Right, a topological space may not have a metric.
A metric is additional structure on a set.
A set with a metric gives rise to a metric topology on that set. (I mean usual non-neg, symmetric, non-deg metric).
I am reading John Lee's Topological Spaces. There he defines a top manifold as:
1) topological space locally homeomorphic to R^n, 2) Hausdorff & 3) second countable.
So, assuming ST is a topological space (by this def), can't we use the local homeomorphism to R^n to define (locally) a metric on ST via pullback of the Euclidean metric on R^n? And likewise any topological manifold is a (local)metric space (but not general topological spaces).
It seems that ST is a topological manifold with a locally Euclidean metric. This describes its topological structure as a metric space. We than add further structure to this metric/topological space by adding the non-Riemanian Lorentz metric. Thus we are using one metric and corresponding open balls to describe the topology, and another metric to describe the 'physical' distance between points. the physical distance between some points is 0, which is a very different topology than the locally Euclidean one; but the Lorentz metric can't be used to describe open balls for a topology?
as Chris Hillman said:
"Lorentzian metrics get their topology from the (locally euclidean) topological manifold structure, not from the bundled indefinite bilinear form."
But it seems odd to me that we impose a locally euclidean topology, then use a quite different metric to describe physical separation of points..
