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.