- #1
nateHI
- 146
- 4
Is it fair to say, when talking about spacetime with a given metric, it would be redundant to state that the associated set has the metric topology placed on it. In other words, let ##M## be a set, ##O## the metric topology, ##\nabla## a connection, ##g## a metric, and ##T## be the direction of time, then
##(M,O,\nabla,g,T)=(M,\nabla,g,T)##
Edit: I suppose what I'm asking is, does the metric ##g## induce the metric topology?
##(M,O,\nabla,g,T)=(M,\nabla,g,T)##
Edit: I suppose what I'm asking is, does the metric ##g## induce the metric topology?
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