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From: rusty <mr.rusty@xxxxxxx>
Newsgroups: sci.math
Subject: Re: Complete metric quotient spaces
> - For a compact metric space, the quotient will be a (compact) metric
> space iff all equivalence classes are closed. (In general for the
> quotient to be metrisable each equivalence class must be closed as the
> inverse image of a closed set under a continuous mapping!)
What metric is that? Certainly not the metric described a bove.
> Yes the quotient metric d(C_1,C_2) given as the infimum of d(x_1,x_2)
> where x_1 runs over C_1 and x_2 over C_2. Since X is compact, there
> is no problem.
[-3,3] is a compact metric space.
The quotient space of identifying -3 and 3 to a point p
is homeomorphic to a circle.
All of the equivalence classes are closed, either a singleton
point {r}, -3 < r < 3, or the point p = { -3,3 }.
Yes, the circle is metrizable. Not however,
by the d you describe, for it isn't a metric.
d({-2}, {2}) = 4
d({-2}, p) = 1 = d({2}, p)
d({-2}, {2}) is not <= d({-2}, p) + d({2}, p)
> - The topology on a subset of a complete separable metric space can be
> given by a complete metric iff the subset is a G_delta.
What's a complete metric?
> I wrote "complete separable metric space". Complete and separable
> are adjectives that qualify "metric space".
I've never heard of a complete metric, only a complete metric space.
[0,1] is a complete separable metric space.
(0,1) is G_delta.
(0,1) is not complete metric space.
So what you claiming? That d:R^2 -> R,
d(x,y) = |x - y| is a complete metric?
As a matter of fact, don't metric subspaces, with the inherited metric,
have the subspace topology? So what's the context of your proposition?
> Do you have a problem?
Only with completely separable. ;-)
--
> - For a compact metric space, the quotient will be a (compact) metric
> space iff all equivalence classes are closed. (In general for the
> quotient to be metrisable each equivalence class must be closed as the
> inverse image of a closed set under a continuous mapping!)
>
What metric is that? Certainly not the metric described above .
> No, I don't think I'd do it in terms of explicit metrics, since I
> usually think of quotients of a compact metric space X in terms of
> unital closed *-subalgebras of C(X).
Algebra has become an annoyingly loose word with different meanings in
different fields. So you taking some subset of C(X) with some topology
and by that topology, the subset is closed. In addition there is some
thing * that does something to functions in C(X).
> I would identify C(X/~) with the subalgebra of C(X) consisting of
> functions constant on equivalence classes. As as subset of C(X),
> C(X/~) is separable and the unit ball B of its dual is compact and
> metrisable in the weak * topology. An explicit metric could be
> written down using a sequence of functions uniformly dense in the
> unit ball of C(X/~). Using point evaluation, X/~ becomes a compact
> subset of B and its topology is given by the restriction of the
> metric. (The explicit construction of functions in C(X/~) is a
> separate issue.)
> The result on quotients appears as Proposition 17 in
> Chapter IX, 2.10 of Bourbaki's General Topology.
No do have.
> - The topology on a subset of a complete separable metric space can be
> given by a complete metric iff the subset is a G_delta.
>
What's a complete metric?
> A metric which makes the space complete. Usually a topological space
> is called Polish (or polonais) if its topology can be given by a
> metric for which it becomes separable and complete.
I don't get this.
> The topology on a subset of a complete separable metric space
> can be given by a complete metric iff the subset is a G_delta.
[0,1] is a complete separable metric space.
(0,1) is G_delta.
(0,1) is not complete metric space.
How can (0,1) have a complete metric?
Do you mean iff
subset is a closed G_delta?
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