506 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

I claim f is continuous. f (x)< a, means there must be some t ∈ D with t < a and x ∈Vt .Thus f−1 ([0,a)) = ∪{Vt : t < a, t ∈ D}, an open set.

Next consider x ∈ f−1 ([0,a]) so f (x) ≤ a. If t > a, then x ∈ Vt from the definition off (x). Thus

f−1 ([0,a])⊆ ∩{Vt : t > a}= ∩{V t : t > a}which is a closed set. If x ∈∩{Vt : t > a}, then f (x)≤ a from the definition and so equalityholds and f−1 ([0,a]) is closed. This is also true if a = 1. In this case, f−1 ([0,1]) = X .Therefore, for a≥ 0,

f−1 ((a,1])∪ f−1 ([0,a]) = X

and so f−1 ((a,1]) is an open set. It follows that f is continuous because f−1 (a,b) =f−1 ([0,b))∩ f−1 ((a,1]) the intersection of two open sets. Since this is so for every inter-val, it follows that the inverse image of any open set is open and so f is continuous. Clearlyf (x) = 0 on H. If x∈VC, then x /∈Vt for any t ∈D so f (x) = 1 on VC. Let g(x) = 1− f (x).■

In any metric space there is a much easier proof of the conclusion of Urysohn’s lemmawhich applies. The following is Lemma 3.12.1 listed here for convenience.

Lemma 19.1.24 Let S be a nonempty subset of a metric space, (X ,d) . Define

f (x)≡ dist(x,S)≡ inf{d (x,y) : y ∈ S} .

Then f is continuous.

In a metric space it is all much easier.

Theorem 19.1.25 Let (X ,τ) be a locally compact metric space in which the clo-sures of balls are compact and let H ⊆U where H is compact and U is open. Then thereexists g : X→ [0,1] such that g is continuous, g(x) = 1 on H and g(x) = 0 if x /∈V for someopen set V whose closure is compact.

Proof: Let δ > 0 be such that for all h ∈ H,dist(h,UC

)> δ . This exists because

h→ dist(h,UC

)is continuous and so achieves its minimum on H which must be positive

because UC is closed. Now consider the balls B(h,δ ). These cover the compact set Hand so there are finitely many which do so. B(h1,δ ) , · · · ,B(hm,δ ) where the closure ofeach of these is compact. Also B(h j,δ )⊆U. Because if x ∈ B(h j,δ ), then d (x,h j)≤ δ <

dist(h,UC

). Let V =∪m

j=1B(h j,δ ) . Thus V =∪mj=1B(h j,δ ) because there are only finitely

many sets. Also V is compact because it is a finite union of compact sets. Now define

g(x)≡dist(x,VC

)dist(x,H)+dist(x,VC)

This is continuous, equals 1 on H and equals 0 off V because the denominator is alwayspositive since both H,VC are closed. ■

A useful construction when dealing with locally compact Hausdorff spaces is the notionof the one point compactification of the space.

Definition 19.1.26 Suppose (X ,τ) is a locally compact Hausdorff space. Then letX̃ ≡ X ∪{∞} where ∞ is just the name of some point which is not in X which is called thepoint at infinity. A basis for the topology τ̃ for X̃ is

τ ∪{

KC where K is a compact subset of X}.

506 CHAPTER 19. HAUSDORFF SPACES AND MEASURESI claim f is continuous. f (x) < a, means there must be some t € D witht <aandx€V,.Thus f—! ((0,a)) =U{V; :t < a,t € D}, an open set.Next consider x € f~!([0,a]) so f (x) <a. Ift >a, then x € V; from the definition off (x). Thusf! ({0,a]) COLY st > ad =A{V; :t > ahwhich is a closed set. If x E N{V; :t > a}, then f (x) <a from the definition and so equalityholds and f~! ([0,a]) is closed. This is also true if a = 1. In this case, f~! ([0,1]) =X.Therefore, for a > 0,fl ((a, 1) Uf ([0, a]) =Xand so f-!((a,1]) is an open set. It follows that f is continuous because f~! (a,b) =f—! ((0,b)) A £7! ((a, 1)) the intersection of two open sets. Since this is so for every inter-val, it follows that the inverse image of any open set is open and so f is continuous. Clearlyf(x) =OonH. IfxeV°, then x ¢ V,; for any t € D so f (x) =1 on VV. Let g(x) =1—f (x).aIn any metric space there is a much easier proof of the conclusion of Urysohn’s lemmawhich applies. The following is Lemma 3.12.1 listed here for convenience.Lemma 19.1.24 Let S be a nonempty subset of a metric space, (Xd) . Definef (x) = dist (x, S) = inf {d (x,y): y € S}.Then f is continuous.In a metric space it is all much easier.Theorem 19.1.25 Ler (X,T) be a locally compact metric space in which the clo-sures of balls are compact and let H C U where H is compact and U is open. Then thereexists g: X — [0,1] such that g is continuous, g(x) = 1 on H and g(x) =O ifx ¢ V for someopen set V whose closure is compact.Proof: Let 5 > 0 be such that for all h € H,dist(h,U©) > 6. This exists becauseh — dist (A, U ©) is continuous and so achieves its minimum on H which must be positivebecause U© is closed. Now consider the balls B(h,5). These cover the compact set Hand so there are finitely many which do so. B(h1,6),---,B(/m,65) where the closure ofeach of these is compact. Also B(h;,6) C U. Because if x € B(h;, 6), then d(x,hj) <6 <dist (n,U°) . Let V =U"_,B(hj,6). Thus V =U", B (hj, 5) because there are only finitelymany sets. Also V is compact because it is a finite union of compact sets. Now definedist (x, vo)dist (x, H) + dist (x,V©)g(x)This is continuous, equals 1 on H and equals 0 off V because the denominator is alwayspositive since both H,V“ are closed. llA useful construction when dealing with locally compact Hausdorff spaces is the notionof the one point compactification of the space.Definition 19.1.26 Suppose (X,7) is a locally compact Hausdorff space. Then letX =X U {oo} where © is just the name of some point which is not in X which is called thepoint at infinity. A basis for the topology T for X isTU {Ke where K is a compact subset of X} .