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I'm reviewing this problem for my analysis qual.

Let $f:X\rightarrow X$ be a continuous mapping from a metric space to itself. Assume $f $ has no fixed points. Prove that, if $X $ is compact, there exists an $\epsilon $ such that $d (x, f (x)) \ge \epsilon $ for every $x\in X $.

I'm having trouble figuring out what to do. I predict having to take neighborhoods of every $x\in X $ and find a finite subcover. But I'm not sure where that gets me in terms of the fixed point thing. I know that $f (x) \ne x $ means each center of the balls comprising the finite subcover will have to move, but I don't know what that gets me. Help?

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Hint: Consider the (continuous!) function $g:X \to \Bbb R$ given by $$ g(x) = d(x,f(x)) $$ Why must $g$ achieve its minimum?


To do this in a manner similar to the way you originally planned: for each $n \in \Bbb N$, define $U_n = \{x \in X: d(x,f(x)) > 1/n\}$. Take a finite subcover.

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  • $\begingroup$ short and elegant $\endgroup$
    – Learnmore
    Jan 8 '15 at 5:11
  • $\begingroup$ Is this really shorter than the other answer that I fleshed out, or are its details just encapsulated in the proofs of the results on achieving minimums necessary for this proof? (not criticism, this is an honest question) $\endgroup$ Jan 8 '15 at 7:03
  • $\begingroup$ Well, you didn't really "flesh out" the proof here; the important detail, were you to "take neighborhoods of every $x \in X$", is which neighborhood you select. $\endgroup$ Jan 8 '15 at 13:51
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Hint Assume by contradiction that this is not true. Then for each $n$ you can find $x_n$ so that $d (x_n, f (x_n)) \le \frac{1}{n}$.

Now $x_n$ has a cluster point $y$ (Why?).

What is $f(y)$?

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  • $\begingroup$ Because $x_n $ is a sequence in a compact set it has a subsequence $x_{n_k}$ that converges to a limit point $x_{n_k} \rightarrow x $. Since continuity preserves convergent sequensequences we have $f (x_{n_k}) \rightarrow f (x) $. But we know that $d (x_n,f (x_{n}))<\frac {1}{n} \rightarrow 0$ so $d (x,f (x))=0$ so $x=f (x) $, contradiction. Is that right? $\endgroup$ Jan 8 '15 at 3:50
  • $\begingroup$ @BurqueWhote Yup :) $\endgroup$
    – N. S.
    Jan 8 '15 at 4:31

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