-k}$. But the sequence $(\gamma_n)$ converges to its supremum $\gamma \in X$, and hence

$$\lvert f(\gamma_{n+1}) - f(\gamma_n)\rvert \leqslant \lvert f(\gamma_{n+1}) - f(\gamma)\rvert + \lvert f(\gamma) - f(\gamma_n)\rvert < 2^{-(k+1)}$$

for $n$ so large that $\lvert f(\gamma_m) - f(\gamma)\rvert < 2^{-(k+2)}$ for all $m \geqslant n$.

The existence of the sequence $(\alpha_n)$ established, let $\alpha = \sup\limits_{n\in\mathbb{N}} \alpha_n$.

Then $f$ is constant on $[\alpha,\Omega)$.

share|improve this answer
    
Thank you for your clear proof! –  Shir Sivroni Feb 14 at 14:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.