Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Equidistance
Replies: 7   Last Post: Jul 13, 2013 12:22 AM

 Messages: [ Previous | Next ]
 quasi Posts: 12,067 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 x-axis (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 y-axis (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 x-axis
>>and q is on the positive y-axis.
>>
>>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
>>
>>
>>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 1-point space.

There no counterexample with S having just 2 points,
but here's a simple 3-point 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

Date Subject Author
7/12/13 William Elliot
7/12/13 quasi
7/12/13 quasi
7/12/13 quasi
7/12/13 quasi
7/12/13 quasi
7/13/13 William Elliot
7/12/13 David C. Ullrich