Is spacetime isomorphic to a metric space? I know that spacetime, as described by General Relativity (GR), is a pseudo-Riemannian manifold. The label "pseudo" is due to the fact that the metric of spacetime entails not only positive distances, but also negative and null distances (resulting in the three kinds of spacetime intervals: space-like, time-like, and light-like). Thus, spacetime has a very different geometry than any metric space.
My question: I wonder whether there can be an isomorphism between spacetime (defined as a pseudo--Riemannian manifold) and at least one metric space? In other words, can the GR spacetime be smoothly transformed into any metric space? If not (as my intuition says), is there proof of this fact?
 A: It is worth noting that oftentimes in mathematical analysis, metric spaces are used to define the topology of a space. Say we are considering Minkowski spacetime in $\mathbb{R}^4$. Then our Minkowski spacetime has almost all of the properties of the Wikipedia page on metric spaces, so long as we keep in mind that the distance function used to define the topology is the usual distance function $d(\vec{x},\vec{y})=\|\vec{x}-\vec{y}\|$. Convergence, limits, completeness, compactness of subsets, continuity, etc., are all defined using "the usual topology" -- the one that comes from this Euclidean metric $d$. So as far as topology is concerned, in general relativity we already have access to a metric space (the metric spaces of subsets of $\mathbb{R}^4$ in each of our coordinate charts). I bring this up because it is a concern I had when I was learning about differential geometry, the topology is not defined using the Minkowski metric.
A separate question is asking what Riemannian manifolds we can define given a manifold in general relativity. You would have to specify what structures you want to preserve on it, and what kind of isomorphism you want to talk about!
A: You can certainly equip a generic spacetime with a metric, if you choose to do so. For example, you could define a Riemannian metric tensor $\gamma$ from the pseudo-Riemannian metric tensor $g$ by simply flipping the relevant sign(s), and then use $\gamma$ to define a metric on the spacetime manifold. Of course, this metric will generically be physically meaningless.
A: Yes it is, but the found structure has no physical meaning in general.
As a matter of fact, as a general result on differentiable (paracompact) manifolds,  every smooth manifold $M$ (without a preferred metric) can be equipped with a smooth Riemannian metric $h$. Next, assuming $M$ is connected, you can define a distance on $M$ whose metric topology is the initial topology on $M$, so that $M$ is homeomorphic to a metric space $(M,d)$, the homeomorphism being the identity map. This distance $d(p,q)$ is nothing but the $\inf$ of the lengths of the smooth paths joining $p$ and $q$ in  $M$ computed with respect to $h$:
$$d(p,q) := \inf\left\{ \left.\int_\gamma \sqrt{h(\dot{\gamma},\dot{\gamma}) }ds \:\right|\gamma \in C^\infty([a,b]; M)\:, \gamma(a)=p\:,\gamma(b)=q \right\}\:.$$
If $(M,g)$ is a time-orientable smooth Lorentzian manifold, there is an even shorter way to construct a (smooth) Riemannian metric $h$ on $M$. If $T$ is an everywhere defined timelike smooth vector field on $(M,g)$ (it exists because the manifold is time orientable) where $g$ is the (smooth) Lorentzian metric, a Riemanniam metric can be defined as
$$h(X,Y):= g(X,Y) - \frac{2g(T,X)g(T,Y)}{g(T,T)}\:.$$
Indeed, $h$ is smooth by construction. Next for any $p\in T_pM$ refer to a $g$-pseudo orthonormal basis of $T_pM$, $e_0,...,e_{n-1}$,  with $T$ parallel to $e_0$. Here, $X=X_p^\mu e_\mu, Y=Y_p^\nu e_\nu \in T_pM$ and the definition above yields, point by point,
$$h_p(X,Y)=  X^0_pY_p^0 +\sum_{a=1}^{n-1}X_p^aY_p^a$$
which is evidently a positive metric.
