The sequence $(\frac{1}{n})$ using the discrete metric is Cauchy?

Question

Considering the sequence $(\frac{1}{n})$ . Using the discrete metric, is it Cauchy?

That's what I thought,

With the adopted metric, we have

$d(x_{m},x_{n})= 0$ , if $x_{m}$ = $x_{n}$

$d(x_{m},x_{n})= 1$ , if $x_{m}$ $\ne$ $x_{n}$

If $\epsilon$ >0, there is $n_{0}(\epsilon)$ such that

$d(x_{m},x_{n})$ < $\epsilon$ $\forall$ $m,n$ > $n_{0}(\epsilon)$

If we use $0 < \epsilon < 1$, and $m\ne n$ then

$d(x_{m},x_{n}) = 1 > \epsilon$

So it is not Cauchy with the adopted metric.

Answer 1

You are correct. I would only change the language slightly in one place to improve the clarity. Instead of

If $\epsilon$ >0, there is $n_{0}(\epsilon)$ such that

$d(x_{m},x_{n})$ < $\epsilon$ $\forall$ $m,n$ > $n_{0}(\epsilon)$

I would say

If $\{ x_n \}$ is a Cauchy sequence, then for every $\epsilon > 0$ there is . . . (same as before)