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On Sat, 16 Sep 2006, William Elliot wrote:
> From: rusty <mr.rusty@xxxxxxx>
> Newsgroups: sci.math
> Subject: Re: Complete metric quotient spaces
>
> rusty wrote:
>
> > 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.
>
Amongst other things.
> > 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 ?
It is needed to give a poor man's version of the slice structure of
orbits.
>
> 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, if
you thought for a minute. No the regularity condition I wrote is a poor
man's version of the slice structure of orbits.
> > 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 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.
Also what I write down defines a metric whereas what
you write down does not. In addition it does not have the uniformity
described below. (I suggest you draw a picture.)
Since for some reason you don't seem to be able to find a proof, 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) \le D(C_1,C_2).
Similarly there is a point x_3\in C_3 such that
d(x_2,x_3) \le D(C_2,C_3).
But then
d(x_1,x_3) \le d(x_1,x_2) + d(x_2,x_3) \le D(C_1,C_2) + D(C_2,C_3).
In particular
d(x_1,C_3) \le D(C_1,C_2) + D(C_2,C_3).
Hence
max_{x_1\in C_1) d(x_1,C_3) \le D(C_1,C_2) + D(C_2,C_3).
Similarly
max_{y_3\in C_3} d(y_3,C_1) \le 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) \le D(C_1,C_2) + D(C_2,C_3)
as required.
>
> > So D(C_1,C_2) <= r iff each point of one class is within a distance r
> > of some point of the other class. It is easy to see that D defines a
> > metric on X/~.
>
> It is not. Let S = [0,1] and P = { 0,1 }
> Consider the quotient space, S/P identifying P to a point.
> That is counter example except for lack of regularity.
This is not a counterexample to what I have written.
In this case the quotient space can naturally be identified as a set with
both (0,1] and [0,1). This gives two metrics on the quotient and the metric I
have defined is in this case the maximum of these two. But the maxixum of
two metrics is again 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.
>
> I don't see how regularity applies. What applies is well-defined.
Regularity is not used here. Who said it had to be?
>
> > (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).
>
> > 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.
In your case it would seem that the possibility of cancelling or
superseding a posted message might bring its advantages.
P.S. This post was written remotely with ssh and Pine from a mathematics
institute in the Canadian Rockies. (I hesitate to post directly from
the computer provided in my room here.)
--
rusty
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