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I'm trying to prove the following:

If $(a_n)$ is a cauchy sequence that does NOT tend to 0, then $\exists N$ s.t. $\forall n > N, a_n \neq 0$.

Here's my proposed proof (please excuse the poor writing):

$(a_n)$ does not tend to 0 implies that there exists a smallest $\epsilon > 0$ s.t. $\forall M \exists k > M$ s.t. $|a_k| > \epsilon$. This smallest $\epsilon$ must exist, else we could choose arbitrarily smaller $\epsilon$ s.t. $|a_n| < \epsilon$ and then $(a_n)$ would tend to 0, which is false.

Since $(a_n)$ is a cauchy sequence, choose $N$ s.t. $|a_n - a_m| < \epsilon$ (the $\epsilon$ above) $\forall n, m > N$.

Then $|a_n - a_k| < \epsilon$ (substituting $a_k$ for $a_m$ since one of the $a_m$ must satisfy the property that $a_k$ has)

$-\epsilon < a_n - a_k < \epsilon$

$a_k - \epsilon < a_n < a_k + \epsilon$

If $a_k > \epsilon$, then $0 < a_k - \epsilon < a_n$

If $a_k < -\epsilon$, then $a_n < a_k + \epsilon < 0$

In either case, $a_n \neq 0$

Is my proposed proof correct? Is there a simpler, more elegant way to do it? The part I'm most wary of is where I claim the existence of a smallest $\epsilon$ since the sequence doesn't tend to 0. It makes sense to me, but I'm not sure if I'm allowed to.

P.S. We haven't proved that all cauchy sequences in the reals converge, so I'm not sure if I'm allowed to use that fact.

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  • $\begingroup$ But $\mathbb{R}$ is Banach, so every Cauchy sequence converges $\endgroup$
    – Juniven Acapulco
    Commented Dec 26, 2016 at 1:06
  • $\begingroup$ @juniven thank you, but we haven't proven that yet so I'm not sure if we're allowed to use that $\endgroup$
    – kevlar
    Commented Dec 26, 2016 at 1:06
  • $\begingroup$ Its a standard result, no need to prove it. The result is known as the "Cauchy Convergence Criterion" $\endgroup$
    – Juniven Acapulco
    Commented Dec 26, 2016 at 1:08
  • $\begingroup$ see proofwiki.org/wiki/Cauchy%27s_Convergence_Criterion $\endgroup$
    – Juniven Acapulco
    Commented Dec 26, 2016 at 1:10
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    $\begingroup$ Also you should omit "smallest" and just say "there exist an $\epsilon>0$. Which is the negation of the definition of the convergence. $\endgroup$
    – Momo
    Commented Dec 26, 2016 at 1:50

1 Answer 1

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Instead of choosing $N$ s.t. $\lvert a_n-a_m\rvert<\epsilon$, you might choose $N$ s.t. $\lvert a_n-a_m\rvert<\epsilon/2$

Then for $n>\max\{M,N\}$ and $k>n$ with $|a_k|>\epsilon$ use $|x-y|\ge |x|-|y|$:

$\lvert a_n\rvert=\lvert a_k-a_k+a_n\rvert\ge \lvert a_k\rvert-\lvert a_k-a_n\rvert> \epsilon-\epsilon/2=\epsilon/2$

This means that $a_n$ cannot be $0$

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  • $\begingroup$ Beautiful, thanks. It seems like every proof needs you to add 0 and use some inequality involving absolute values :) $\endgroup$
    – kevlar
    Commented Dec 26, 2016 at 2:38
  • $\begingroup$ One issue that still confuses me: if we don't choose a smallest $\epsilon$ (which I agree seems hokey), then it doesn't appear to me that the claim that $\exists \epsilon$ s.t. $\forall M \exists k$ s.t. $|a_k| > \epsilon$ is true anymore. Instead, it seems you must first choose an $M$, from which $\epsilon$ falls out. In other words, it seems the proper claim is $\forall M \exists \epsilon > 0$ and $\exists k$ s.t. $|a_k| > \epsilon$. It seems to me that $\epsilon$ is a function of $M$, which is not how I originally worded it. Or can we choose $\epsilon$ first? If so, why? $\endgroup$
    – kevlar
    Commented Dec 26, 2016 at 2:49
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    $\begingroup$ $$a_n\rightarrow0\iff \forall\epsilon>0\ \exists M\ \forall n>M\ |a_n|<\epsilon$$ $$a_n\nrightarrow0\iff \exists\epsilon>0\ \forall M\ \exists n>M\ |a_n|\ge\epsilon$$ $\endgroup$
    – Momo
    Commented Dec 26, 2016 at 2:59
  • $\begingroup$ That clears up everything, thanks $\endgroup$
    – kevlar
    Commented Dec 26, 2016 at 3:49

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