Another Analysis question: continuity and compactness


by TaylorWatts
Tags: analysis, compactness, continuity
TaylorWatts
TaylorWatts is offline
#1
Oct27-08, 12:35 AM
P: 16
Let I = [0,1] be the closed unit interval. Suppose f is a continuous mapping from I to I. Prove that for one x an element of I, f(x) = x.

Proof:

Since [0,1] is compact and f is continuous, f is uniformly continuous.

This is where I'm stuck. I'm wondering if I can use the fact that since max {|x-y| = 1} if |x-y| = 1 then f(x) - f(y) < max {episolin}. This of course only occurs when WLOG x=1 y=0.

Stuck as far as the rest of it goes.
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Moo Of Doom
Moo Of Doom is offline
#2
Oct27-08, 03:14 AM
P: 367
Consider the function [itex]g : I \to \mathbb{R}[/itex] given by [itex]g \left( x \right) = f \left( x \right) - x[/itex]. What happens if [itex]g \left( x \right) = 0[/itex] for some [itex]x[/itex]? How can you show that this has to happen in [itex]I[/itex]? I'm sure you must know a nice theorem that you can use here.


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