When every continuous function on a metric space is also uniformly continuous.

Question

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.

Answer 1

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.

Answer 2

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.