10.1. METRIC SPACES 249

Proof: Consider the following.

d (x,y)≤ d (x,xn)+d (xn,y)≤ d (x,xn)+d (xn,yn)+d (yn,y)

so d (x,y)−d (xn,yn)≤ d (x,xn)+d (yn,y) . Similarly

d (xn,yn)−d (x,y)≤ d (x,xn)+d (yn,y)

and so|d (xn,yn)−d (x,y)| ≤ d (x,xn)+d (yn,y)

and the right side converges to 0 as n→ ∞. ■First are some simple lemmas featuring one dimensional considerations. In these, the

metric space is R and the distance is given by d (x,y) ≡ |x− y| .First recall the nested in-terval lemma. You should have seen something like it in calculus, but this is often not thecase because there is much more interest in trivialities like integration techniques.

Lemma 10.1.20 Let [ak,bk]⊇ [ak+1,bk+1] for all k = 1,2,3, · · · . Then there exists a pointp in ∩∞

k=1 [ak,bk].

Proof: We note that for any k, l,ak ≤ bl . Here is why. If k ≤ l, then ak ≤ al ≤ bl .If k > l, then bl ≥ bk ≥ ak. It follows that for each l, supk ak ≤ bl . Hence supk ak isa lower bound to the set of all bl and so it is no larger than the greatest lower bound. Itfollows that supk ak ≤ infl bl . Pick x∈ [supk ak, infl bl ]. Then for every k,ak ≤ x≤ bk. Hencex ∈ ∩∞

k=1 [ak,bk] . ■

Lemma 10.1.21 The closed interval [a,b] is compact. This means that if there is a collec-tion of open intervals of the form (a,b) whose union includes all of [a,b] , then in fact [a,b]is contained in the union of finitely many of these open intervals.

Proof: Let C be a set of open intervals the union of which includes all of [a,b] andsuppose [a,b] fails to admit a finite subcover. That is, no finite subset of C has unionwhich contains [a,b]. Then this must be the case for one of the two intervals

[a, a+b

2

]and[ a+b

2 ,b]. Let I1 be the one for which this is so. Then split it into two equal pieces like

what was just done and let I2 be a half for which there is no finite subcover of sets ofC . Continue this way. This yields a nested sequence of closed intervals I1 ⊇ I2 ⊇ ·· · andby the above lemma, there exists a point x in all of these intervals. There exists U ∈ Csuch that x ∈U. Thus x ∈ (a,b) ∈ C . However, for all n large enough, the length of In isless than min(|x−a| , |x−b|). Hence In is actually contained in (a,b) ∈ C contrary to theconstruction. Hence [a,b] is compact after all. ■

As a useful corollary, this shows that R is complete.

Corollary 10.1.22 The real line R is complete.

Proof: Suppose {xk} is a Cauchy sequence in R. Then there exists M such that

{xk}∞

k=1 ⊆ [−M,M] .

Why? If there is no convergent subsequence, then for each x ∈ [−M,M] , there is an openset (x−δ x,x+δ x) which contains xk for only finitely many values of k. Since [−M,M] iscompact, there are finitely many of these open sets whose union includes [−M,M]. This isa contradiction because [−M,M] contains xk for all k ∈ N so at least one of the open setsmust contain xk for infinitely many k. Thus there is a convergent subsequence. Therefore,using Theorem 10.1.16, the original Cauchy sequence converges to some x ∈ [−M,M]. ■