|
|
From: rusty <mr.rusty@xxxxxxx>
Newsgroups: sci.math
Subject: Re: Complete metric quotient spaces
On Sat, 16 Sep 2006, William Elliot wrote:
>> 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!)
>
>> Here is a simple direct proof using an explicit quotient metric in
>> an important special case.
>
>> Let (X,d) be a *compact* metric space and R in X x X be a *closed*
>> subset defining an equivalence relation ~.
>
> Thus R/~ is Hausdorff space.
>> In addition assume that the equivalence relation satisfies the
>> following *regularity* condition: if x ~ y and x_n -> x then
>> there exist y_n ~ x_n such that y_n -> y.
>
> Is that the condition needed to make well defined the definition
> [x_n]_n -> [x] when (x_n)_n -> x ?
> No, that would be well-defined
> for all j, xj ~ yj, (xj)_j -> x, (yj)_j -> y ==> x ~ y
| This statement is a consequence of the fact that
| R is closed in X x X,
It is? Oh of course, as X/R is Hausdorff, a sequence ([xj])_j within X/R
will have only one limit. So the quotient map/projection mapping (xj)_j
to ([xj])_j preserves limits x and [x]. Thus [x] = [y] and x ~ y
> > For equivalence classes C_1 and C_2, define D(C_1,C_2) to be the
> > least non-negative number d such that d(x_1,C_2) <= d for all x_1
> > in C_1 and d(x_2,C_1) <= d for all x_2 in C_2.
>
> How is this any different than
> D(C1, C2) = inf{ d(x,y) | x in C1, y in C2 } ?
> It is not as continuous maps on compact sets attain inf.
| Well what I define is completely different. I am not sure whether
| you are deliberately trying to be obtuse, but I ask that EVERY
I'm deliberating. Have you been deliberated?
| point of C_1 be within r of some point of C_2 and vice versa.
| In your definition D(C_1,C_2) <= r iff at least one point of C_1
| is within r of some point of C_2.
D(C1,C2) =
inf{ r | for all x in C1, y in C2, d(x,C2), d(y,C1) <= r }
D(C,C) = 0.
If D(C1,C2) = 0, then some x in C1 with d(x,C2) = 0
x in cl C2 = C2 = C1 (since X/R is Hausdorff)
D(C2,C1) =
inf{ r | for all x in C2, y in C1, d(x,C1), d(y,C2) <= r }
= D(C1,C2)
> Since for some reason you don't seem to be able to find a proof,
But look, I can make your TeX legible.
> here is a proof of the triangle inequality. Let x_1 in C_1.
> By compactness there is a point x_2 in C_2 such that
> d(x_1,x_2) = d(x_1,C_2) <= D(C_1,C_2).
> Similarly there is a point x_3 in C_3 such that
> d(x_2,x_3) <= D(C_2,C_3).
> But then
> d(x_1,x_3) <= d(x_1,x_2) + d(x_2,x_3) <= D(C_1,C_2) + D(C_2,C_3).
> In particular
> d(x_1,C_3) <= D(C_1,C_2) + D(C_2,C_3).
> Hence
By generalizing on x_1
> max_{x_1 in C_1) d(x_1,C_3) <= D(C_1,C_2) + D(C_2,C_3).
> Similarly
Let's go do the twist.
> max_{y_3 in C_3} d(y_3,C_1) <= D(C_1,C_2) + D(C_2,C_3).
> Since D(C_1,C_3) is by definition the maximum of the
> two quantities on the LHS, it follows that
> D(C_1,C_3) <= D(C_1,C_2) + D(C_2,C_3)
> as required.
QED.
> But the maxixum of two metrics is again a metric.
Interesting, the sup of a collection of metrics is a metric.
the inf of a collection of metrics is a pseudo-metric.
and the min a collection of metrics is a metric.
>> To check that the map (X,d) -> (X/~,D) is continuous,
>> suppose x_n -> x. Let C_n = [x_n], C = [x]. If D(C_n,C) -/->0
>> then, passing to a subsequence if necessary :
>
>> /either/ (1) there are points y_n in C_n with inf d(y_n,C) > 0
>> /or/ (2) there are points z_n in C such that inf d(z_n,C_n) > 0
>
>> (1) Passing to a subsequence if necessary we may assume that y_n ->
>> y. Since R is closed and y_n ~ x_n, y must lie in C, contradicting
>> d(y,C)= lim d(y_n,C) >0.
>
> What applies is well-defined.
>> (2) Passing to a subsequence if necessary we may assume that
>> z_n -> z. Since x ~ z and x_n -> x, there exist w_n ~ x_n such
>> that w_n -> z by regularity. Thus d(z_n,w_n) -> 0, contradicting
>> inf d(z_n,C_n) > 0.
>
> Ok, regularity applies.
| Yes, this is where it is needed (it is not required
| to check that D is a metric).
Nope, compact metric with Hausdorff quotient is required.
>> Hence D(C_n,C) -> 0 and the map is continuous. But then there is
>> a continuous 1-1 map of the topological quotient X/~ onto (X/~,D).
>> Since the topological quotient is compact Hausdorff and (X/~,D) is
>> Hausdorff, this map is necessarily a homeomorphism, as required.
Hm. Because the map f:X -> (X/~,D) is continuous surjection and X
compact, f is closed continuous surjection, hence quotient map, the
fundamental theorem of quotients applies. Consequently X/~ and (X/~,D)
are homeomorphic.
----
|
|