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The Fix Is In (Posted on 2006-11-17) Difficulty: 4 of 5
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.)

See The Solution Submitted by Richard    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution | Comment 4 of 12 |
(In reply to re: Solution by Richard)

Back to the drawing board.
  Posted by Bractals on 2006-11-17 18:01:24

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