1. ## Cauchy Sequences

I am working on a problem and was just reading a thread where a user posted :

$(X,d)$ is metric space. for sequence $(x_n)_{n\in \mathbb{N}} \subset X$ we say that is Cauchy sequence if

$(\forall \varepsilon >0) (\exists n_0=n_0(\varepsilon) \in \mathbb{N}) (\forall n,m \in \mathbb{N}) (n,m\ge n_0 \Rightarrow d(x_n,x_m) < \varepsilon )$

meaning that, sequence is Cauchy sequence if

$\displaystyle \lim _{n,m \to \infty } d(x_n,x_m) = 0$

Is it sufficient to show just

$\displaystyle \lim _{n,m \to \infty } d(x_n,x_m) = 0$

for a sequence to be cauchy? Or should it be done using the epsilon definition (which i am struggling with).

Basically i'm trying to construct a metric on the real numbers so that $(R,d)$ is not complete. I have created such a metric (i think) with
$d(x,y)= |exp(-x) - exp(-y)|$ but i am struggling to show with the epsilon definition that the sequence of natural numbers is cauchy on it.

2. If $n\le m$ (both positive integers) then $d(n,m)\le e^{-n}$.

3. $n\le m$ (both positive integers) then $d(n,m)\le e^{-n} < e^{-N} < epsilon$
?

So i should take N > ln(1/epsilon)?

4. Originally Posted by Scopur
$n\le m$ (both positive integers) then $d(n,m)\le e^{-n} < e^{-N} < epsilon$
?

So i should take N > ln(1/epsilon)?
I would just note that $e^{-n}\to 0$, so an N can be found for any positive epsilon.