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?
Best Answer
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.