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More properties -- what do we know about partial metric spaces?

Not every quasi metric space is weightable [Mat94].

For each partial metric space (X,p), p'(x,y) = p(x,y)/(1+p(x,y)) is a partial metric inducing the same topology and ordering as p [KV94].

Each T0 space having a sigma-disjoint base admits a weighted quasi metric, equivalently, is pmetrizable [Kv94].

Each weighted quasi-metric space is quasi developable [KV94].

The necessity of the small self-distances axiom p(x,x)<=p(x,y) is questionable. As if, following Heckmann, we take a weak partial metric to be a partial metric less the small self-distances axiom then if p is a weak partial metric, then p'(x,y) = max(p(x,x), p(x,y), p(y,y)) is a partial metric inducing the same topology and thus the same preorder as p [He99, Theorem 2.19].

The necessity of restricting partial metric distances to only the non-negative reals is questionable. In dropping this requirement O'Neil has shown that the partial metric topology T[p] can be constructed as before, that there is a dual partial metric p*(x,y) = p(x,y) - p(x,x) - p(y,y), that T[d] = T[p]\/T[p*] (where T[d] is the usual metric topology associated with p), and that (S, T[p], T[d]) is a suggestive bi-topology for studying domains [O95, O96].

To describe every Scott domain by a partial metric requires a generalisation of the notion of metric beyond that of the partial metric. Kopperman et.al. have shown that if the non-negative reals are generalised to a value quantale then partial metrics can describe any domain [KMP04].

Every omega-algebraic domain D has a comptaible partial metric p such that p(x,x)=0 exactly when x is a constructive maximal [Sm05].