# Are some discrete (and all finite) metric spaces complete?

For example, it seems to me from the definition of complete that $\mathbb{N}$ with (say) the Euclidean metric would be complete, since any Cauchy sequence on $\mathbb{N}$ must converge to an integer. (That is, it would look like 5,1,4,2,3,3,3,3,3,3....) So is it actually complete?

Similarly, it seems that any metric space on a finite set would be complete as well. Is this correct, and if not, why not?

If you have suggestions for texts or online resources that would help me with these types of definitions, that would also be awesome. Thanks very much!

-
....and some discrete metric spaces are not complete. For example $\{1/n : n\in\mathbb{N}\}$ with the usual metric. – Michael Hardy Mar 30 '12 at 2:38

You are correct in your reasoning. More generally, if $(M, d)$ is a metric space such that there exists $\epsilon > 0$ such that for any $x, y \in M$, we have $d(x,y) > \epsilon$ whenever $x\neq y$, then $M$ is complete.

To see this, say $\{x_n\}$ is a Cauchy sequence. Then there exists $N\in\mathbb{N}$ such that for all $n, m \geq N$, $d(x_n, x_m) < \epsilon$ (same $\epsilon$ as above). But if $x_n\neq x_m$, then $d(x_n, x_m) > \epsilon$. In other words, for all $n, m \geq N$, $x_n = x_m$ - that is, the "tail" of the sequence is constant. Hence the sequence obviously converges.

-
Great, thanks very much! – usul Mar 30 '12 at 2:18

Yes, this is correct. As WNY points out in the comment below, not every discrete metric space is complete. Every discrete metric space is completely metrizable, however. This means that we can change the metric to an equivalent one, so that the space becomes complete in the new metric.

Equivalence of metrics is best seen from the point of view of general topology, as it means that the two metrics yield the same topology, but can also be described purely in terms of metrics: two metrics $d_1$ and $d_2$ are equivalent if every open ball $B_1$ in the first metric contains $B_2$, an open ball in the second metric and vice versa.

The following example should be helpful:

Let $M$ be an arbitrary set. For $x,y\in M$, define $$d(x,y)=\begin{cases}0; &\textrm{if }x=y\\1; &\textrm{if }x\neq y\end{cases}$$ Then $(M,d)$ is a complete metric space. Such a metric is usually referred to as $the$ discrete metric. (But note that not every discrete metric space is of this form.)

-
Is $\{x: x = 1/k, k\in\mathbb{N}\}$ with the Euclidean distance a discrete metric space? If by "discrete" we mean that any singleton set is open, then the answer is yes. But this space isn't complete. – William Mar 30 '12 at 2:10
@WNY: You are right. I mixed it up with completely metrizable. I shall edit appropriately. – Dejan Govc Mar 30 '12 at 2:14
I see, thank you! – usul Mar 30 '12 at 2:18
@usul: I apologize for the mistake before. I hope the new version of my answer is more helpful. – Dejan Govc Mar 30 '12 at 2:39
It is, no problem! – usul Mar 30 '12 at 3:09

Let $(X,d)$ be a complete metric space, and let $C$ be a subset of $X$ such that the intersection of $C$ with every ball is closed, then $C$ is complete as well.

We now try to prove the statement above. Let $(x_n)$ be a Cauchy sequence in $C$ , then $(x_n)$ is bounded. Hence $(x_n)$ is contained in some ball of $(X,d)$. By the above property $(x_n)$ is a cauchy sequence in a closed subset of a complete metric space, and hence $C$ is a complete metric space.

To see how the above result applies to your particular example of $\mathbb{N}$, consider the intersection of $\mathbb{N}$ with any ball $B(x,r)$ for $x\in\mathbb{R}$ and $r>0$. Then $\mathbb{N}\cap B(x,r)$ is a finite subset of $\mathbb{R}$ and hence closed. By the above result $\mathbb{N}$ is hence a complete metric space (seen as a subspace of $\mathbb{R}$ with its standard topology).

-