quasi
Posts:
10,968
Registered:
7/15/05


Re: Equidistance
Posted:
Jul 12, 2013 7:33 AM


quasi wrote: >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.
Well, the most trivial counterexample would be to let S be a 1point space.
There no counterexample with S having just 2 points, but here's a simple 3point counterexample ...
Let S = {0,1,3} with the usual distance function inherited from R.
The S is a compact metric space.
Let f:S > S be defined by
f(0) = 1 f(1) = 3 f(3) = 0
Then the map f:S > S is a homeomorphism.
But this gives
d(f(0),f(1)) = 2 whereas d(0,1) = 1 d(f(1),f(3)) = 3 whereas d(1,3) = 2 d(f(3),f(0)) = 1 whereas d(3,0) = 3
It follows that there do not exist distinct p,q in S such that
d(f(p),f(q)) = d(p,q)
validating the counterexample.
Similar counterexamples can be found for all positive finite cardinalities of S except for S = 2.
quasi

