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