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.