Topology of Minkowski Space Is the topology of Minkowski space the same as that of $\mathbb{R}^4$? My thoughts would be no, because of the very different inner products define very different metrics, and because the metric determines the open balls, it determines the topology. 
 A: The topology is $M^{1,3}~=~\mathbb R\times\mathbb R^3$ which is the product of the number line for time and the three space. The Lorentz group is a system of transformations between $\mathbb R$ and any direction within $\mathbb R^3$. Since light cones or null rays are invariant these transformations can't change the topology of the spacetime. Hence the topology of $M^{1,3}$ is not changed by Lorentz transformations.
This carries over to general relativity as well. The diffeomorphsims of spacetime are such that the topology of the spacetime is not changed. This has a parallel with quantum mechanics. If you could change the topology of the spatial part $\mathbb R^3$ so it is multiply connected then a Lorentz boost can convert this into a time loop. This means one could clone quantum states. This suggests that $no~topology~change~=~no~cloning$. That general relativity prevents topology change seems to prevent an attack of the quantum clones.
A: The topologies coincide since Minkowski spacetime is strongly causal.
https://en.wikipedia.org/wiki/Spacetime_topology 
A: gj255, ACuriousMind, Balu, Lawrence B. Crowell, Jackson Burzynski 
$\mathbb{R}^4$ is not the metric of space-time. It is just a space we begin with to indicate a coordinate system, in terms of which the metric or pseudo-metric may be described. The true topology (or pseudo-topology!) of space-time is to be found from the metric,  or more precisely, pseudo-metric, defined by the invariant interval. It is not necessary to set up a topology to define a metric. One can use a metric to define a topology.
Similar examples: (1) If we begin with a rectangular strip defined by $0<\phi<2\pi$ and $0<\theta<\pi$, and use the metric $ds^2=d\theta^2+sin^2\theta d\phi^2$, we get the metric for the surface of a sphere, which implies the topology of that surface.
(2) If we begin with a  rectangular strip defined by $0<\phi<2\pi$ and $0<\psi<2\pi$, and use the metric $ds^2= d\phi^2+d\psi^2$, we get the metric for the surface of a torus, which implies the topology of that surface.
Use of variables conventionally employed for angles is a way of implying that they are assumed to be periodic with period $2\pi$. For the sphere, values of $\theta>\pi$ will result in a second (alternative) designation of points on the sphere, so they are usually omitted.
