Let $(M,d)$ be a metric space such that every function of $C(M,\mathbb{R})$ is also uniformly continuous.

a) Does this imply $M$ is complete?

My guess is yes. By considering a Cauchy sequence in $M$, then its image will be Cauchy in $\mathbb{R}$, so it will converge, and I guess proving the original sequence converges will be easy.

b) Does M need be compact that a) holds?

My guess is no. Meaning that, if $M$ is uncountable and the metric is such that {x} is also open, then the covering $\cup_x \{x\}$ has no finite subcover.

c) If $M$ has a finite number of isolated points, then M is compact.

This seems harder. Any ideas welcome.

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    $\begingroup$ b is correct ,,, $\endgroup$ – Guy Fsone Dec 27 '17 at 15:09
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    $\begingroup$ For what it's worth, Norman Levine studied such sets $M$ in Uniformly continuous linear sets [Amer. Math. Monthly 62 #8, October 1955, pp. 579-580] (incidentally, the title at the JSTOR front page has "set" as singular, which is incorrect), and later Norman Levine and William G. Saunders continued the study of this notion for continuous functions from one metric space to another metric space in Uniformly continuous sets in metric spaces [Amer. Math. Monthly 67 #2, February 1960, pp. 153-156]. $\endgroup$ – Dave L. Renfro Dec 27 '17 at 18:46
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    $\begingroup$ For b. it suffices to use a countable set in the discrete $0-1$-metric, uncountable is overkill. $\endgroup$ – Henno Brandsma Dec 27 '17 at 19:00

You posted three questions. Here's an answer to the second one. No, $M$ doesn't need to be compact. Consider the space $\ell^2$ with its usual metric. For each $n\in\mathbb N$, let $e_n\in\ell^2$ be the sequence such that its $n$th term is $1$ and all others are $0$. Let$$M=\left\{\frac{e_n}m\,\middle|\,m,n\in\mathbb N\right\}\cup\{0\}.$$ Then every continuous function from $M$ into $\mathbb R$ is uniformly continuous. However, $M$ is not compact. For instance, the sequence $(e_n)_{n\in\mathbb N}$ has no convergent subsequence.

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For (c) we can use the fact that $(X, d)$ is compact if and only if it's complete and totally bounded (see here).

We argue by contradiction. If $(X, d)$ is not compact, then it is not totally bounded and there is $\epsilon >0$ so that $(X, d)$ is not covered by finitely many balls of radius $\epsilon$. By induction, that means there is a sequence $\{x_n\}$ in $X$ so that

$$d(x_n , x_m) \ge \epsilon, \ \ \ \forall n\neq m.$$

Since $(X, d)$ has only finitely many isolated points, by taking a subsequence if necessary, we may assume that none of $x_n$ is isolated. So there is $\epsilon> \delta(n) \to 0$ and $y_n \in B(x_n, \delta (n)), y_n \neq x_n$.

Then define $f \in C(X, \mathbb R)$ by

$$ f (x) = \sum_{n=1}^\infty \max\left\{ 0, 1-\frac{ d(x, x_n)}{d(y_n,x_n)}\right\}.$$

Then $f(x_n) =1, f(y_n) = 0$. Since $d(y_n, x_n) \le \delta (n)$, this $f$ cannot be uniform continuous. Thus $(X, d)$ has to be totally bounded.

Remark I've used implicitly that (a) is true. To see this, one can argue similarly: if $(X, d)$ is not complete, then there is $y\in \overline X$ in the completion of $(X, d)$ and $a_n \to y, a_n \in X$ (abusing by thinking $X\subset \overline X$). But the function

$$g: X\to \mathbb R ,\ \ g(x) = \frac{1}{d_{\overline X} (x, y)}$$

is continuous and is not uniform continuous: if it is, the function $g$ can be extended to $\overline X$.

For (b), your intuition is completely correct: Just take any infinite set $X$ and give it the discrete metric. All functions are continuous and uniform continuous, but $(X, d)$ is not compact.

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