quasi
Posts:
11,911
Registered:
7/15/05


Re: Equidistance
Posted:
Jul 12, 2013 6:47 AM


quasi wrote: >quasi wrote: >>William Elliot wrote: >> >>>Let (S,d) be a compact Hausdorff space and f:S > S a >>>homeomorphism. How can it be show that there are some distinct >>>x and y for which d(f(x),f(y)) = d(x,y)? >> >>Presumably you meant compact _metric_ space. >> >>Still, I don't believe the claim. >> >>What is your source for the problem? > >As I suspected, the claim is false. > >A counterexample can be constructed as follows ... > >Let t_1,t_2,t_3, ... be a sequence of pairwise distinct positive >real numbers such that t_n > 0 as n > oo > >Define two infinite sequences > > a_1,a_2,a_3, ... > > b_1,b_2,b_3, ... > >of points in R^2 by > > a_n = (t_n,0) for all n in N > > b_n = (0,2*(t_n)) for all n in N > >and let S be the subset of R^2 defined by > > S = A U B U {(0,0)} > >where A = {a_n  n in N} and B = {b_n  n in N}. > >Let the distance function d on S be the usual Euclidean distance >function on R^2. > >Then (S,d) is a compact metric space. > >Finally, define a function f:S > S by > > f(a_n) = b_n for all n in N > > f(b_n) = a_n for all n in N > > f((0,0)) = (0,0) > >It's easily seen that the function f:S > S is continuous. > >Also, f o f is the identity on S, so f is bijective with >f^(1) = f. > >It follows that the map f: S > S is a homeomorphism. > >Claim there do not exist distinct points p,q in S such that >d(f(p),f(q)) = d(p,q). > >Suppose otherwise. Thus, suppose p,q are distinct points in S >such that d(f(p),f(q)) = d(p,q). > >For convenience, define > > t_0 = 0 > a_0 = (0,0) > b_0 = (0,0) > >Consider 3 cases ... > >Case (1): p,q are both on the nonnegative xaxis (it's possible >that one of them is (0,0)). > >Assume p,q are given by > > p = a_i > q = a_j > >for some distinct nonnegative integers i,j. > >Then by definition of f, > > f(p) = b_i > f(q) = b_j > >But then > > d(f(p),f(q)) = 2*abs(t_i  t_j) > >whereas > > d(p,q) = abs(t_i  t_j) > >and those distances can't be equal. > >Case (2): p,q are both on the nonnegative yaxis (it's possible >that one of them is (0,0)). > >Assume p,q are given by > > p = b_i > q = b_j > >for some distinct nonnegative integers i,j. > >Then by definition of f, > > f(p) = a_i > f(q) = a_j > >But then > > d(f(p),f(q)) = abs(t_i  t_j) > >whereas > > d(p,q) = 2*abs(t_i  t_j) > >and just as in case (1), those distances can't be equal. > >Case (3): p,q are not on the same axes. > >Without loss of generality, assume p is on the positive xaxis >and q is on the positive yaxis. > >Assume p,q are given by > > p = a_i > q = b_j > >for some distinct i,j in N. > >Then by definition of f, > > f(p) = b_i > f(q) = a_j > >But then > > d(f(p),f(q)) = sqrt((t_j)^2 + 4*(t_i)^2) > >whereas > > d(p,q) = sqrt((t_i)^2 + 4*(t_j)^2) > >Then > > d(f(p),f(q)) = d(p,q) > > => sqrt((t_j)^2 + 4*(t_i)^2 = sqrt((t_i)^2 + 4*(t_j))^2 > > => 3*(t_i)^2 = 3*(t_j)^2 > > => t_i = t_j > >contradiction. > >Thus, the counterexample is validated.
Oops.
I take back my counterexample.
In case (3) we can have i = j.
Thus, for example, letting
p = a_1 q = b_1
we get
f(p) = b_1 f(q) = a_1
which yields d(f(p),f(q)) = d(p,q).
Hence, I withdraw my proposed counterexample.
Still, I suspect the claim is false.
quasi

