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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 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 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.

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_(n-2) 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 3-element set

{t_n,t_(n+1),t_(n+2)}

a 3-cycle.

Note that for any 3-cycle 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 3-cycle.

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 3-cycles.

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 3-cycles,

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

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