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?


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.

  • $\begingroup$ 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

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 $f(y)$?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.