sequentially compact metric spaces are equivalently compact metric spaces




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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



Using excluded middle and countable choice, then:

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

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

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


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

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

F nCl({x k|kn}) F_n \;\coloneqq\; Cl(\left\{ x_k \,\vert\, k \geq n \right\})

and write

U nX\F n U_n \coloneqq X \backslash F_n

for their open complements.

Assume now that the intersection of all the F nF_n were empty

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

or equivalently that the union of all the U nU_n were all of XX

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

hence that {U nX} n\{U_n \to X\}_{n \in \mathbb{N}} were an open cover. By the assumption that XX is compact, this would imply that there is a finite subset {i 1<i 2<<i k}\{i_1 \lt i_2 \lt \cdots \lt i_k\} \subset \mathbb{N} with

X =U i 1U i 2U i k =U i k. \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=F_{i_k} = \empty, which contradicts the construction of F i kF_{i_k}. Hence we have a proof by contradiction that assumption (*)(\ast) is wrong, and hence that there must exist an element

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

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

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

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

This proves that compact implies sequentially compact for metric spaces.

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

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

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

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



In the proof of prop. 1 the implication that a compact topological space is sequentially compact requires less of (X,d)(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.

Revised on June 21, 2017 04:29:34 by Urs Schreiber (