Prove: Let f be a nondecreasing, but not necessarily continuous, mapping of the closed interval [0,1] into itself. Then f has a fixed point, i.e., there is some x in [0,1] such that f(x)=x.

(Just to avoid any misunderstanding, [0,1] is the set of all real numbers between 0 and 1, with both 0 and 1 also included.)

(In reply to No Subject by Joel)

There appears to be a problem with your method. If f(x)=1 for all x in [0,1], then S(1)=1 and S(2) cannot be found because S(1)+1/2=3/2 which lies outside of [0,1].

Posted by Richard on 2006-11-17 19:58:09