7.10. THE TIETZE EXTENSION THEOREM 155

By equicontinuity, there exists δ > 0 such that if d (x,y) < δ , then d ( f (x) , f (y)) < ε

8for all f ∈ A . Let {xi}m

i=1 be a δ/2 net for X . Since there are only finitely many xi, itfollows from pointwise compactness that there exists a subsequence, still denoted by { fn}which converges at each xi. There exists xmn such that

ρ ( fn, fm)−ε

8< d ( fn (xmn) , fm (xnm))

≤ d ( fn (xnm) , fn (xi))+d ( fn (xi) , fm (xi))+d ( fm (xi) , fm (xnm))

<ε

8+d ( fn (xi) , fm (xi))+

ε

8(7.9.9)

where here xi is such that xnm ∈ B(xi,δ ). From the convergence of fn at each xi, there existsN such that if m,n > N, then for all xi,

d ( fn (xi) , fm (xi))<ε

8.

Now 7.9.9 results in the contradiction,

ε− ε

8<

ε

8+

ε

8+

ε

8

It follows that A and hence A is totally bounded. This proves the more important direc-tion.

Next suppose A is compact. Why must A be pointwise compact and equicontinuous?If it fails to be pointwise compact, then there exists x ∈ X such that { f (x) : f ∈A } is notcontained in a compact set of Y . Thus there exists ε > 0 and a sequence of functions in A{ fn} such that d ( fn (x) , fm (x)) ≥ ε . But this implies ρ ( fm, fn) ≥ ε and so A fails to betotally bounded, a contradiction. Thus A must be pointwise compact. Now why must it beequicontinuous? If it is not, then for each n ∈ N there exists ε > 0 and xn,yn ∈ X such thatd (xn,yn) < 1/n but for some fn ∈ A , d ( fn (xn) , fn (yn)) ≥ ε. However, by compactness,there exists a subsequence

{fnk

}such that limk→∞ ρ

(fnk , f

)= 0 and also that xnk ,ynk →

x ∈ X . Hence

ε ≤ d(

fnk

(xnk

), fnk

(ynk

))≤ d

(fnk

(xnk

), f(xnk

))+d(

f(xnk

), f(ynk

))+d(

f(ynk

), fnk

(ynk

))≤ ρ

(fnk , f

)+d(

f(xnk

), f(ynk

))+ρ

(f , fnk

)and now this is a contradiction because each term on the right converges to 0. The middleterm converges to 0 because f

(xnk

), f(ynk

)→ f (x).

7.10 The Tietze Extension TheoremIt turns out that if H is a closed subset of a metric space, (X ,d) and if f : H → [a,b] iscontinuous, then there exists g defined on all of X such that g = f on H and g is continuous.This is called the Tietze extension theorem. First it is well to recall continuity in the contextof metric space.