3.13. COMPLETION OF METRIC SPACES 85

Wi Ui Vi

By Lemma 3.12.4, let U i ≺ φ i ≺Vi, ∪ni=1W i ≺ γ ≺ ∪n

i=1Ui. Define

ψ i(x) ={

γ(x)φ i(x)/∑nj=1 φ j(x) if ∑

nj=1 φ j(x) ̸= 0,

0 if ∑nj=1 φ j(x) = 0.

If x is such that ∑nj=1 φ j(x) = 0, then x /∈ ∪n

i=1U i. Consequently γ(y) = 0 for all y near xand so ψ i(y) = 0 for all y near x. Hence ψ i is continuous at such x. If ∑

nj=1 φ j(x) ̸= 0, this

situation persists near x and so ψ i is continuous at such points. Therefore ψ i is continuous.If x ∈ K, then γ(x) = 1 and so ∑

nj=1 ψ j(x) = 1. Clearly 0≤ψ i (x)≤ 1 and spt(ψ j)⊆Vj. ■

3.13 Completion of Metric SpacesLet (X ,d) be a metric space X ̸= /0. Perhaps this is not a complete metric space. In otherwords, it may be that Cauchy Sequences do not converge. Of course if x ∈ X and if xn = xfor all n then {xn} is a Cauchy sequence and it converges to x.

Lemma 3.13.1 Denote by x a Cauchy sequence x being short for {xn}∞

n=1. Then if x,yare two Cauchy sequences, limn→∞ d (xn,yn) exists.

Proof: Let ε > 0 be given and let N be so large that whenever n,m≥ N, it follows thatd (xn,xm) ,d (yn,ym)< ε/2. Then for such n,m

|d (xn,yn)−d (xm,ym)| ≤ |d (xn,yn)−d (xn,ym)|+ |d (xn,ym)−d (xm,ym)|≤ d (yn,ym)+d (xn,xm)< ε

by Lemma 3.12.1. Therefore, {d (xn,yn)}n is a Cauchy sequence in R and so it converges.■

Definition 3.13.2 Let x∼ y when limn→∞ d (xn,yn) = 0.

Lemma 3.13.3 ∼ is an equivalence relation.

Proof: Clearly x∼ x and if x∼ y then y ∼ x. Suppose then that x∼ y and y ∼ z. Isx∼ z?

d (xn,zn)≤ d (xn,yn)+d (yn,zn)

and both of those terms on the right converge to 0. ■

Definition 3.13.4 Denote by [x] the equivalence class determined by the Cauchysequence x. Let d ([x] , [y])≡ limn→∞ d (xn,yn) .

Theorem 3.13.5 Denote by X̂ the set of equivalence classes. Then d defined aboveis a metric, X̂ with this is a complete metric space, and X can be considered a dense subsetof X̂ .