# Showing that $d(m,n)=|m^{-1}-n^{-1}|$ is not a complete metric on $\mathbb Z^+$

Let $X$ be a set of all positive integers and define metric $d$ on $X$ by $d(m,n)=|m^{-1} - n^{-1}|$. I'm required to show $(X,d)$ is not a complete space.

SOLUTION:

Let $\{x_n\}$ be any Cauchy sequence in $X$. Then choose $\epsilon=0.5$; there exists a number $N$ such that $d(x_n,x_m)<0.5$ for all $m,n>N$.

$d(x_n,x_m)=|x_n^{-1} - x_m^{-1}|<0.5$.

I'm not sure if $\{x_n\}$ may converge here, I beg help please.

kind regards

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I guessed that what you wrote as $n-1$ and $m-1$ were intended to be $n^{-1}$ and $m^{-1}$; if that’s not right, please let me know. – Brian M. Scott Apr 14 '12 at 10:58
If it is right, consider the sequence $x_n=n$. Is it Cauchy? Does it converge? – Brian M. Scott Apr 14 '12 at 11:01
A way of coming up with a counterexample here is to ask yourself the question: when are two integers $m$ and $n$ close to each other with respect to this metric? Obviously this happens, if $m=n$, but that's the case with every metric space. Another automatic way for the distance to be small in this case is, when both $m$ and $n$ are large (as then their reciprocals are small). There we have more elbow room, and a chance to come up with a Cauchy-sequence that does not have a limit. – Jyrki Lahtonen Apr 14 '12 at 11:31

The sequence $x_n = n \in \mathbb{Z}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.
Using the deﬁnition of $d$, it is easy to verify that the sequence $x_n=(1,2,3,\ldots)$ is a Cauchy sequence in $X=Z^+$. We see that $\forall k \in X, d(x_n,k)\rightarrow k^{(-1)}$ as $n \rightarrow \infty$. Hence $(x_n)$ does not converge in $X$ and so $X$ is not complete.