proof of cauchy's first therom on limit
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Answer:
here is your answer
Step-by-step explanation:
Theorem 1 Every Cauchy sequence of real numbers converges to a limit.
Proof of Theorem 1 Let {an} be a Cauchy sequence. For any j, there is a natural
number Nj so that whenever n, m ≥ Nj , we have that |an − am| ≤ 2
−j
. We now consider
the sequence {bj} given by
bj = aNj − 2
−j
.
Notice that for every n larger than Nj , we have that an > bj . Thus each bj serves
as a lower bound for elements of the Cauchy sequence {an} occuring later than Nj . Each
element of the sequence {bj} is bounded above by b1 + 1, for the same reason. Thus the
sequence {bj} has a least upper bound which we denote by L. We will show that L is the
limit of the sequence {an}. Suppose that n > Nj . Then
|an − L| < 2
−j + |an − bj | = 2−j + an − bj ≤ 3(2−j
).
For every > 0 there is j() so that 21−j < and we simply take N() to Nj()
.
The idea of the proof of Theorem 1 is that we recover the limit of the Cauchy sequence
by taking a related least upper bound. So we can think of the process of finding the limit
of the Cauchy sequence as specifying the decimal expansion of the limit, one digit at a
time, as this how the least upper bound property worked.