240 CHAPTER 13. SOME FUNDAMENTALS∗

Proof: Let ε = 1 in the definition of a Cauchy sequence and let n > n1. Then from thedefinition,

|an−an1 |< 1.

It follows that for all n > n1,|an|< 1+ |an1 | .

Therefore, for all n,

|an| ≤ 1+ |an1 |+n1

∑k=1|ak| . ■

Theorem 13.3.6 If a sequence {an} in Rp converges, then the sequence is a Cauchy se-quence. Also, if some subsequence of a Cauchy sequence converges, then the originalsequence converges.

Proof: Let ε > 0 be given and suppose an → a. Then from the definition of conver-gence, there exists nε such that if n > nε , it follows that

|an−a|<ε

2

Therefore, if m,n≥ nε +1, it follows that

|an−am| ≤ |an−a|+ |a−am|<ε

2+

ε

2= ε

showing that, since ε > 0 is arbitrary, {an} is a Cauchy sequence. It remains to that the lastclaim.

Suppose then that {an} is a Cauchy sequence and a = limk→∞ank where{ank

}∞

k=1is a subsequence. Let ε > 0 be given. Then there exists K such that if k, l ≥ K, then|ak−al |< ε

2 . Then if k > K, it follows nk > K because n1,n2,n3, · · · is strictly increasingas the subscript increases. Also, there exists K1 such that if k > K1,

∣∣ank −a∣∣ < ε

2 . Thenletting n > max(K,K1), pick k > max(K,K1). Then

|a−an| ≤∣∣a−ank

∣∣+ ∣∣ank −an∣∣< ε

2+

ε

2= ε.

Therefore, the sequence converges. ■

Definition 13.3.7 A set K in Rp is said to be sequentially compact if every sequence in Khas a subsequence which converges to a point of K.

Theorem 13.3.8 If I0 = ∏pi=1 [ai,bi] where ai ≤ bi, then I0 is sequentially compact.

Proof: Let {ak}∞

k=1 ⊆ I0 and consider all sets of the form ∏pi=1 [ci,di] where [ci,di]

equals either[ai,

ai+bi2

]or [ci,di] =

[ai+bi

2 ,bi

]. Thus there are 2p of these sets because

there are two choices for the ith slot for i = 1, · · · , p. Also, if x and y are two points in oneof these sets,

|xi− yi| ≤ 2−1 |bi−ai| .