6.7. THE LIMIT OF A SEQUENCE 107

Such a number exists because of the definition of limit. Therefore, let

nε > max(n1,n2) .

For n≥ nε ,

|an ·bn−a ·b| ≤ (|a|+1) |bn−b|+ |b| |an−a|

< (|a|+1)ε

2(|a|+1)+ |b| ε

2(|b|+1)≤ ε.

This proves 6.7.9. The proof of 6.7.10 is entirely similar and is left for you.

6.7.1 Sequences And CompletenessRecall the definition of a Cauchy sequence.

Definition 6.7.6 {an} is a Cauchy sequence if for all ε > 0, there exists nε such that when-ever n,m≥ nε ,

|an−am|< ε.

A sequence is Cauchy means the terms are “bunching up to each other” as m,n getlarge.

Theorem 6.7.7 Let {an}∞

n=1 be a Cauchy sequence in Fp. Then there exists a unique a∈Fp

such that an→ a.

Proof: Let an =(an

1, · · · ,anp). Then

|ank−am

k | ≤ |an−am|

which shows for each k = 1, · · · , p, it follows{

ank

}∞

n=1 is a Cauchy sequence in F. Thisrequires that both the real and imaginary parts of an

k are Cauchy sequences in R whichmeans the real and imaginary parts converge in R. This shows

{an

k

}∞

n=1 must converge tosome ak. That is limn→∞ an

k = ak. Letting a =(a1, · · · ,ap) , it follows from Theorem 6.7.3that

limn→∞

an = a.

This proves the theorem.

Theorem 6.7.8 The set of terms in a Cauchy sequence in Fp is bounded in the sense thatfor all n, |an|< M for some M < ∞.

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| .

This proves the theorem.