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

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quasi

Posts: 10,477
Registered: 7/15/05
Re: Equidistance
Posted: Jul 12, 2013 6:24 AM
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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.

quasi



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