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Is there a general condition to tell whether a topological space is metrisable? (Without having to find the metric explicitly?)


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The following Wikipedia link contains some very relevant information: – Asaf Karagila Oct 10 '11 at 9:57
But summarizing it here: if second countable and T3, then metrisable. – abatkai Oct 10 '11 at 10:27
Ah, thanks guys! – marc Oct 10 '11 at 10:29
up vote 2 down vote accepted

Historically the oldest theorem that gives a complete characterization is the (Bing-)Nagata-Smirnov theorem: a space $X$ is metrizable iff it is regular, $T_1$ and it has a $\sigma$-locally-finite base $\mathcal{B}$; the latter means that $\mathcal{B} = \cup_{n \in N} \mathcal{B}_n$, where each $\mathcal{B_n}$ is locally finite (where a family of subsets $\mathcal{F}$ is locally finite when every $x$ in $X$ has a neighborhood that intersects at most finitely many members of $\mathcal{F}$). Bing's variant is that he uses a $\sigma$-discrete base (as above, where every $\mathcal{B_n}$ is a discrete family, so that every point in $X$ has a neighborhood that intersects at most one member from the family), which was discovered independently around the same time that Smirnov and Nagata did their (also separate) discoveries; it was "in the air", one could say.

It's inspired by the study of covering properties like paracompactness; there are other characterizations in terms of special bases too. Many other metrization theorems exist; the main ones are covered in e.g. Engelking's "general topology".

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