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 21, 2017 at 04:29:34. See the history of this page for a list of all contributions to it.