If $A$ is not totally bounded then $A$ has an infinite subset $B$ homeomorphic to a discrete space.

My approach since $A$ is not totally bounded we can find $\epsilon>0$ and a sequence $x_n$ suchthat $d(x_m, x_n)> \epsilon$. Then what to do? any idea?


1 Answer 1


Just show that $\{x_n:n\in\mathbb{N}\}$ is discrete, i.e. every singleton is an open set.

For each $n\in\mathbb{N}$, show that $B(x_n,\epsilon)$ only contains $x_n$.

  • $\begingroup$ how to show this? $\endgroup$
    – 00012 suxn
    Nov 26, 2015 at 15:01

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