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.)
This is a specialization (to the real interval [0,1], a complete lattice with respect to the order <=) of the Knaster-Tarski Fixed Point Theorem that holds for any complete lattice. See the Wikipedia article entitled Knaster–Tarski theorem. For a link to a .pdf of the classic article that contains Tarski's original proof, see http://www.projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.pjm/1103044538
Let A be the set of all a in [0,1] such that a <= f(a). Since f takes values only in [0,1], 0 <= f(0) and therefore 0 must be in A so that A is not empty. If a is in A, then f(a) is also in A, since f is nondecreasing. Let x be the least upper bound of the elements of A. Then for any a in A, a <= x so that a <= f(a) <= f(x) which makes f(x) an upper bound of the elements of A. Since x is the least upper bound of the elements of A, it must then be the case that x <= f(x). Hence A contains x. But x being in A means that f(x) is in A and x being an upper bound of the elements of A then implies that f(x) <= x. Hence f(x)=x, and the result is proved.
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