quasi
Posts:
9,903
Registered:
7/15/05


Re: Equidistance
Posted:
Jul 12, 2013 8:42 AM


quasi wrote: >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.
The finite counterexample with 3 points can be amplified to get a counterexample where S is infinite ...
Let t_1,t_2,t_3, ... be a decreasing sequence of positive real numbers such that
(1) t_n > 0 as n > oo
(2) The set {t_n  n in N} is linearly independent over Q.
Let S = {0} U {t_n  n in N}
Let d be the distance function on S inherited from R.
Then (S,d) is a compact metric space.
Define f:S > S by
f(0) = 0
f(t_n) = t_(n+1) if n != 0 (mod 3)
f(t_n) = t_(n2) if n = 0 (mod 3)
For example,
f(t_1) = t_2 f(t_2) = t_3 f(t_3) = t_1
f(t_4) = t_5 f(t_5) = t_6 f(t_6) = t_4
...
It's easily seen that the map f:S > S is a homeomorphism.
If n in N with n = 1 (mod 3), call the 3element set
{t_n,t_(n+1),t_(n+2)}
a 3cycle.
Note that for any 3cycle T, f(T) = T.
Now suppose p,q are distinct elements of S such that d(f(p),f(q)) = d(p,q).
Without loss of generality, assume p > q.
d(f(p),f(q)) = d(p,q)
=> f(p)  f(q) = p  q
=> f(p)  f(q) = p  q
If q = 0 the above simplifies to
f(p) = p,
contradiction since the only fixed point of f is 0.
Hence, p > q > 0.
Consider two cases ...
Case (1): Suppose p,q belong to the same 3cycle.
Then there are 3 subcases ...
Case (1a): p=t_n and q=t_(n+1)
But then
d(f(t_n),f(t_(n+1))) = t_(n+1)  t_(n+2)
whereas
d(t_n,t_(n+1)) = t_n  t_(n+1)
Hence d(f(p),f(q)) = d(p,q) yields
t_n  2*(t_(n+1)) + t_(n+2) = 0
contrary to the assumption of linear independence.
Case (1b): p=t_n and q=t_(n+2)
But then
d(f(t_n),f(t_(n+2))) = t_n  t_(n+1)
whereas
d(t_n,t_(n+2)) = t_n  t_(n+2)
Hence d(f(p),f(q)) = d(p,q) yields
t_(n+1) = t_(n+2)
contrary to t_(n+1) > t_(n+2).
Case (1c): p=t_(n+1) and q=t_(n+2)
But then
d(f(t_(n+1)),f(t_(n+2))) = t_n  t_(n+2)
whereas
d(t_(n+1),t_(n+2)) = t_(n+1)  t_(n+2)
Hence d(f(p),f(q)) = d(p,q) yields
t_n = t_(n+1)
contrary to t_n > t_(n+1).
Case (2): Suppose p,q belong to distinct 3cycles.
Let p = t_m and let q = t_n with m < n.
Let f(p) = t_i and let f(q) = t_j.
Since f has no fixed points other than 0,
t_m != t_i t_n != t_j
and since p > q with p,q in distinct 3cycles,
t_m > t_n t_m > t_j
t_i > t_n t_i > t_j
Thus, t_m,t_n,t_i,t_j are pairwise distinct.
Then
d(f(p),f(q)) = t_i  t_j = t_i  t_j
and
d(p,q) = t_m  t_n = t_m  t_n
so
d(f(p),f(q)) = d(p,q)
=> t_i  t_j = t_m  t_n
contrary to the assumption of linear independence.
The counterexample is thus validated.
quasi

