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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)

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re(2): No Subject | Comment 6 of 12 |
(In reply to re: No Subject by Richard)

Whoops, in all cases that I used (1/2)^n I should have used (1/2)^(n+1)

so if f(x)=1 for all x then the sequence should be 0, 1/2, 3/4, 7/8,...


  Posted by Joel on 2006-11-17 20:45:20

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