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http://en.wikipedia.org/wiki/Complete_metric_space says that a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M.

There is no Cauchy sequence in the set of integers $Z$, or any discrete set for that matter. Hence, is it Cauchy complete, as all the Cauchy sequences it has (a null set) have a limit inside $Z$?

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    $\begingroup$ But! If we modify the metric such that for example $d(n,m):=|\frac1n-\frac1m|$, then it will still have the discrete topology, but it will also have a Cauchy sequence that wants to tend to $+\infty$.. $\endgroup$ – Berci Apr 10 '13 at 10:37
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Cauchy sequences exist in such metric spaces (ie. when there exists $\alpha>0$ such that $d(x,y) \geq \alpha$ for all $x \neq y$): they are just eventually constant.

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  • $\begingroup$ What do you call 'discrete metric space' ? If it is a metric space whose topology is discrete, then Cauchy sequences are NOT eventually constant (cf Berci's comment, or consider the space $\{ \tfrac{1}{n} : n \in {\mathbb{N}} \} \subset {\mathbb{R}})$. $\endgroup$ – user10676 Apr 10 '13 at 11:39
  • $\begingroup$ @user10676: Indeed, it is not really clear; I edited my answer. $\endgroup$ – Seirios Apr 10 '13 at 12:26
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$\Bbb Z$ is indeed Cauchy complete, but not for the reason you state.

Unravelling the definition for a Cauchy sequence, we get that:

$$\forall \epsilon > 0: \exists N: \forall m,n > N: d(a_n, a_m) < \epsilon$$

and for $\epsilon = \frac 12$ we note that this must mean $d(a_n,a_m) = 0$ (since otherwise it exceeds $1$) i.e. $a_n = a_m$ for some $N$ and all $m,n \ge N$. That is, $a_n$ is eventually constant.

Now there is an obvious guess for the limit of eventually constant sequences, and we conclude that $\Bbb Z$ with the Euclidean metric is complete.

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