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Chapter 3: Problem 8

Show directly that a bounded, monotone increasing sequence is a Cauchy sequence.

Short Answer

Expert verified
To show that a bounded, monotone increasing sequence is a Cauchy sequence, you first express the difference of the \(m\)th and \(n\)th terms, use properties of the sequence to rewrite this difference, then use the properties of \(\epsilon\) and \(N\) to show this difference is less than \(\epsilon\). This completes the proof.

Step by step solution

01

Reiterate the Problem

We want to show that any bounded, monotone increasing sequence is a Cauchy sequence. That means for any \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(m,n > N\), we have \(|a_m - a_n| < \epsilon \). We need to consider a general bounded, increasing sequence and proceed accordingly.
02

Define a Bounded, Monotonic Increasing Sequence

Define a bounded monotonic increasing sequence, \((a_n)\), such that \(a_n \leq a_{n+1}\) for every \(n \in \mathbb{N}\). Moreover, there exists \(M \in \mathbb{R}\) such that \(a_n \leq M\) for all \(n \in \mathbb{N}\).
03

Take an Arbitrary Epsilon

Take an arbitrary \(\epsilon > 0\). By the archimedean principle, there exists an integer \(N\) such that \(N\epsilon > M - a_1\). So, \(N\) is the natural number we want.
04

Express the Difference

Express the difference \(|a_m - a_n|\) where \(m>n>N\). Since the sequence is increasing, this difference can be written as \(a_m - a_n\).
05

Use the Properties of the Sequence

Since \(a_m\) can be at most \(M\), and \(a_n\) is at least \(a_{N+1}\), the difference \(a_m - a_n\) is at most \(M - a_{N+1}\).
06

Apply the Property of N

Recall that we have chosen \(N\) such that \(N\epsilon > M - a_1\). Since the sequence is increasing, \(a_1 \leq a_{N+1}\) and \(M - a_{N+1} \leq M - a_1 < N\epsilon\). Consequently, for any \(n > N\), \(M - a_n < N\epsilon\).
07

Finalize the Proof

Plug this back into the equation from step 4. If \(n,m > N\), then we have \(|a_m - a_n| = a_m - a_n \leq M - a_n < N\epsilon.\) Simply choosing \(m > N\) and \(n = N + 1\) suffices because then, for \(m, n > N\), \(a_m - a_n < \epsilon\). Thus, the sequence \((a_n)\) is a Cauchy sequence.

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