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Hi,

I am trying to prove that every convergent sequence is Cauchy - just wanted to see if my reasoning is valid and that the proof is correct.

Thanks!

Prove that every convergent sequence is Cauchy

Suppose (s

By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

Since (s

Now, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

But then this means ∃s

i.e. s

So M - ε < M - φ < s

This is the same as M - ε < M - φ < M - α < M

But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

I.e. the sequence is Cauchy.

I am trying to prove that every convergent sequence is Cauchy - just wanted to see if my reasoning is valid and that the proof is correct.

Thanks!

1. Homework Statement1. Homework Statement

Prove that every convergent sequence is Cauchy

## Homework Equations

/ Theorems[/B]**Theorem 1:**Every convergent set is bounded**Theorem 2:**Every non-empty bounded set has a supremum (through the completeness axiom)**Theorem 3:**Limit of sequence with above properties = Sup S (proved elsewhere)**Incorrect - not taken as true in second attempt of proof**## The Attempt at a Solution

Suppose (s

_{n}) is a convergent sequence with limit L. Let S = {s_{n}: n∈ℕ}By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

Since (s

_{n}) is convergent, for every ε > 0, ∃N such that ∀n > N, | s_{n}- L | < ε, where L = MNow, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

But then this means ∃s

_{n}such that M - φ < s_{n}< M.i.e. s

_{n}+ α = M for some α > 0So M - ε < M - φ < s

_{n}< M.This is the same as M - ε < M - φ < M - α < M

But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

I.e. the sequence is Cauchy.

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