# Are the set of integers not complete?

• Sep 11th 2009, 05:14 PM
Are the set of integers not complete?
Define the metric spaces with $( \mathbb {Z} , \rho )$, the set of integers with the metric $\rho (x,y) = \mid x - y \mid$

Find sequence $\{ x_n \}$ in this metric space such that it is cauchy but do not converge.

I understand I must find a sequence that takes only integer values but converge to non-integer. But how should I get it to be cauchy?

Thank you.
• Sep 11th 2009, 05:28 PM
Taluivren
By setting Cauchy constant $\varepsilon = 1/2$ you easily see how all Cauchy sequences in this space must look like. And to what values they converge.
• Sep 11th 2009, 07:13 PM
So suppose that $\{ x_n \}$ is a cauchy sequence in $\mathbb {Z}$, then $\forall \epsilon > 0$, there exists $N \in \mathbb {N} \ s.t. \ \forall n,m \geq N$, we have $\mid x_n - x_m \mid < \epsilon$

Set $\epsilon = \frac {1}{2}$, we then have $\mid x_n - x_m \mid < \frac {1}{2}$

Am I on the right track here?
• Sep 11th 2009, 08:09 PM
ThePerfectHacker
Quote:

So suppose that $\{ x_n \}$ is a cauchy sequence in $\mathbb {Z}$, then $\forall \epsilon > 0$, there exists $N \in \mathbb {N} \ s.t. \ \forall n,m \geq N$, we have $\mid x_n - x_m \mid < \epsilon$
Set $\epsilon = \frac {1}{2}$, we then have $\mid x_n - x_m \mid < \frac {1}{2}$
It is not possible for $|x_n-x_m| < \tfrac{1}{2}$ unless $x_n = x_m$ because $x_n,x_m$ are integers and so their difference is always an integer. Thus, the only way you can have an integer less than 1/2 is when that integer is zero.