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proof the following:

suppose ($s_n$) is a Cauchy sequence of real numbers. There exists a real number s such that lim n→∞ $s_n$ = s.

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    $\begingroup$ You cannot prove this without a definition of $\mathbb R.$ Definitions are tools to be used. There are different (but equivalent) def'ns of $\mathbb R.$ Which one are you using? $\endgroup$ Apr 9, 2017 at 3:00
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    $\begingroup$ @DanielWainfleet You are right. I see many strange answers which don't seem to care what results are allowed to be used. $\endgroup$ May 30, 2018 at 5:48

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For a sequence $\left\{s_n\right\}_{n=1}^{\infty}$ to be Cauchy, then: $\forall \epsilon > 0, \exists$ a cut off point $N\in \mathbb{N} \ s.t.$

for $m>n$, $$m,n>N\Rightarrow |s_m-s_n|<\epsilon$$

This means, given two sequences such that $m>n$, their tails approach each other arbitrarily close, for $m,n$ big enough.

So prove if $s_n$ is Cauchy, then its limit $\exists$ and converges to some real value $L$.

So suppose $\left\{s_n\right\}_{n=1}^{\infty}$ is a Cauchy sequence.

Then by definition, $\forall \epsilon > 0, \exists$ a cut off point $N\in \mathbb{N} \ s.t.$

$$m,n>N\Rightarrow |s_m-s_n|<\epsilon$$

Consider $m>n$, let's say $m=N+5$ (this choice is arbitrary), then by the triangle inequality we have:

$$|s_n-s_m|=|s_n-s_{N+5}|\Rightarrow |s_n|-|s_{N+5}|\leq|s_n-s_{N+5}|<5$$ upon choosing $\epsilon = 5$ (Why? Any $\epsilon>0$ will do).

So, we see: $|s_n|\leq5+|a_{N+5}|\Rightarrow \left\{S_n\right\}_{n=1}^{\infty}$ is bounded. (Why is this bounded, and what is it? Hint: it is the maximum of some set, including $|A_{N+5}|+5$. Left as an exercise for you).

Well because $s_n$ is bounded, that implies it has at least one convergent subsequence, call it $\left\{{s_{n_i}}\right\}_{i=1}^{\infty}$. This is true via the Bolzano-Weierstrass theorem which states, a non-empty bounded sequence contains at least one convergent subsequence.

So back to our original statement, $\forall \epsilon>0$, $\exists \ N \in \mathbb{N}$ for $m>n, \ s.t. \ $

$$m,n>N\Rightarrow |s_m-s_n|<\frac {\epsilon}{2}$$

Why do we choose $\frac {\epsilon}{2}?$

Because our subsequence $\left\{{s_{n_i}}\right\}_{i=1}^{\infty}$ converges to some number, let's say $L_i$, then $\forall \epsilon>0$, $\exists$ a cutoff $I \in \mathbb{N} \ s.t. \ $

$$i>I\Rightarrow|{s_{n_i}}-L_i|<\frac {\epsilon}{2}$$

So we have:

for $$n>N\Rightarrow m\geq i>I\geq N\Rightarrow |s_n-{s_{n_i}}|<\frac {\epsilon}{2}$$

Finally we see,

$\forall \epsilon>0$, $\exists$ a cutoff point $N \in \mathbb{N} \ s.t. \ $

$$n>N\Rightarrow |s_n-L|\leq |s_n-{s_{n_i}}|+|{s_{n_i}}-L_i|<\frac {\epsilon}{2} + \frac {\epsilon}{2} = \epsilon$$

Thus, we note by the $\epsilon,N$ definition of a limit, the sequence $\left\{s_{n}\right\}_{i=1}^{\infty}$ converges to $L_i=L$ (limit).

In fact, your claim holds for the converse as well, that is, if the sequence is convergent, the sequence is Cauchy. Can you prove this?

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  • $\begingroup$ Can you prove Bolzano-Weierstarss Theorem without knowing that $\mathbb R$ is complete? $\endgroup$ May 30, 2018 at 5:49
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    $\begingroup$ Yes, consider the proof of Bolzano–Weierstrass using nested intervals. Why down vote? $\endgroup$ Jul 23, 2018 at 7:54
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Hint: Cauchy sequences are bounded. Bounded sequences of real numbers have convergent subsequences. This subsequence will pull along the entire sequence because it is cauchy.

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Well this may be ham fisted.

Fix an $\epsilon > 0$. Then there exist an $N $ so that $n,m >N $ mean $|s_n-s_m|<\epsilon $. For each $n > N $ define the set $K_n=\{s_i|i \ge n\} $. Note that for every $s_i\in K_n $ that $s_n-\epsilon < s_i < s_n+\epsilon $ so each $K_n $ is bounded below. Define $b_n=\inf K_n $.

Notice further that each of $b_n $ that $s_{N+1}-\epsilon < b_n <s_{N+1}+\epsilon $ so the set of $b_n $ is bounded above. Let $L=\sup \{b_n\} $.

Okay, now for any $\gamma >0$ let $M $ be such that for all $m,n >M $,$|s_n-s_m|< .9\gamma$.

Let $n>\max(M,N)$. Then if $L \le s_n - \gamma $ then $b_n \le L $ and there is an $s_{i>n }$ so that $b_n \le s_i \le s_n-\gamma <s_n-.9\gamma$. That's a contradiction.

So $s_n -\gamma < L $. Now all $s_{i>n} $ are such that $s_i < s_n +.9\gamma $ so $b_i \le s_i < s_n - .9\gamma $. So $L=\sup \{b_i\}\le s_n +.9\gamma <s_n +\gamma$.

So $|s_n - L|<\gamma $ and so $\{s_n\} $ converges to $L$.

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  • $\begingroup$ I think the penultimate paragraph has a mistake. It should be $s_n + .9\gamma$ $\endgroup$
    – Kapol
    Oct 29, 2017 at 22:24

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