From: Michael Barr <barr@barrs.org>
To: CATEGORIES LIST <categories@mta.ca>,
Subject: Re: Generalization of Browder's F.P. Theorem?
Date: Thu, 16 Jan 2003 18:05:39 -0500 (EST) [thread overview]
Message-ID: <Pine.LNX.4.10.10301161755400.3423-100000@triples.math.mcgill.ca> (raw)
In-Reply-To: <3E26BBF3.2D413C6A@cs.bham.ac.uk>
In Errett Bishop's Constructive Analysis (anyone who is interested in
analysis over a topos absolutely must know this book), he proves that for
any continuous endomorphism f of a disk and for every eps > 0, there is a
point x in the disk for which |f(x) - x| < eps. A couple of points should
be made. First f has to be constructible and eps has to be provably
positive. For Bishop, a real number is an equivalence class of pairs of
RE sequences of integers a_n and b_n such that for all m,n |a_n/b_n -
a_m/b_m| < 1/m + 1/n and a function is constructive if it is a machine for
turning one such sequence into another. To be continuous, it there must
be a function delta(eps) that produces for each eps > 0, a delta(eps) such
that |x - y| < delta(eps) implies that |f(x) - f(y)| < eps. (In fact,
there is a non-constructive proof that every constructive function is
continuous.)
Bishop then claims, without proof as far as I can see, that there is a
fixed point free endomorphism of the disk. What this means is that when
you extend this function to all reals, any fixed point is not a
constructible real number.
On Thu, 16 Jan 2003, Steven J Vickers wrote:
>
> I'm intrigued by Peter McBurney's question [below]. It looks rather like a
> question of the constructive content of Brouwer's fixed point theorem.
>
> Suppose S is (homeomorphic to) an n-cell. Then in the internal logic of the
> topos of sheaves over [0,1], f is just a continuous endomap of S. If
> Brouwer's theorem were constructively true then f would have a fixpoint,
> and that would come out as a continuous section of the projection [0,1]xS
> -> [0,1]. More precisely, it would be a map g: [0,1] -> S such that f(x,
> g(x)) = g(x) for all x. If this existed then the set A = {(x, g(x))| x in
> [0,1]} would be as required.
>
> However, the proof of Brouwer that I've seen is not constructive - it goes
> by contradiction. So maybe the requirements on A are a way of getting
> constructive content in Brouwer's result.
>
> What is known constructively about Brouwer's fixed point theorem?
>
> Steve Vickers.
>
> Peter McBurney wrote:
>
> > Hello --
> >
> > Does anyone know of a generalization of Browder's Fixed Point Theorem
> > from R^n to arbitrary topological spaces, or to categories of same?
> >
> > *****************
> >
> > Theorem (Browder, 1960): Suppose that S is a non-empty, compact, convex
> > subset of R^n, and let
> >
> > f: [0,1] x S --> S
> >
> > be a continuous function. Then the set of fixed points
> >
> > { (x,s) | s = f(x,s), x \in [0,1] and s \in S }
> >
> > contains a connected subset A such that the intersection of A with {0} x
> > S is non-empty and the intersection of A with {1} x S is non-empty.
> >
> > *****************
> >
> > Many thanks,
> >
> > -- Peter McBurney
> > University of Liverpool, UK
>
>
>
>
next prev parent reply other threads:[~2003-01-16 23:05 UTC|newest]
Thread overview: 21+ messages / expand[flat|nested] mbox.gz Atom feed top
2003-01-15 14:00 Peter McBurney
2003-01-16 14:04 ` Steven J Vickers
2003-01-16 23:00 ` Prof. Peter Johnstone
2003-01-16 23:05 ` Michael Barr [this message]
2003-01-21 18:11 ` Andrej Bauer
2003-01-22 10:13 ` Cauchy completeness of Cauchy reals Martin Escardo
2003-01-22 23:33 ` Dusko Pavlovic
2003-01-23 19:56 ` Category Theory in Biology Peter McBurney
2003-01-24 8:51 ` Cauchy completeness of Cauchy reals Martin Escardo
2003-01-25 2:21 ` Dusko Pavlovic
2003-01-25 16:24 ` Prof. Peter Johnstone
2003-01-27 3:57 ` Alex Simpson
2003-01-23 6:29 ` Vaughan Pratt
2003-02-04 0:47 ` Vaughan Pratt
2003-02-05 16:06 ` Prof. Peter Johnstone
2003-01-23 9:50 ` Mamuka Jibladze
2003-01-24 1:34 ` Ross Street
2003-01-24 16:56 ` Dusko Pavlovic
2003-01-24 19:48 ` Dusko Pavlovic
2003-01-17 16:19 Generalization of Browder's F.P. Theorem? Carl Futia
2003-01-18 12:39 ` S Vickers
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