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?


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 $[\alpha,\Omega)$.

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

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