Distances in general relativity Three dimensional distance doesn't exist in general relativity because the integration of $ d \vec l $ depends on the integration path. Well, maybe there is no mininum but it must be a maximum lower bound. Why can't we define the threedimensional distance this way?
 A: If you do not assign a 3D spacelike submanifold where to compute the variations you say (and there are infinite inequivalent choices), the infimum is always $0$. In fact, you can always continuously deform every given spacelike curve joining two spacelike related points  to a light-like curve. This fact implies that the spatial distance between two spacelike related events has no non-ambiguous meaning. To define it,  you have to choose a spacelike 3-surface containing the points and to take the infimum of the length of the curves joining the points an belonging to the 3-surface (referring to the metric induced from that of the spacetime). This notion of distance, though depending on a arbitrary choice, will satisfy all requirements of a distance function as the triangular inequality and all that.
A: The distance is already observer dependend in special relativity, for an observer flying from $\rm A$ to $\rm B$ the distance between $\rm A$ and $\rm B$ is also shorter than for an observer at rest relative to $\rm A$ and $\rm B$ because of regular length contraction. In general relativity we have the same situation: if we take an observer freely falling from infinity (meaning with the negative escape velocity) into a Schwarzschild black hole and let him measure the distance from the horizon to the singularity, he finds the distance to be $d=2$ (in natural units of $\rm GM/c^2$):
Approach 1: Droste coordinates
The radial component of the metric is $g_{rr}=-1/(1-2/r)$, and the escape velocity $v=\sqrt{2/r}$, with the gamma factor $\gamma=1/\sqrt{1-v^2}$. The distance our free faller measures between the singularity is therefore
$$d=\int_2^0 \frac{\sqrt{-g_{rr}}}{\gamma } \, {\rm d}r = 2$$
Approach 2: Raindrop coordinates
In raindrop coordinates we are already measuring the distances in the frame of our infalling observer, which gives us a velocity of $v=0$ and $\gamma=1$. The radial component of the metric tensor in this coordinates is simply $g_{rr}=-1$, so we again get
$$d=\int_2^0 \frac{\sqrt{-g_{rr}}}{\gamma } \, {\rm d}r = 2$$
If you start free fall from rest at an infinitesimal distance above the horizon, the numerically integrated distance in your frame appoaches $d=\pi$. On the other hand, if you try to calculate the distance in the frame of a stationary far away observer you get $d=-i \pi$, which is unphysical for the reason that there can't be any stationary observers between the horizon and the singularity. But if you transform into a physically meaningful frame of reference, it is no problem to calculate distances. The caveat that times and distances are relative are not a thing that only happens in general relativity, the same already happens in special relativity with the only difference that the metric tensor is Minkowski.
