quasi
Posts:
10,660
Registered:
7/15/05


Re: Equidistance
Posted:
Jul 12, 2013 6:24 AM


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.
quasi

