Let {X,T} be a topology, T the set of open subsets of X.




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Definition: Three points x, y, z of X are in *relation N* (Nxyz, read "x is nearer to y than to z") iff 


1. there is a basis **B** of T and **b** in **B** such that x and y are in **b** but z is not and
2. there is no basis **C** of T and **c** in **C** such that x and z are in **c** but not y.




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For some topologies there are no points x, y, z in relation N, for example if T = {Ø,X} or T = P(X), but for others there are (e.g. for ones induced by a metric [my claim]).




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Definition: A topology has *property M1* iff 


(x)(y) ((z) (z ≠ x & z ≠ y) → Nxyz) → x = y


(This is an analogue of d(xy) = 0 → x = y, the best one I can imagine).




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Definition: A topology has *property M2* iff


(x)(y)(z) Nxyz & Nyzx → Nzyx


(This is a kind of an analogue of d(xy) = d(yx), the best one I can imagine)




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First (bunch of) question(s): 


1. Properties *M1* and *M2* do not capture the whole of the corresponding conditions of a metric. Can anyone figure out "better" definitions (e.g. an analogon of x = y → d(xy) = 0)?
2. Can anyone figure out a *property M3* that is an analogue of the triangle equality?


If it can be shown that no such property M3 is definable, the following becomes obsolete. 


If such a definition can be made, we define:




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Definition: A topology has *property M* (read "induces a metric") iff it has properties M1, M2, M3.




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Second question: 


Which topologies have property M, i.e. induce a metric? Are these "accidentally" exactly those that are induced by a metric?



Can topologies induce a metric?
Your condition 1 is satisfied for all triples $x,y,z\in X$ such that $z\not\in\{x,y\}$ if the space is [$T\_1$](http://en.wikipedia.org/wiki/T1_space).


Maybe reading a bit about uniform spaces and the corresponding metrizability results will be of help.


