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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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Solution Solution | Comment 2 of 12 |
 
Let a_0 = f(0) and define
    a_(n+1) = f(a_n) for n = 0,1,2,...
------------------------------------------------
a_1 = f(a_0) >= f(0) = a_0 since a_0 = f(0) >= 0.
Assume a_(n+1) >= a_n. Then
a_(n+2) = f(a_(n+1)) >= f(a_n) = a_(n+1).
Therefore, by induction we see that {a_n} is a
nondecreasing sequence.
------------------------------------------------
Since {a_n} is bounded above by 1, it converges.
Let a = lim{a_n}. Then,
f(a) = f(lim{a_n}) = lim{f(a_n)} = lim{a_(n+1)} = a
 

  Posted by Bractals on 2006-11-17 14:24:02
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