gn.general topology - Can topologies induce a metric? (revised)

This is a revised version of a question I already posted, but which patently was ill posed. Please give me another try.

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For comparison's sake, the axioms of a metric:

Axiom A1: $(forall x) d(x,x) = 0$

Axiom A2: $(forall x,y) d(x,y) = 0 rightarrow x = y$

Axiom A3: $(forall x,y) d(x,y) = d(y,x)$

Axiom A4: $(forall x,y,z) d(x,y) + d(y,z) geq d(x,z)$

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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 $B$ ($N_Bxyz$) 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 $(forall x,y) x neq y rightarrow N_Bxxy$

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 $(forall x,y,z) N_Bzyx rightarrow (N_Byxz rightarrow N_Bxyz)$

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Definition: $T$ is pre-metric_i iff $(exists B) B$ is pre-metric_i (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 $(exists B) B$ 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.

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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 $(exists B) B$ 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.")