I tried to prove this but my proof did not match with my book's. So I want to verify whether my proof is correct or not.

Theorem: Every Cauchy sequence in $\mathbb R$ has a limit.

Let us assume the contrary that there is a sequence $(a_n)$ which is Cauchy but not convergent.

1.Since the sequence is not convergent,for all real $a$, there must be an $\epsilon$ such that for all $n \in \mathbb N$, $\exists n_0 \geq n$ such that $|a_{n_0}-a|\geq \epsilon$.

2.Since $(a_n)$ is Cauchy, we can show that that particular $\epsilon$ we talked above,there is a natural $N$ such that for all $n,m\geq N$, $|a_n-a_m|<\epsilon$.

3.Go and look $(1)$. I can thus find $m_0 \geq N$ such that $|a_{m_0}-a| \geq \epsilon >|a_{m_0}-a_n|$ for all $n \geq N$. Since $a$ is arbitrary, putting $a=a_n$,we get contradiction.Thus, the proof.

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    $\begingroup$ Your $\epsilon $ should be dependent on $a$ so you should state "for every $a\in\mathbb {R} $ there exists an $\epsilon >0$ corresponding to it such that..." $\endgroup$
    – Paramanand Singh
    Sep 11, 2021 at 13:00

2 Answers 2


In step 1, you're fixing $n_0$ which has the property $|a_{n_0} - a| \geq \epsilon$, but in step 3, you're using that $|a_{m_0} - a| \geq \epsilon$ for your $m$ large enough.

You can't first fix $n_0$ and then replace it. That's already assuming that your sequence $a_n$ converges to $a$, which is what you're trying to prove.

The standard argument constructs (using Bolzano-Weierstrass) a convergent subsequence of $a_n$. It can then be shown that any Cauchy sequence with a convergent subsequence is convergent.

A quicker way to see that your argument doesn't work is the following: it doesn't rely on the "completeness" of $\mathbb{R}$. You will learn that - in more general settings - convergent sequences are always Cauchy, but not all Cauchy sequences converge. For this you really need the completeness (which is used in the proof of Bolzano Weierstrass).

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    $\begingroup$ I'd add to your final paragraph: the theorem is false in $\mathbb{Q}$, but no line of the proof cares whether we're in $\mathbb{Q}$ or $\mathbb{R}$, so the proof must be wrong. $\endgroup$ Sep 11, 2021 at 10:50
  • $\begingroup$ I do agree that my proof is wrong. But the part of argument that you have stated to be wrong is correct. I guess there is mistake in other part of my answer. $\endgroup$
    – user953078
    Sep 11, 2021 at 11:50
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    $\begingroup$ That part is correct since,all I did was to choose $N$ as $n$ of (1). There is nothing wrong in that. $\endgroup$
    – user953078
    Sep 11, 2021 at 11:51
  • $\begingroup$ I tried to find the fault and am writing an answer to this. Please let me know if my answer is correct. $\endgroup$
    – user953078
    Sep 11, 2021 at 11:52

From other answers, I became completely sure that my proof must be wrong because if I imitate my proof for a Cauchy sequence in $\mathbb Q$,the same result would follow. But here is what went wrong in my proof:
I claimed to replace $a_n$ in my proof in (3) and then,tried to get a contradiction. But to replace a particular $a_n$, I must first know $N$ in the above proof.
Note that $\epsilon$ is dependent upon $a$,the real number. Now,when I do choose $a=a_n$ in my proof, I get that it must give a particular $\epsilon$ which in turn would create a new value of $N$ and I can not make sure that the value of $a_n$ I have choosen is such that $n \geq N$.

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    $\begingroup$ Yes, this is correctly locating the error. To put it another way: In step (2), you say “…for that particular ε”, and subsequently you work with that particular ε. But before you can fix a specific such ε, you would have to say “Fix some a…” — only then can you take some ε which has the property from (1) with respect to your given a. $\endgroup$ Sep 11, 2021 at 19:17

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