Contents

# Contents

## Statement

###### Proposition

Using excluded middle and dependent choice then:

Let $(X,d)$ be a metric space which is sequentially compact. Then it is totally bounded metric space.

###### Proof

Assume that $(X,d)$ were not totally bounded. This would mean that there existed a positive real number $\epsilon \gt 0$ such that for every finite subset $S \subset X$ we had that $X$ is not the union of the open balls of radius $\epsilon$ around the elements of this finite subset

$X \neq \underset{s \in S}{\cup} B^\circ_s(\epsilon) \,.$

This would mean that we could inductively choose elements

• $x_0 \in X$

• $x_1 \in X\backslash B^\circ_{x_0}(\epsilon)$

• $x_{n+1} \in X \backslash \left( \underoverset{i = 0}{n}{\cup} B^\circ_{x_i}(\epsilon)\right)$

This constituted a sequence $(x_i)$, and by assumption of sequential compactness this would have a converging sub-sequence. But this would then be a contradiction, since by the above assumption that $X$ is not totally bounded we have $d(x_j, x_k) \gt \epsilon$ for all $j \neq k$. Hence we have a proof by contradiction that $(X,d)$ in fact is totally bounded.

Last revised on June 26, 2019 at 07:20:05. See the history of this page for a list of all contributions to it.