Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.

(i) Show that there are functions f

x ∈ C(X, [0, 1]) with f

x(x) = 1 and f

x = 0 on C, where x ∈ K.

(ii)Use the functions f

x to make an open cover of K

(iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.

(iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.

NOTE: X doesn't have to be normal (T4)!

ANSWERS:

X is completely regular and C ⊂ X is closed so for every x not in C there is a f

x ∈

C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and deﬁne

. U

x is open and contains x. So {U

x | x ∈ K} is an open cover of K. K is compact, so there are x

1, . . . , x

n ∈ K such that U

x1 ∪ · · · ∪ U

xn ⊃ K. We define g(x) = f

x1(x) + · · · + f

xn(x) or g(x) = max{f

x1(x), . . . , f

xn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ U

i for some i, so f

i(x) > 1/2.

So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.

Question: Why is U

x open? What are the other possibilities of making an open cover of K by means of using the f

x? Is this the most natural choice?