LAUSR

Let $${(X, m, \leq)}$$(X,m,≤) be a partially ordered metric space, that is, a metric space (X, m) equipped with a partial order $${\leq}$$≤ on X. We say that a T0-quasi-metric d on X is m-splitting provided that $${d\vee d^{-1}=m}$$d∨d-1=m. Furthermore d is said to be $${(X, m, \leq)}$$(X,m,≤)-producing provided that d is m-splitting and the specialization partial preorder of d is equal to $${\leq}$$≤. It is known and easy to see that if $${(X, m, \leq)}$$(X,m,≤) is a partially ordered metric space that is produced by a T0-quasi-metric and $${\leq}$$≤ is a total order, then there exists exactly one producing T0-quasi-metric on X. We first will give an example that shows that a partially ordered metric space can be uniquely produced by a T0-quasi-metric although $${\leq}$$≤ is not total.Then we present solutions to the following two problems: Let $${(X, m, \leq)}$$(X,m,≤) be a partially ordered metric space and A a subset of X. (1) Suppose that d is a T0-quasi-metric on A which is $${m\vert ({A \times A})}$$m|(A×A)-splitting. When can d be extended to an m-splitting T0-quasi-metric $${\widetilde{d}}$$d~ on X?(2) Suppose that d is a T0-quasi-metric on A which is $${(A, m\vert (A\times A)}$$(A,m|(A×A), $${{\leq}\vert ({A\times A}))}$$≤|(A×A))-producing. When can d be extended to a T0-uasi-metric $${\widetilde{d}}$$d~ on X that produces $${(X, m, \leq)?}$$(X,m,≤)?