
Re: homeomorphisms of compact sets and the hausdorff distance
Posted:
Nov 8, 2010 1:07 PM


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=qeuclidean space, A=msubmanifold, and the B_n are unions of open qballs 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 >

