# Thread: Cauchy <=> convergent

1. ## Cauchy <=> convergent

So I am trying to prove that a sequence is cauchy if and only if it is convergent using only the Bolzano Weierstrass therom for sequences
[Any bounded sequence must have at least one convergent subsequence]

So far, all I have gotten in the forward direction (cauchy => convergent) is that a cauchy sequence is bounded, and hence by the Bolzano Weierstrass theorem, there must be at least ONE convergent subsequence. I have a theorem in the book which states:
[A sequence converges to A iff each of its subsequences converge to A.]
Is this the proper theorem to use in my problem and if so how do I show that ALL of the subsequences converge, or even that more than one subsequence exists at all?

Also in the backwards direction (convergent => cauchy), I am completely at a loss. Any help would be greatly appreciated.
Thanks!

2. Originally Posted by dannyboycurtis
Also in the backwards direction (convergent => cauchy), I am completely at a loss. Any help would be greatly appreciated.
$\displaystyle \text{If }\left( {x_n } \right) \to L\text{ then }\varepsilon > 0\, \Rightarrow \,\left( {\exists N} \right)\left[ {n \geqslant N\, \Rightarrow \,\left| {x_n - L} \right| < \frac{\varepsilon}{2} } \right]$
If $\displaystyle n,m\ge N$ then $\displaystyle \left| {x_n - x_m } \right| \leqslant \left| {x_n - L} \right| + \left| {L - x_m } \right| < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon$

3. [QUOTE=dannyboycurtis;380174]So I am trying to prove that a sequence is cauchy if and only if it is convergent using only the Bolzano Weierstrass therom for sequences
[Any bounded sequence must have at least one convergent subsequence]

So far, all I have gotten in the forward direction (cauchy => convergent) is that a cauchy sequence is bounded, and hence by the Bolzano Weierstrass theorem, there must be at least ONE convergent subsequence. I have a theorem in the book which states:
[A sequence converges to A iff each of its subsequences converge to A.]
Is this the proper theorem to use in my problem and if so how do I show that ALL of the subsequences converge, or even that more than one subsequence exists at all?[quote]
Let $\displaystyle \{a_n\}$ be your sequence and $\displaystyle \{a_{n_k}\}$ be the subsequence that converges to A. Given any $\displaystyle \epsilon> 0$ there exist $\displaystyle N_1$ such that if $\displaystyle n> N$ and $\displaystyle a_n$ is in the convergent subsequence then $\displaystyle |a_n- A|< \epsilon/2$. Also, because $\displaystyle \{a_n\}$ is a Cauchy sequence, there exist $\displaystyle N_2$ such that if m and n are both larger than $\displaystyle N_2$, [tex]|a_m- a_n|< \epsilon/2.

Now, let N be the larger of $\displaystyle N_1$ and $\displaystyle N_2$ so that if n> N, both are true. Since $\displaystyle \{a_{n_k}\}$ is an infinite sequence, there certainly exist $\displaystyle n_k> N$ also. Then, for n> N, $\displaystyle |a_n- A|\le |a_n- a_{n_k}|+ |a_{n_k}- A|$ and both are less than $\displaystyle \epsilon/2$.

Also in the backwards direction (convergent => cauchy), I am completely at a loss. Any help would be greatly appreciated.
Thanks!
Plato showed this. And, by the way, the proof in this direction does NOT require "Bolzano Weierstrasse" or any form of "completeness". For example, in the rational numbers Cauchy sequences are not necessarily convergent, but any convergent sequence is a Cauchy sequence.