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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: ( You must be logged in to post comments.)
  Subject Author Date
re: Silly corollaryRobby Goetschalckx2006-11-19 20:47:50
Silly corollaryJLo2006-11-19 05:54:54
SolutionBrief proofJLo2006-11-19 05:51:03
re(5): No SubjectRichard2006-11-18 01:55:42
re(4): No SubjectJoel2006-11-18 00:14:27
re(3): No SubjectRichard2006-11-17 21:35:44
re(2): No SubjectJoel2006-11-17 20:45:20
re: No SubjectRichard2006-11-17 19:58:09
re(2): SolutionBractals2006-11-17 18:01:24
re: SolutionRichard2006-11-17 16:45:58
SolutionSolutionBractals2006-11-17 14:24:02
SolutionNo SubjectJoel2006-11-17 13:37:39
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