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**CSM** 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 fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.

(ii)Use the functions fx 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 fx ∈

C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and deﬁne $\displaystyle U_x =f_x^{−1}((\frac{1}{2}, 1])\subset X$. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ · · · ∪ Uxn ⊃ K. We define g(x) = fx1(x) + · · · + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(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 Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

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