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This question may be a little too easy for this site, but I'll ask it anyway: when is a Hausdorff topological space metrisable?

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I think this site is a little too hung up about avoiding easy questions. I understand they don't want the site to degrade into a homework Q&A, but I think questions like yours are fine. – John D. Cook Nov 20 '09 at 16:00
One word.. "Nagata" – Jose Capco Nov 21 '09 at 1:04

The wikipedia page here is probably of interest. It contains links to a few metrization theorems, including results of Nagata-Smirnov's and Bing's. There is also the Smirnov metrization theorem that appears in Munkres' topology: it states that A space X is metrizable if and only if it is paracompact, Hausdorff, and locally metrizable.

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This Wikipedia page is an excellent answer to the question. Urysohn's theorem (second-countable, regular, Hausdorff is metrizable) and Smirnov's theorem (locally metrizable, Hausdorff, paracompact is metrizable) really are what people use all the time. – Greg Kuperberg Nov 20 '09 at 15:32

I think it's also worth pointing out the Wikipedia page on Moore spaces. It turns out that the full answer to your question (if you don't allow conditions like "locally metrizable") has something to do with the Continuum Hypothesis.

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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! – Kevin Buzzard Nov 20 '09 at 20:05
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). – HJRW Nov 20 '09 at 21:24

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