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}.