3
$\begingroup$

It is 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 $f\colon X \rightarrow \mathbb R$ be a continuous, real-valued function on $X$.

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

$\endgroup$
4
$\begingroup$

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 $[\alpha,\Omega)$.

$\endgroup$
  • $\begingroup$ Thank you for your clear proof! $\endgroup$ – topsi Feb 14 '14 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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