504 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

Definition 19.1.16 Let (Xi,τ i) be topological spaces. ∏ni=1 Xi is the Cartesian

product. Define a product topology as follows. Let B = ∏ni=1 Ai where Ai ∈ τ i. Then

B is a basis for the product topology.

Theorem 19.1.17 The set B of Definition 19.1.16 is a basis for a topology asclaimed.

Proof: Suppose x ∈ (∏ni=1 Ai)∩ (∏n

i=1 Bi) where Ai and Bi are open sets. Supposethat x= (x1, · · · ,xn) . Then xi ∈ Ai ∩Bi for each i. Therefore, x ∈ ∏

ni=1 Ai ∩Bi ∈B and

∏ni=1 Ai∩Bi ⊆ ∏

ni=1 Ai. ■

The definition of compactness is also considered for a general topological space. Thisis given next.

Definition 19.1.18 A subset, E, of a topological space (X ,τ) is said to be compactif whenever C ⊆ τ and E ⊆ ∪C , there exists a finite subset of C ,{U1 · · ·Un}, such thatE ⊆ ∪n

i=1Ui. (Every open covering admits a finite subcovering.) E is precompact if Eis compact. A topological space is called locally compact if it has a basis B, with theproperty that B is compact for each B ∈B.

In general topological spaces there may be no concept of “bounded”. Even if there is,closed and bounded is not necessarily the same as compactness. However, in any Hausdorffspace every compact set must be a closed set.

Theorem 19.1.19 If (X ,τ) is a Hausdorff space, then every compact subset mustalso be a closed set.

Proof: Suppose p /∈K a compact set. For each x∈ X \{p}, there exist open sets, Ux andVx such that x∈Ux, p∈Vx, and Ux∩Vx = /0. If K is assumed to be compact, there are finitelymany of these sets, Ux1 , · · · ,Uxm which cover K. Then let V ≡ ∩m

i=1Vxi . It follows that V isan open set containing p which has empty intersection with each of the Uxi . Consequently,V contains no points of K and is therefore not a limit point of K. ■

Definition 19.1.20 If every finite subset of a set P whose elements are sets hasnonempty intersection, the set P is said to have the finite intersection property.

Theorem 19.1.21 Let K be a set whose elements are compact subsets of a Haus-dorff topological space, (X ,τ). Suppose K has the finite intersection property. Then/0 ̸= ∩K .

Proof: Suppose to the contrary that /0 = ∩K . Then consider C ≡{

KC : K ∈K}. It

follows C is an open cover of K0 where K0 is any particular element of K . But then thereare finitely many K ∈K , K1, · · · ,Kr such that K0 ⊆ ∪r

i=1KCi implying that

K0∩ (∩ri=1Ki)⊆

(∪r

i=1KCi)∩ (∩r

i=1Ki) = (∩ri=1Ki)

C ∩ (∩ri=1Ki) = /0,

contradicting the finite intersection property. ■There is a fundamental theorem, called Urysohn’s lemma which is valid for locally

compact Hausdorff spaces which is presented next.

Lemma 19.1.22 Let X be a locally compact Hausdorff space and let K ⊆V ⊆ X whereK is compact and V is open. Then there exists an open set Uk containing k such that Uk iscompact and Uk ⊆Uk ⊆V. Also there exists U such that U is compact and K ⊆U ⊆U ⊆V .