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/
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