Let $(X, d)$ be a complete metric space without isolated points. If each function continues is uniformly continuous, show that $X$ is compact.

Question

Let $(X, d)$ be a complete metric space without isolated points. If each function continues $f: X \to \mathbb{R}$ is uniformly continuous, show that X is compact.

I was reviewing this question Compact iff all continuous functions are uniformly cont.. I am not very clear in the proof why the connectedness is necessary. in this case I was thinking if I could prove that $X$ is totally bounded (how could i prove it?) and use the fact that complete and totally bounded implies that $X$ is compact. The truth is I am not very clear on how to do this proof.

Answer 1

OUTLINE:

• Show that if $X$ is not totally bounded, it has a countably infinite closed discrete subset $\{x_n:n\in\Bbb Z^+\}$. For each $n\in\Bbb Z^+$ let $$\epsilon_n=\min\{d(x_n,x_m):n\ne m\in\Bbb Z^+\}\,;$$ then $\epsilon_n>0$.
• Show that for each $n\in\Bbb Z^+$ there is a $y_n\in X$ such that $$0<d(x_n,y_n)<\min\left\{\frac1n,\frac{\epsilon_n}4\right\}\,.$$
• Show that for each $n\in\Bbb Z^+$ there is a continuous $f_n:X\to[0,1]$ such that $f(x_n)=1$, and $f(x)=0$ if $d(x,x_n)\ge d(x_n,y_n)$.
• Let $f(x)=\sum_{n\in\Bbb Z^+}f_n(x)$ for each $x\in X$. Show that $f$ is continuous but not uniformly continuous.