Continuous function on a compact set with no fixed points

Question

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?

Answer 1

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 $f(y)$?

Answer 2

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

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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.