A Continuous Function on a Specified Interval

Date: 11/13/2003 at 12:31:42
From: Bogdan
Subject: A continuous function

Hello, Dr. Math.

Could you please help me with the following problem?

Does there exist a partition of the set X = [0,1) into two subsets A
and B and a continuous function f(x) such that:

x in A ==> f(x) in B
x in B ==> f(x) in A.

I've tried, but I haven't gotten anywhere.

Date: 11/14/2003 at 02:42:31
From: Doctor Jacques
Subject: Re: A continuous function

Hi Bogdan,

Note first that if, instead of X = [0,1), we had Y = [0,1] (the closed 
interval), there would be no solution.  This is a consequence of the 
"fixed point theorem" that states that for every continuous function
f : Y -> Y there is a point x such that f(x) = x.  You can read more 
about this on:

  Analysis
    http://mathforum.org/library/drmath/view/51915.html 

  Fixed Point Theorem
    http://mathworld.wolfram.com/FixedPointTheorem.html 

Now, if X is the half-open interval [0,1), we must find two things:

  * A suitable function f
  * The subsets A and B

A necessary condition for f is that there be no point in X such that 
f(x) = x.  To find such a function f, we may look at the proof of the 
fixed point theorem and try to "defeat" it.

This means that the graph of f must remain above or below the line
y = x in the interval X.  A possible example would be:

  f(x) = (x + 1)/2

  (note that x = 1 does not belong to X).

Now, to find the subsets A and B, consider the following half-open 
intervals:

  [1,1/2), [1/2,3/4), [3/4, 7/8),...,[1 - 1/2^n, 1 - 1/2^(n+1)),...

and look at the graph of f over those intervals.

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/