Is there a classification of Metric Spaces? As in group theory, there is a concept of isomorphism between metric spaces called isometry.
Two metric spaces $X$ and $Y$ are isometric if there is a function that perserves the distance of two elements. That function is called isometry.
The thing is that properties of metric spaces (completeness, compactness, connectedness, etc.) are preserved  under isometry. 
So, thinking about the classification of finite simple groups, I was wondering if there is any classification of Metric Spaces up to isometry, or at least a specific category of Metric Spaces (like finite simple groups in Group Theory). Also, I am interested if there is a more general topological classification of Metric Spaces up to homeomorphism (isomorphism in topology).
 A: That's too ambitious, there is an enormous variety of metric spaces. But if you restrict to specific classes, then a lot has been done. I can think of two examples: Hilbert spaces and geodesic surfaces. These are completely classified. 
Hilbert spaces are vector spaces equipped with a scalar product, which induces a norm, hence a distance and so a structure of metric space, which is required to be complete. Hilbert spaces are completely characterized up to isomorphism (a stronger form of isometry); in particular, any separable real Hilbert space is isomorphic to 
$$
\ell^2:=\left\{\boldsymbol{x}=(x_1, x_2, x_3, \ldots)\ :\ x_j\in\mathbb R,\ \sum_{j=1}^\infty x_j^2<\infty\right\},$$
where the scalar product is given by 
$$
\langle \boldsymbol{x}, \boldsymbol{y}\rangle = \sum_{j=1}^\infty x_j y_j.$$ 
(Non-separable Hilbert spaces are classified in terms of cardinality of their orthonormal bases. Also, the same result holds in the complex case, with obvious modifications). 
Geodesic surfaces are something I know less. I will refer to this beautiful note of Etienne Ghys, section 3. It is in French, but I am sure I have seen an English translation on the net.
A: I don't think so, but two metrics are equivalent iff there are $m, M$ such that $1/md'(x,y)\le d(x,y)\le Md'(x,y)$ for all $x,y\in X$.
