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Nov 20, 2009 at 21:24 comment added HJRW Fair point. I suppose by a full answer I meant an equivalent condition to metrisability that doesn't use the notion of metrisability anywhere (eg in locally metrisable). I think that the condition provided by Jones' Theorem is necessary and sufficient (subject to some assumption about the Continuum Hypothesis).
Nov 20, 2009 at 20:05 comment added Kevin Buzzard Whatever is a "full" answer? Isn't the "full" answer "A top space is metrisable iff there exists a metric inducing the topology"? What did you have in mind Henry? Probably not that!
Nov 20, 2009 at 15:43 history answered HJRW CC BY-SA 2.5