If every continuous function $f:X\to\mathbb{R}$ (where $X$ is a subset of a compact metric space), is uniformly continuous, then am I right to assume that $X$ is compact as well?

I think it should but I'm not sure if I am correct. Like if $X$ is a subset of a compact metric space then shouldn't $X$ itself the domain be compact as well if $f$ is uniformly continuous?


Let $K$ be a compact metric space such that $X \subseteq K$. Suppose $X$ is not compact, then $X$ is not closed in $K$, hence there is a point $k \in \bar X \setminus X$. Consider the function $$ f \colon X \to \mathbb R, \quad x \mapsto \frac 1{d(x,k)} $$ ($d$ denoting the metric of $K$). Then $f$ is continuous, but not uniformly continuous, as in the letter case, $f$ would have a continuous extension to $\bar X$, which is impossible.

  • $\begingroup$ So I'm wrong afterall? oh boy $\endgroup$ – user97994 Oct 1 '13 at 11:33
  • 2
    $\begingroup$ You aren't wrong. Why do you think so? $\endgroup$ – martini Oct 1 '13 at 11:34
  • $\begingroup$ @user97994 He's shown "$X$ not compact $\implies$ there exists a continuous, not uniformly continuous $f$". This is equivalent to what you wanted. $\endgroup$ – Arthur Oct 1 '13 at 11:35

You might be interested in the following paper:

Is Every Continuous Function Uniformly Continuous? by Ray F. Snipes Mathematics Magazine, Vol. 57, No. 3 (May, 1984), pp. 169-173 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2689666 .


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