# Proving every Cauchy sequence is convergent. Does the following proof work?

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.

• 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..." Sep 11 at 13:00

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).

• 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. Sep 11 at 10:50
• 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.
– user953078
Sep 11 at 11:50
• That part is correct since,all I did was to choose $N$ as $n$ of (1). There is nothing wrong in that.
– user953078
Sep 11 at 11:51
• I tried to find the fault and am writing an answer to this. Please let me know if my answer is correct.
– user953078
Sep 11 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 $$n \geq N$$.

• 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. Sep 11 at 19:17