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

3 Answers 3


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?

  • $\begingroup$ 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

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.


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 $L$.

  • $\begingroup$ 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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