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William Elliot wrote:
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.
I would have preferred it if you had been able to do the whole of this
undergraduate exercise on your own.
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.
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rusty
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