I am trying to understand better Cauchy sequence in the book that I am reading he is trying to provide counter argument that in general metric spaces we can find sequence that are Cauchy but aren't convergent for example $\frac{1}{n}$ is Cauchy but doesn't converge in the space T = (0,1].

I am trying to prove that the sequence is Cauchy to begin with if we consider $n \geq m > 0$ Consider $|a_n - a_m| = \frac{m - n}{mn} < \frac{m}{mn} = 1/n < \epsilon$. So if we choose N = $\frac{1}{\epsilon}$ that should work since for $n, m > N = \frac{1}{\epsilon}$

we have the following:

$|a_n - a_m| = \frac{m - n}{mn} < \frac{m}{mn} = \frac{1}{n} < \epsilon$.

what do you guys think any the proof is good or is there anything to make it better ? It is always good to learn from the extra bit of details as it makes my brain understand the subject more !

  • 2
    $\begingroup$ You want to say that $|a_n-a_m|=\frac{|n-m|}{mn}$ and that $|n-m|\lt n$. Note that $|n-m|\lt m$ is not true, $m=100$, $n=5000$. $\endgroup$ Jul 25, 2015 at 19:19
  • $\begingroup$ first, it should be written as $|a_n - a_m|= | \frac{m-n}{mn} |=\frac{ |m-n|}{mn} $ absolute value sign here is important as $m-n \leq 0 $. $\endgroup$
    – Shreya
    Jul 25, 2015 at 19:20
  • $\begingroup$ $N$ is usually taken to be an integer, while $1/\epsilon$ generally isn't. This isn't a big problem, though, and is easily remedied by saying, for instance, "let $N\in \Bbb N$ be larger than $1/\epsilon$", or something to that effect. However, while it is common practice, it is not necessary, and what you've done works just fine. $\endgroup$
    – Arthur
    Jul 25, 2015 at 20:03

1 Answer 1


Take an arbitrary $\epsilon > 0$. I think you can say that \begin{equation*} |a_n - a_m| = \left| \frac{1}{n} - \frac{1}{m} \right| \leqslant \frac{1}{n} + \frac{1}{m}. \end{equation*} If you choose an index $N \in \mathbb{N}$ greater than $\dfrac{2}{\epsilon}$, then you have \begin{equation*} |a_n - a_m| \leqslant \frac{1}{n} + \frac{1}{m} \leqslant \frac{1}{N} + \frac{1}{N} < \epsilon \quad \text{for all}\ m, n \geqslant N. \end{equation*} Therefore, $\left\{\dfrac{1}{n}\right\}$ is a Cauchy sequence.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .