Cauchy sequence

**particlejohn** · July 10th 2008, 11:55 AM

Prove that a sequence converges iff it is a Cauchy sequence.

So the sequence $a_n \to L$ . For $\varepsilon > 0$ there is an $N \in \bold{N}$ such that $j \geq N \implies |a_{n}-L| < \varepsilon/2$ . And so $m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon$ (by triangle inequality).

Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$ . Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$ . So $a_N - 1/2 < a_n < a_N+1/2$ .

Is this the correct way of proving this?

**Plato** · July 10th 2008, 12:17 PM

Originally Posted by particlejohn

Prove that a sequence converges iff it is a Cauchy sequence.
So the sequence $a_n \to L$ . For $\varepsilon > 0$ there is an $N \in \bold{N}$ such that $j \geq N \implies |a_{n}-L| < \varepsilon/2$ . And so $m,n \geq N \implies |a_{m}-a_{n}| = |a_{m}-L +L-a_{n}| \leq |a_{m}-L|+|L-a_{n}| < \varepsilon$ (by triangle inequality).

That part is certainly correct and well done. What are you doing with the converse?

If $a_n$ is a Cauchy sequence, then you need to find a number to which it converges.
Again I don’t know what theorems you have.

Here is one way to proceed.
1. Every sequence contains a monotone subsequence.
2. Every Cauchy sequence is bounded.
3. Every bounded monotone sequence converges.
4. Prove the limit of that subsequence is the limit of the whole sequence.

**ThePerfectHacker** · July 10th 2008, 01:28 PM

Originally Posted by particlejohn

Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$ . Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$ . So $a_N - 1/2 < a_n < a_N+1/2$ .

You proved the forward direction correctly. (In fact in any metric space a convergent sequence is Cauchy. However, not necessarily the other way around. When the converse hold we say the metric space is complete).

The way I like to prove this fact is by introducing the notion of limsup's and liminf's. Let $\{a_n\}$ be a sequence. If $\{a_n\}$ is not bounded from above then we define $\limsup a_n = \infty$ . Otherwise, we contruct the following sequence: $s_1 = \sup\{ s_k|k\geq 1\} = \sup\{ s_1,s_2,...\}$ , $s_2 = \sup\{ s_k|k\geq 2\} = \sup\{ s_2,s_3,...\}$ , in general $s_n = \sup\{ s_k | k\geq n\}$ . Of couse these sequences make sense because we assumed the sequence was bounded from above. Let us define $s_n$ 's to be the superior sequence. Note that $s_1\geq s_2\geq s_3 \geq ...$ . Thus, $\{ s_n\}$ is a non-increasing sequence and it must have a limit (or $-\infty$ ). Now we define $\limsup a_n = \lim s_n$ this is the limsup of $a_n$ . The beauty of this is that the sequence $a_n$ does not need to have a limit, not even $\pm \infty$ divergence, but yet the limit superior always exists. Now we can define limit inferior in an analogus fashion. If $\{a_n\}$ is not bounded from below we define $\liminf a_n = -\infty$ . Otherwise construct $i_n = \inf\{ s_k | k\geq n\}$ just like above. Now confirm that $i_1 \leq i_2 \leq i_3 \leq ...$ . This means $\{ i_n\}$ the inferior sequence is a non-decreasing sequence. Thus, it must have a limit (or $+\infty$ ). We define $\liminf a_n = \lim i_n$ . Finally if $\{a_n\}$ is bounded (meaning both from above and below) then $s_n$ has a lower bound (i.e. $i_1$ ) which automatically means $\limsup a_n$ is a real number. Likewise if $\{a_n\}$ is bounded then $i_n$ has an upper bound (i.e. $s_1$ ) which automatically means $\liminf a_n$ is a real number.
There are a number of nice properties that are not hard to prove, let $L = \mathbb{R}\cup \{\pm \infty\}$ :

if $\lim a_n = L$ then $\liminf a_n = \limsup a_n = L$
if $\lim a_n = L$ then $\liminf a_n \leq L \leq \limsup a_n$
$\liminf a_n \leq \limsup a_n$
if $\liminf a_n = \limsup a_n = L$ then $\lim a_n = L$

We are now able to elegantly prove that Cauchy sequences are convergent. The first fact we will need is that is $\{a_n\}$ is a Cauchy sequence then $\{a_n\}$ is a bounded sequence. I leave that for you to prove, it happens to be easy. Since $a_n$ is bounded it means $\liminf a_n$ and $\limsup a_n$ are real numbers. To show $\{a_n\}$ is convergent is sufficies to use property # 4 from that list. Let $\epsilon > 0$ . Since $\{a_n\}$ is Cauchy it means there is $N\in \mathbb{N}$ such that $|a_n - a_m | < \epsilon$ for all $n,m\geq N$ . Thus, $a_m - \epsilon < a_n < a_m + \epsilon$ for all $n,m\geq N$ . Thus, $a_m + \epsilon$ is an upper bound for $a_n$ , so $s_N = \sup\{ s_k | k\geq N\} \leq a_m + \epsilon$ . But this show $s_N - \epsilon < a_m$ i.e. $s_N - \epsilon$ is a lower bound for $a_m$ . Thus, $s_N - \epsilon \leq \inf \{ a_k | k\geq N \} = i_N$ . This means $\limsup a_n \leq s_N \leq i_N + \epsilon \leq \liminf a_n + \epsilon$ . This statement is true for any $\epsilon > 0$ so $\limsup a_n \leq \liminf a_n$ . But by property #3 we have $\liminf a_n \leq \limsup a_n$ . Thus, $\liminf a_n = \limsup a_n$ . This prove that $\{a_n\}$ is convergent.

The reason why limsup's and liminf's are helpful in proving convergence is that they are always convergent if the sequence is bounded. And so dealing with them is always possible.

**particlejohn** · July 10th 2008, 02:33 PM

Originally Posted by Plato

That part is certainly correct and well done. What are you doing with the converse?

If $a_n$ is a Cauchy sequence, then you need to find a number to which it converges.
Again I don’t know what theorems you have.

Here is one way to proceed.
1. Every sequence contains a monotone subsequence.
2. Every Cauchy sequence is bounded.
3. Every bounded monotone sequence converges.
4. Prove the limit of that subsequence is the limit of the whole sequence.

Now suppose $a_n$ is a Cauchy sequence. Choose $\varepsilon = 1/2$ . Then there is an $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < 1/2$ . So $a_N - 1/2 < a_n < a_N+1/2$ for all $n \geq N$ .

So $\{a_{n}: n \geq N \}$ is bounded. Also, $\{a_{n}: n < N \}$ is also bounded. The union, $\{a_{n}: n \geq N \} \cup \{a_{n}: n < N \} = \{a_{n}: n \in \bold{N}\}$ is thus bounded. So it has a convergent subsequence $c_n \to L$ . For $\varepsilon _1 > 0$ we choose $N \in \bold{N}$ such that $n,m \geq N \implies |a_{n}-a_{m}| < \varepsilon_{1}/2$ . So $c_{k} = a_{m_{k}}$ such that $m_{k} \geq N$ and $|c_{m_{k}}-L| < \varepsilon_{1}/2$ . For $n \geq N$ , we have $|a_{n}-L| = |a_{n}-c_{k}+c_{k}-L| \leq |a_{n}-c_{k}| + |c_{k}-L| < |a_{n}-a_{m_{k}}| + \varepsilon_{1}/2 < \varepsilon_1$ . So $a_n \to L$ . $\square$

**Plato** · July 10th 2008, 02:39 PM

That's good. It works.
You say what the $b_n$ in the proof comes from.

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