Contents

# Contents

## Idea

For general topological spaces the condition of being compact neither implies nor is implied by being sequentially compact. However for metric spaces the two conditions happen to be equivalent

## Statement

###### Proposition

Using excluded middle and countable choice, then:

If $(X,d)$ is a metric space, regarded as a topological space via its metric topology, then the following are equivalent:

1. $(X,d)$ is a compact topological space.

2. $(X,d)$ is a sequentially compact topological space.

###### Proof

Assume first that $(X,d)$ is a compact topological space. Let $(x_k)_{k \in \mathbb{N}}$ be a sequence in $X$. We need to show that it has a sub-sequence which converges.

Consider the topological closures of the sub-sequences that omit the first $n$ elements of the sequence

$F_n \;\coloneqq\; Cl(\left\{ x_k \,\vert\, k \geq n \right\})$

and write

$U_n \coloneqq X \backslash F_n$

for their open complements.

Assume now that the intersection of all the $F_n$ were empty

$(\star) \phantom{AA} \underset{n \in \mathbb{N}}{\cap} F_n \;= \; \emptyset$

or equivalently that the union of all the $U_n$ were all of $X$

$\underset{n \in \mathbb{N}}{\cup} U_n \;=\; X \,,$

hence that $\{U_n \to X\}_{n \in \mathbb{N}}$ were an open cover. By the assumption that $X$ is compact, this would imply that there is a finite subset $\{i_1 \lt i_2 \lt \cdots \lt i_k\} \subset \mathbb{N}$ with

\begin{aligned} X & = U_{i_1} \cup U_{i_2} \cup \cdots \cup U_{i_k} \\ & = U_{i_k} \end{aligned} \,.

This in turn would mean that $F_{i_k} = \empty$, which contradicts the construction of $F_{i_k}$. Hence we have a proof by contradiction that assumption $(\ast)$ is wrong, and hence that there must exist an element

$x \in \underset{n \in \mathbb{N}}{\cap} F_n \,.$

By definition of topological closure this means that for all $n$ the open ball $B^\circ_x(1/(n+1))$ around $x$ of radius $1/(n+1)$ must intersect the $n$th of the above subsequence:

$B^\circ_x(1/(n+1)) \,\cap\, \{x_k \,\vert\, k \geq n \} \;\neq\; \emptyset \,.$

Picking one point $(x'_n)$ in the $n$th such intersection for all $n$ hence defines a sub-sequence, which converges to $x$.

This proves that compact implies sequentially compact for metric spaces.

For the converse, assume now that $(X,d)$ is sequentially compact. Let $\{U_i \to X\}_{i \in I}$ be an open cover of $X$. We need to show that there exists a finite sub-cover.

Now by the Lebesgue number lemma, there exists a positive real number $\delta \gt 0$ such that for each $x \in X$ there is $i_x \in I$ such that $B^\circ_x(\delta) \subset U_{i_x}$. Moreover, since sequentially compact metric spaces are totally bounded, there exists then a finite set $S \subset X$ such that

$X \;=\; \underset{s \in S}{\cup} B^\circ_s(\delta) \,.$

Therefore $\{U_{i_s} \to X\}_{s \in S}$ is a finite sub-cover as required.

## Remarks

###### Remark

In the proof of prop. the implication that a compact topological space is sequentially compact requires less of $(X,d)$ than being a metric space. Actually, the proof works for any first-countable space that is a countably compact space, i. e. any countable open cover admits a finite sub-cover. Hence countably compact metric spaces are equivalently compact metric spaces.

Last revised on June 5, 2022 at 15:26:35. See the history of this page for a list of all contributions to it.