This is a revised version of a question I already posted, but which patently was ill posed. Please give me another try.
For comparison's sake, the axioms of a metric:
Axiom A1: (forallx)d(x,x)=0
Axiom A2: (forallx,y)d(x,y)=0rightarrowx=y
Axiom A3: (forallx,y)d(x,y)=d(y,x)
Axiom A4: (forallx,y,z)d(x,y)+d(y,z)geqd(x,z)
Let T = {X,T} be a topology, B a base of T, x,y,z in X
Definition D0: x is nearer to y than to z with respect to B (NBxyz) iff (exists b in B) x, y in b & z notin b & (nexists b in B) x, z in b & y notin b
Definition D1: B is pre-metric1 iff (forallx,y)xneqyrightarrowNBxxy
Definition D2: B is pre-metric2 iff (forall x,y,z) ((z neq x & z neq y) rightarrow N_Bxyz) rightarrow x = y
Definition D3: B is pre-metric3 iff (forallx,y,z)NBzyxrightarrow(NByxzrightarrowNBxyz)
Definition: T is pre-metrici iff (existsB)B is pre-metrici (i = 1,2,3).
Definition: B is pre-metric iff B is pre-metric1, pre-metric2 and pre-metric3.
Definition: T is pre-metric iff (existsB)B is pre-metric.
Remark: D1 is an analogue of axiom A1, D2 of axiom A2, D3 of axiom A3.
Remark: T is pre-metric1 iff T is T1[not quite sure].
Remark: If T is induced by a metric, then T is pre-metric.
Question: Can a property pre-metric4 be defined such that T induces a metric iff T is induced by a metric with
Definition: B is metric iff B is pre-metric and pre-metric4.
Definition: T induces a metric iff (existsB)B is metric.
Remark: Property pre-metric4 should be an analogue of A4 (the triangle inequality).
If provably no such property can be defined does this shed a light on the difference (an asymmetry) between topologies and metric spaces? ("It's the triangle inequality, that cannot be captured topologically.")
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