150 CHAPTER 7. METRIC SPACES AND TOPOLOGICAL SPACES

Theorem 7.8.4 Suppose K is a nonempty compact subset of Rn and A ⊆C (K,X) is uni-formly bounded and uniformly equicontinuous. Then if { fk} ⊆ A, there exists a function,f ∈C (K,X) and a subsequence, fkl such that

liml→∞

ρK(

fkl , f)= 0.

To give a proof of this theorem, I will first prove some lemmas.

Lemma 7.8.5 If K is a compact subset of Rn, then there exists D≡ {xk}∞

k=1 ⊆ K such thatD is dense in K. Also, for every ε > 0 there exists a finite set of points, {x1, · · · ,xm} ⊆ K,called an ε net such that

∪mi=1B(xi,ε)⊇ K.

Proof: For m∈N, pick xm1 ∈K. If every point of K is within 1/m of xm

1 , stop. Otherwise,pick

xm2 ∈ K \B(xm

1 ,1/m) .

If every point of K contained in B(xm1 ,1/m)∪B(xm

2 ,1/m) , stop. Otherwise, pick

xm3 ∈ K \ (B(xm

1 ,1/m)∪B(xm2 ,1/m)) .

If every point of K is contained in B(xm1 ,1/m)∪B(xm

2 ,1/m)∪B(xm

3 ,1/m), stop. Other-

wise, pickxm

4 ∈ K \ (B(xm1 ,1/m)∪B(xm

2 ,1/m)∪B(xm3 ,1/m))

Continue this way until the process stops, say at N (m). It must stop because if it didn’t,there would be a convergent subsequence due to the compactness of K. Ultimately all termsof this convergent subsequence would be closer than 1/m, violating the manner in whichthey are chosen. Then D = ∪∞

m=1 ∪N(m)k=1

{xm

k

}. This is countable because it is a countable

union of countable sets. If y ∈ K and ε > 0, then for some m, 2/m < ε and so B(y,ε) mustcontain some point of

{xm

k

}since otherwise, the process stopped too soon. You could have

picked y.

Lemma 7.8.6 Suppose D is defined above and {gm} is a sequence of functions of A havingthe property that for every xk ∈ D,

limm→∞

gm (xk) exists.

Then there exists g ∈C (K,X) such that

limm→∞

ρ (gm,g) = 0.

Proof: Define g first on D.

g(xk)≡ limm→∞

gm (xk) .

Next I show that {gm} converges at every point of K. Let x ∈ K and let ε > 0 be given.Choose xk such that for all f ∈ A,

d ( f (xk) , f (x))<ε

3.