2.5. COMPACTNESS AND CONTINUOUS FUNCTIONS 45

2.5 Compactness and Continuous FunctionsAs usual, we are not worrying about empty sets. Fussing over these is usually a waste oftime. Thus if a set is mentioned, the default is that it is nonempty.

Definition 2.5.1 A metric space K is compact if whenever C is an open cover of K,there exists a finite subset of C {U1, · · · ,Un} such that K ⊆ ∪n

k=1Uk. In words, every opencover admits a finite sub-cover.

Directly from this definition is the following proposition.

Proposition 2.5.2 If K is a closed, nonempty subset of a nonempty compact set H, thenK is compact.

Proof: Let C be an open cover for K. Then C ∪{

KC}

is an open cover for H. Thusthere are finitely many sets from this last collection of open sets, U1, · · · ,Um which coversH. Include only those which are in C . These cover K because KC covers no points of K. ■

This is the real definition given above. However, in metric spaces, it is equivalent toanother definition called sequentially compact.

Definition 2.5.3 A metric space K is sequentially compact means that whenever{xn} ⊆ K, there exists a subsequence

{xnk

}such that limk→∞ xnk = x ∈ K for some point x.

In words, every sequence has a subsequence which converges to a point in the set.

There is a fundamental property possessed by a sequentially compact set in a metricspace which is described in the following proposition. The special number described iscalled a Lebesgue number.

Proposition 2.5.4 Let K be a sequentially compact set in a metric space and let C bean open cover of K. Then there exists a number δ > 0 such that whenever x ∈ K, it followsthat B(x,δ ) is contained in some set of C .

Proof: If C is an open cover of K and has no Lebesgue number, then for each n∈N, 1n is

not a Lebesgue number. Hence there exists xn ∈K such that B(xn,

1n

)is not contained in any

set of C . By sequential compactness, there is a subsequence{

xnk

}such that xnk → x ∈ K.

Now there is r > 0 such that B(x,r)⊆U ∈ C . Let k be large enough that 1nk

< r2 and also

large enough that xnk ∈ B(x, r

2

). Then B

(xnk ,

1nk

)⊆ B

(xnk ,

r2

)⊆ B(x,r) contrary to the

requirement that B(

xnk ,1nk

)is not contained in any set of C . ■

In any metric space, these two definitions of compactness are equivalent.

Theorem 2.5.5 Let K be a nonempty subset of a metric space (X ,d). Then it iscompact if and only if it is sequentially compact.

Proof: ⇐ Suppose K is sequentially compact. Let C be an open cover of K. ByProposition 2.5.4 there is a Lebesgue number δ > 0. Let x1 ∈ K. If B(x1,δ ) covers K, thenpick a set of C containing this ball and this set will be a finite subset of C which covers K.If B(x1,δ ) does not cover K, let x2 /∈ B(x1,δ ). Continue this way obtaining xk such thatd (xk,x j)≥ δ whenever k ̸= j. Thus eventually {B(xi,δ )}n

i=1 must cover K because if not,you could get a sequence {xk}which has every pair of points further apart than δ and hence