Let $M$ be a metric space. If every continuous function $f : M \rightarrow \mathbb{R}$ has compact range then $M$ is compact.

Question

I got this question from "Real Mathematical Analysis" by Charles Chapman Pugh. Specifically Exercise 118 in Chapter 2. I'm not sure how to go about proving this question, but it seems useful for another equivalence of what it means for a metric space to be compact.

EDIT: I don't see how this question is a duplicate of the other link in question. Or I at least don't understand how it is a duplicate.

If there exists an injective function, you are done by the open mapping theorem. — Andres Mejia, Dec 15 '17 at 00:33
@Darkdub your first sentence is what you're trying to prove! — Andres Mejia, Dec 15 '17 at 00:33

score 4 · Accepted Answer · edited Dec 15 '17 at 18:09

4

A stronger result is this: if every real continuous function on $M$ is bounded then $M$ is compact. To prove this suppose M is not compact. Then there is a sequence $\{x_n\}$ of distinct points with no convergent subsequence. Let $A=\{x_1,x_2,\ldots\}$. Then $A$ has no limit points in $M$ and hence it is closed. Define $f:A \to \mathbb R$ by $f(x_n)=n$. Then $f$ is continuous and Tietze Extension Theorem shows that this function can be extended to a continuous function $F$ on $M$. Clearly, $F$ is continuous but not bounded.

edited Dec 15 '17 at 18:09

José Carlos Santos

427,504

answered Dec 15 '17 at 07:58

Kavi Rama Murthy

311,013

This is great! Thank you. – Darkdub Dec 15 '17 at 12:33

Let $M$ be a metric space. If every continuous function $f : M \rightarrow \mathbb{R}$ has compact range then $M$ is compact.

1 Answers1