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

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quasi

Posts: 10,205
Registered: 7/15/05
Re: Equidistance
Posted: Jul 12, 2013 6:47 AM
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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
>
>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



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