I now have an answer to my own question (and thanks to Alexander Wauth of Bielefeld who independently found the same answer and sent it to me). For each x \in X, consider the topology T_x in which any set not containing x is open, and a set containing x is open iff it's a (not necessarily open!) neighbourhood of x in (X,T). It's easy to verify that all such topologies are sober; and their intersection is clearly T. I'd still be interested to know if anyone has considered this question before. Peter Johnstone On Mon, 2 Aug 2004, Prof. Peter Johnstone wrote:
Does anyone know of any work that might be relevant to the following question:
Given a (non-sober) topology T on a set X, is T always expressible as the intersection of the sober topologies which contain it? Equivalently, given a set S \subseteq X which is not open in T, can we find a sober topology T' \supseteq T in which S is still not open?