2.5. COMPACTNESS AND CONTINUOUS FUNCTIONS 49

Proof: The given sequence is contained in some set of the form ∏pi=1 [−r,r] which is

a compact set as shown in Corollary 2.5.13. Hence the given sequence has a convergentsubsequence. Here we regard Cp as R2p. ■

It is always the case that a compact set in a metric space is a closed set. In fact, this istrue for any Hausdorff space. What is done in general is to axiomatize the idea of a metricspace to define a general topological space as follows. Here X is a nonempty set.

1. Let τ be the collection of open sets called the topology, τ ⊆P (X). Then if C ⊆ τ ,∩C ∈ τ.

2. If Ui ∈ τ for i = 1,2, · · · ,n, then ∩ni=1Ui ∈ τ

Definition 2.5.15 Hausdorff space is a general topological space which has theproperty that if x ̸= y, then there exist open sets Ux and Uy containing x,y respectively suchthat Ux∩Uy = /0.

Proposition 2.5.16 If K is a compact subset of a Hausdorff space, then it is closed. Inparticular, this holds for any metric space.

Proof: Let K be a nonempty compact set and suppose p /∈K. Then for each x∈K, thereare open sets Ux,Vx such that x ∈Vx and p ∈Ux and Ux∩Vx = /0. Then since V is compact,there are finitely many Vx which cover K say Vx1 , · · · ,Vxn . Then let U = ∩n

i=1Uxi . It followsp ∈U and U has empty intersection with K. In fact U has empty intersection with ∪n

i=1Vxi

because it is contained in each Uxi . Since U is an open set and p∈KC is arbitrary, it followsKC is an open set. ■

The following is a very important property pertaining to compact sets. It is a surprisingresult. However, it follows from the definition of compactness.

Proposition 2.5.17 Suppose F is a nonempty collection of nonempty compact setswith the finite intersection property. This means that the intersection of any finite subset ofF is nonempty. Then ∩F ̸= /0.

Proof: If the conclusion were not so, ∪{

FC : F ∈F}= X and so, in particular, pick-

ing some F0 ∈F ,{

FC : F ∈F}

would be an open cover of F0. A point in F0 is not in FC0

so it must be in one of the above sets F ̸= F0. Since F0 is compact, some finite subcover,FC

1 , · · · ,FCm exists, F0 ⊆ ∪m

k=1FCk . Therefore, the finite intersection property is violated be-

cause

F0∩ (∩mk=1Fk)⊆

(∪m

k=1FCk)∩ (∩m

k=1Fk) = (∩mk=1Fk)

C ∩ (∩mk=1Fk) = /0 ■

Note that absolutely no mention was made of context. This is because this finite in-tersection property is always true whenever you have a set of compact sets. Of course, inthis book, we typically have in mind a metric space. I am just pointing out that all of itgeneralizes.

2.5.1 Continuous FunctionsThe following is a fairly general definition of what it means for a function to be continuous.It includes everything seen in typical calculus classes as a special case.