# Another Analysis question: continuity and compactness

by TaylorWatts
Tags: analysis, compactness, continuity
 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.
 P: 367 Consider the function $g : I \to \mathbb{R}$ given by $g \left( x \right) = f \left( x \right) - x$. What happens if $g \left( x \right) = 0$ for some $x$? How can you show that this has to happen in $I$? I'm sure you must know a nice theorem that you can use here.

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