How to show that$(X,d)$ is totally bounded

Question

How to show that a metric space $(X,d)$ is totally bounded $\iff$ every infinite subsets of $X$ contains distinct points which distinct points that arbitrarily close to each other.

I don't know how to prove $\Leftarrow.$ Please help.

BTW I'm looking for a little hint !

$X$ is totally bounded iff any sequence(net) in it has a Cauchy subsequence(subnet). — , May 14 '13 at 10:13

score 2 · Answer 1 · answered May 14 '13 at 18:48

2

Suppose that $\langle X,d\rangle$ is not totally bounded. Then there is an $\epsilon>0$ such that for all finite $F\subseteq X$, $\{B(x,\epsilon):x\in F\}$ does not cover $X$, i.e., $X\setminus\bigcup_{x\in F}B(x,\epsilon)\ne\varnothing$. This implies that we can recursively construct a set $A=\{x_n:n\in\Bbb N\}\subseteq X$ such that

$$x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$$

for each $n\in\Bbb N$. $A$ is an infinite subset of $X$; does it contain distinct points that are arbitrarily close to each other?

answered May 14 '13 at 18:48

Brian M. Scott

616,228

A is infinite? is it because there are infinitely many finite F? also, could you please your conclusion remark? – 104078 Nov 24 '13 at 20:21
@user104078: $A$ is infinite because the map $n\mapsto x_n$ is an injection: if $n>m$, $x_n\notin B(x_m,\epsilon)$, so $x_n\ne x_m$. In fact, $d(x_n,x_m)\ge\ldots;$ what? – Brian M. Scott Nov 24 '13 at 20:26
1

\epsilon. so the contrapositive is proved. – 104078 Nov 24 '13 at 20:50
@user104078: Exactly right. – Brian M. Scott Nov 24 '13 at 20:51

How to show that$(X,d)$ is totally bounded

1 Answers1