From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Metrization (Was: Re: Non-standard analysis) Date: 15 Feb 1995 18:40:20 GMT In article <3ht6s3$bg6@hamilton.maths.tcd.ie>, Timothy Murphy wrote: >In fact this question seems to me to be answered in many cases >by the theorem that a topological space is metrizable >if it is regular (a closed set and a point not in the set >can always be separated by disjoint open sets) >and the topology has a countable basis. Not to be too picky here, but this isn't quite right. I assume this is a reference to Urysohn's metrization theorem which states these three statements are equivalent: (1) X is a _separable_ metric space (2) X is regular and has a countable basis (3) X is homeomorphic to a subspace of the cube [0,1]^\omega (the last a concept which dovetails with a recent discussion of Tychonoff's theorem). A separable metric space is one with a countable dense subset. The best characterization of metrizability per se of which I am aware is the Nagata-Smirnov metrization theorem: X is metrizable iff it is regular and has a basis which is sigma-locally finite (that is, it has a basis which is a countable union of families B_k of open sets such that any point of X has a neighborhood which meets only finitely many sets in each B_k). Doesn't exactly roll off the tip of your toungue, but that's life. We should be appreciative of _any_ characterization which looks at a space just topologically and says, "yup, there is a metric somewhere which defines this topology". dave ============================================================================== From: duvall@stat.ohio-state.edu (Don Duvall) Newsgroups: sci.math Subject: Re: topology question Date: Fri, 06 Nov 1998 14:53:13 GMT On Fri, 06 Nov 1998 10:52:37 +0100, Brandsma wrote: > What is true: second countable + > regular implies metrisability. (not reversely of course). I believe you also need to assume that the space is T1, although I don't have an counterexample offhand. A T1, regular, second countable topological space is metrizable (Urysohn's theorem). ============================================================================== From: scott@math.csuohio.edu (Brian M. Scott) Newsgroups: sci.math Subject: Re: topology question Date: Sat, 07 Nov 1998 17:02:02 GMT On Fri, 06 Nov 1998 14:53:13 GMT, duvall@stat.ohio-state.edu (Don Duvall) wrote: >On Fri, 06 Nov 1998 10:52:37 +0100, Brandsma >wrote: >> What is true: second countable + >> regular implies metrisability. (not reversely of course). >I believe you also need to assume that the space is T1, although I >don't have an counterexample offhand. A 2-point space with the indiscrete topology. But a regular T_0 space is Hausdorf, so it's sufficient to assume T_0. Brian M. Scott