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William Elliot wrote:
> 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?
This time you have replied to a message posted at 13:40 and cancelled at
13:44 yesterday. Please see the two subsequent posts. By the way (0,1) is a
Polish space since it is homeomorphic to the complete metric space
(-\infty, \infty). I suggest you look up the references in Chapter IX of
Bourbaki's General Topology for your own eduaction. And no I do not mean a
closed G_delta. Please see my two other messages and please check carefully
before replying to a cancelled message.
--
rusty
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