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Re: homeomorphisms of compact sets and the hausdorff distance
Posted:
Nov 8, 2010 1:07 PM
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Thanks everyone for your answers, now i got a refinement of my question:
Take X=a TOP manifold, A=a compact TOP submanifold of X and suppose
that we have a sequence B_n of compact subsets of X such that each B_n
is a union of topological (open) balls in X, such that for any e > 0,
H(A, B_n) < e for n sufficiently far out in the secuence.
It is not hard to see that it is not always true that for n large
enough, B_n is homeomorphic to A (X=q-euclidean space, A=m-submanifold,
and the B_n are unions of open q-balls in X with q>n).
But, Can it be proven that for n large enough, A and B_n have the same
homotopy type ? examining some examples, this would seem plausible. It
would even seem that B_n can be some kind of "special neighborhood" for A
Thanks in advance...
Cheers...
El 18/10/2010 04:00 p.m., Dan Luecking escribi?:
> On Mon, 18 Oct 2010 14:00:01 +0000 (UTC), Rodolfo Conde
> <rcm@gmx.co.uk> wrote:
>
>>
>> Hi everyone...
>>
>> If given a compact subset A of a metric space (X,d), it is true
>> that given e> 0 there exists a compact subset B_e of X such that H(A,B_e)<
>> e (where H is the hausdorff distance).
>
> Perhaps you have phrased this wrong. As stated, it is
> trivially true: take B_e = A since H(A,A) = 0< e.
>
> It is even true if you require B_e to be finite:
> The compact set A can be covered by all the balls
> with centers in A and radius e/2. By compactness, a
> finite number will also cover A. Take B_e to be the
> centers of those balls and you have H(A,B_e)< e.
>
>>
>> It is true that for e> 0 sufficiently small, B_e will be homeomorphic to A
>> ??
>
> Generally no. Take A to be a line segment and B_e to be a finite
> set of points in A spaced a distance e apart.
>
>
> Dan
> To reply by email, change LookInSig to luecking
>
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