74 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

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 Theorem 2.4.8. Therefore, {d (xn,yn)}n is a Cauchy sequence in R and so it converges.■

Definition 2.11.2 Let x∼ y when limn→∞ d (xn,yn) = 0.

Lemma 2.11.3 ∼ 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 2.11.4 Denote by [x] the equivalence class determined by the Cauchysequence x. Let d ([x] , [y])≡ limn→∞ d (xn,yn) .

Theorem 2.11.5 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 .

Proof: That d just defined is a metric is obvious from the fact that the original metricd satisfies the triangle inequality. It is also clear that d ([x] , [y]) ≥ 0 and that if [x] = [y] ifand only if d ([x] , [y]) = 0.

It remains to show that(X̂ ,d

)is complete. Let {[x]n}nbe a Cauchy sequence. From

Theorem 2.3.3 it suffices to show the convergence of a subsequence. There is a subse-quence, denoted as {[xn]} where xn is a representative of [x]n such that d

([xn] ,

[xn+1

])<

4−n. Thus there is an increasing sequence {kn} such that d(xn

k ,xn+1l

)< 2−n if k, l ≥ kn

where kn is increasing in n. Let y ={

xnkn

}∞

n=1. For m≥ kn and the triangle inequality,

d (xnm,ym) = d

(xn

m,xmkm

)≤ d

(xn

m,xnkn

)+d(xn

kn,xm

km

)≤ 2−n +

m−1

∑j=n

d(

x jk j,x j+1

km

)< 2−n +

m−1

∑j=n

2− j < 2−n +2−(n−1) < 2−(n−2)

Then y is a Cauchy sequence since it is a subsequence of one and also d ([xn] , [y])→ 0.To show that X is dense in X̂ , let [x] be given. Then for m large enough, d (xk,xm)< ε

whenever k≥m. It suffices to let y be the constant Cauchy sequence always equal to xm. ■

2.12 Exercises1. Explain carefully why in Rn, B∞ (p,r) = ∏

ni=1 (pi− r, pi + r)