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I wonder if my idea of proof is correct.

Prove that $\{s_n\}_n\subset\mathbb{R}$ is convergent if it is cauchy sequence.

idea of pf: By definition, for every $\epsilon> 0\exists N$ s.t.$|s_m-s_n|<\epsilon \forall m,n >N$ so $|s_n|<|s_m|+\epsilon$ and by completeness we have a supremum $s_0=\sup\{s_n:n\in\mathbb{N}\}$ and then proving $s_n$ converge to $s_0$.

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this is at most a sketch of what might become a proof. Are you asking whether the idea in your formulation is correct or whether what you wrote will be accepted as correct proof in any reasonable course? The answer to the latter is 'no!'. –  Ittay Weiss Jan 23 '13 at 10:48
    
@IttayWeiss ya, i mean the idea –  user56876 Jan 23 '13 at 10:51
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then, no, the sequence does not have to converge to $s_0$ (the sequence is not given to be monotonically increasing). –  Ittay Weiss Jan 23 '13 at 10:56
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@IttayWeiss Oh, i see the problem, i should rather prove it converges to $\limsup$ instead of $\sup$, is it better now??? –  user56876 Jan 23 '13 at 10:58
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@Freddy, i think considering increasing or decreasing is not work as a sequence may converge to a point with oscillating between positive and negative values. So may be what you mean is considering the sequence $\{t_m\}_m$ where $t_m=|s_m|$ right? –  user56876 Jan 23 '13 at 11:03

2 Answers 2

A hint:

It seems that you are allowed to use sup and inf. Assume for a moment that the theorem is true. How could you express $\lim_{n\to\infty} s_n$ in terms of sup's and inf's of certain sets connected with the $s_n$? Maybe you also have a theorem about bounded monotone sequences at your disposal.

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$(x_n)\in\mathbb{R}^\mathbb{N}$

$(x_n)$ converges to $x$ $\Leftrightarrow \forall \epsilon > 0, \exists n_\epsilon\in\mathbb{N}, \forall n \ge n_\epsilon, |x-x_n|<\epsilon$

$(x_n)$ is Cauchy $\Leftrightarrow \forall \alpha > 0, \exists n_\alpha\in\mathbb{N}, \forall n \ge n_\alpha,\forall m \ge n_\alpha, |x_m-x_n|<\alpha$



The idea behind this proof is that you can take two sequences, the sup and the inf of the terms after some $n$ and then proving that these two sequences converge because they are adjacent sequences (after proving they exists by proving the original sequence is bounded). And since the original sequence is between the sup sequence and the inf sequence for all $n$, you get the convergence of your original sequence at the same time.



Suppose $(x_n)$ converges to $x$

Take some $\alpha>0$

Use the convergence property with $\epsilon=\cfrac{\alpha}{2}$

Take some $n_\epsilon$ such as $\forall n \ge n_\epsilon, |x-x_n|<\epsilon$

Then $\forall n \ge n_\epsilon,\forall m \ge n_\epsilon, |x_m-x_n|\le|x-x_n|+|x-x_m|=\cfrac{\alpha}{2}+\cfrac{\alpha}{2} = \alpha$

So $(x_n)$ is Cauchy



Suppose So $(x_n)$ is Cauchy

Use the fact it is Cauchy with $\alpha=1$ to take a $n_1\in\mathbb{N}$ such as $\forall n \ge n_1,\forall m \ge n_1, |x_m-x_n|<1$

Now let $m=n_1$

$\forall n \ge n_1,|x_{n_1}-x_n|<1$

This means that $\forall n \ge n_1, x_n \in ]x_{n_1}-1, x_{n_1}+1[$ so $\{x_n, n\ge n_1\}$ is bounded

$\{x_n, n< n_1\}$ is finite and hence bounded

So $\{x_n, n\in\mathbb{N}\}=\{x_n, n< n_1\}\cup\{x_n, n\ge n_1\}$ is bounded

$\forall n \in\mathbb{N},$ let $S_n=\{x_m, m\ge n\}$

It is a subset of $\{x_n, n\in\mathbb{N}\}$ so it is bounded


Let $(i_n)=(\inf S_n)$ and $(s_n)=(\sup S_n)$

$(i_n)$ is increasing and majored by $s_0$ so it converges

$(s_n)$ is decreasing and minored by $i_0$ so it converges


Suppose $d=\lim\limits_{n\to\infty}s_n-i_n > 0$

Then $\forall \epsilon>0, \exists n_0 \in \mathbb{N}, \forall n\ge n_0, |(s_n-i_n)-d|<\epsilon$

Take $\epsilon=\cfrac{d}{2}$ and chose $n_0 \in \mathbb{N}$ such as

$\forall n\ge n_0, |(s_n-i_n)-d|<\cfrac{d}{2}$

$\forall n\ge n_0, d-\cfrac{d}{2}=\cfrac{d}{2}<(s_n-i_n)$

But $\forall n\ge n_0, \exists p \ge n, \exists q \ge n, i_n=x_p, s_n=x_q$

So you get $\forall n\ge n_0, \exists p \ge n, \exists q \ge n, \cfrac{d}{2}<(x_p-x_q)$

But since your sequence is Cauchy, with $\alpha = \cfrac{d}{2}$

$\exists n_0\in\mathbb{N}, \forall p \ge n_0,\forall q \ge n_0, |x_p-x_q|<\cfrac{d}{2}$

And these two statements contradict each other

So you get $d=\lim\limits_{n\to\infty}s_n-i_n = 0$


Now you have

  • $i_n$ increasing and bounded above
  • $s_n$ decreasing and bounded below
  • $\lim\limits_{n\to\infty}s_n-i_n=0$

You must have a theorem that proves that both converge and

$\lim\limits_{n\to\infty}i_n=\lim\limits_{n\to\infty}s_n$

And since you have $\forall n \in \mathbb{N}, i_n \ge x_n \ge \s_n$, you get that $(x_n)$ converges and $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}i_n=\lim\limits_{n\to\infty}s_n$

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