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rusty wrote:
> 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.
If you can't find them on your news reader the other two messages are:
<news:4506cb3b$0$11850$636a55ce@xxxxxxxxxxxx>
<news:4507c708$0$11852$636a55ce@xxxxxxxxxxxx>
--
rusty
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