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For the metric space defined by distance formula,

$$d(p,q) = \begin{cases}1,&\text{if }p \ne q\\ 0,&\text{if }p = q\;,\end{cases}$$ does the notion of number of dimensions exist?

If yes, then is dimension of the above metric space uncountable?

Can a discrete metric space exist in finite dimensions?

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From "Introduction to Topological Manifolds" by John M. Lee one learns that a discrete space (metric, or equivalently topological) is a zero-dimensional topological manifold if it is countable. That is since a topological manifold is second-countable by definition, and any discrete space is second countable iff it is countable. I don't know if the dimension of an uncountable discrete space is defined. – Ilya Mar 2 '12 at 15:14
@Ilya: Some people require a manifold to be second countable; others do not. And there are many, many notions of dimension. – Brian M. Scott Mar 2 '12 at 15:31
@BrianM.Scott: thanks, do you think that my comment may be confusing then? – Ilya Mar 2 '12 at 15:46
@Ilya: Not really, since you’ve specified the sense in which you’re using the term manifold. – Brian M. Scott Mar 2 '12 at 15:48
up vote 6 down vote accepted

There are several different notions of dimension for topological spaces in general and hence metric spaces in particular, including small and large inductive dimension and covering dimension; all discrete metric spaces have dimension $0$ in all of these senses. There is also a notion of Hausdorff dimension, that applies specifically to metric spaces. If your discrete metric space is countable, its Hausdorff dimension is also $0$; if it’s uncountable, its Hausdorff dimension is $\infty$. There are other notions of dimension for metric spaces, but these are the most familiar.

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+1, but how the Hausdorff dimension of a uncountable discrete space is $1$? When I was following the construction here I've obtained that $C^d = 1$ for all $d\geq 0$ (though it maybe not the best source to rely on) – Ilya Mar 2 '12 at 15:50
@Ilya: Thanks for catching that: it should be $\infty$. – Brian M. Scott Mar 2 '12 at 16:22
then everything is clear - I thought I should have missed some point. – Ilya Mar 2 '12 at 16:23

As a manifold, this object is zero dimensional, as every point is open, the set of all points is an open cover, and the unique map $\{p\}\to\mathbb{R}^0=\{0\}$ is a homeomorphism for each $p$.

(Edited after Ilya's comment: This assumes the metric space is countable, otherwise it is not second countable and thus not a manifold.)

There are also other notions of dimension for metric spaces though, such as Hausdorff dimension, and they may not agree with this one. For a start many metric spaces are not manifolds, so this approach won't give you an answer at all.

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