Contractible metric space
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From: Stuart M Newberger on 27 Feb 2005 04:47 Stuart M Newberger wrote: > Julien Santini wrote: > > Hello, > > > > I have a problem of terminology. What is meant by "starlike" in the > > assertion: "every starlike metric space is contractible" ? > > > > The definition given by the book is: "A metric space M with metric d > is > > starlike in that metric if there is a point p in M such that each > other > > point x in M can be joined to p by a unique arc congruent in the > metric of M > > to a line segment". > > > > What is meant by "unique" ? What is meant by arc ? (traditionally an > arc is > > a continuous map defined on [0,1] right ?) And what about "congruent > [...] > > to a line segment ?" > > > > I would appreciate if someone could translate this definition using a > > mathematical formalism only. > > > > Thank you > > > > -- > > Julien Santini > > Here is my guess."Arc congruent in the metric M means an isometry from > [0,1] (usual metric) into M. So for each q in M there is a unique map > f=f_q :[0,1]->M > with f(0)=p and f(1)=q and d(f(x),f(y))=|x-y| for each x,y in [0,1].Try > the problem now. Regards,Stuart M Newberger Correction The domain of f_q is not [0,1] since then d(p,q)=|1-0|=1. I should have said f_q: [0,D]->M ,D=d(p,q).I think you are going to need (*) d(f_q(t),f_u(t)) is an increasing function of t where u,q in M and t at most the smaller of d(p,q) and d(p,u) .Note f_q(t) and f_u(t) are both at distance t from p.This may help with the joint continuity of f.I do not know whether this can be proved from other geometric axioms for geometry in metric space.I do not see how you can possibly prove it.But I don't know .Regards,Stuart M Newberger
From: Michael Barr on 27 Feb 2005 10:44 "Julien Santini" <santini.julien(a)wanadoo.fr> wrote in message news:<4220809d$0$1244$8fcfb975(a)news.wanadoo.fr>... > > I will take a guess at this. Unique means unique (what else could it > > mean). I take it that congruent means isometric. That is there is a > > point * such that for each other point x, there is one and only one > > map u_x: [0,1] --> X s.t. u(0) = *, u_x(1) = x and dist(u_x(s),u_x(t)) > > = |s-t| for any s,t in [0,1]. I guess this allows you to prove that > > the function h: X x [0,1] --> X defined by h(x,t) = u_x(t) contracts X > > to *. > > That's what I had thought, but my problem is that although I can show h_x: > [0,1] -> X, h_x(t) = h(x,t) is continuous (using this definition of > starlike), I cannot show h_t: X->X, h_t(x) = h(x,t) is continuous. Working > in R^2 for instance, and considering an arc (*p), with a point x on this > arc, and a neighborhood U of x, why couldn't U intersect (no matter how > "small" U is) another arc at point y so that for some fixed t, and for some > fixed e>0, abs(h_t(y)-h_t(x))>e (i.e h_t is not continuous) ? That's what > bothers me. Actually, what I said was obviously nonesense anyway, since it would imply that every point was exactly 1 unit from * and then what about the intermediate points? The only interpretation that makes sense is that u_x is an isometry of [0,dist(*,x))] to X. On the other hand, the continuity of h seems much harder, the more I think about it. I will think about it.
From: Lasse on 27 Feb 2005 14:08 > That's what I had thought, but my problem is that although I can show h_x: > [0,1] -> X, h_x(t) = h(x,t) is continuous (using this definition of > starlike), I cannot show h_t: X->X, h_t(x) = h(x,t) is continuous. Working > in R^2 for instance, and considering an arc (*p), with a point x on this > arc, and a neighborhood U of x, why couldn't U intersect (no matter how > "small" U is) another arc at point y so that for some fixed t, and for some > fixed e>0, abs(h_t(y)-h_t(x))>e (i.e h_t is not continuous) ? That's what > bothers me. How about the following (it would seem that this can be done in R^2, but perhaps it is easier to define on the abstract level): How about the following: in R^2, take the union of the line segment from (0,0) to (1,0); the line segments {1/n} x [1,0] for all natural numbers n, and a half-circle (or any other curve) from (0,0) to (0,1) lying in the left half plane (i.e., x < 0) apart from these two endpoints. Then it would seem that this set is "star-like" by your definition, but it is clearly not contractible. You could assume that your space is compact and this problem would go away, but perhaps you don't want to do that. Hope this helps, Lasse --- (@remove.for.spam.maths.warwick.ac.uk)
From: Lasse on 27 Feb 2005 14:23 > > How about the following (it would seem that this can be done in R^2, > but perhaps it is easier to define on the abstract level): > Hmm. I first wanted to do it at an abstract level, but then figured I might as well just describe it in R^2. Clearly I forgot to remove this line - apologies. > Hope this helps, > > Lasse > --- > (@remove.for.spam.maths.warwick.ac.uk)
From: Julien Santini on 27 Feb 2005 14:53
> How about the following (it would seem that this can be done in R^2, > but perhaps it is easier to define on the abstract level): > The theorem I mentioned is theorem 4.8 (Hocking & Young) and the proof is "left to the reader" (it is clearly implied it must be more or less trivial). However, it seems clear to me that, using this interpretation of starlike, a counterexample can be found easily. I really can't tell. -- Julien Santini |