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 Solution by Bractals)

There seems to be a problem with your method. When I define f(x)= 1/4+x/2 for 0<=x<1/2 and f(x)=1 for 1/2<=x<=1 and apply your method, I get 1/2 for the alleged fixed point and this is clearly wrong.  Your assumption f(lim{a_n})=lim({f(a_n}) requires continuity of f at lim{a_n}, which is not the case for this f.

Posted by Richard on 2006-11-17 16:45:58