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Please help! I am stuck at this question: Let X be compact and $f : X \rightarrow X$ be continuous. If $f$ has no fixed point then there exist $\delta > 0$ such that $d(a,f(a)) \geq \delta$ for each $a \in X$.

What I thought: I was trying to prove it using contra-position, but then the problem seems to be trivial. Please help me.

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  • $\begingroup$ Write out your solution using the contrapositive. If it's correct, then we can tell you so! If it's not, then we can try to fix it. $\endgroup$ – Ian Coley Dec 8 '13 at 7:31
  • $\begingroup$ Let's suppose that there doesn't exist any such $\delta > 0$, for some $a \in X$. Then for that $a$, $f(a) = a$. Am I missing something? $\endgroup$ – Andrew Ray Dec 8 '13 at 7:34
  • $\begingroup$ Perhaps I'm mistaken but I think you need $X$ to be closed. Otherwise take $X=(0,1)$ and $f(x)=x^{2}$. $\endgroup$ – user71352 Dec 8 '13 at 7:38
  • $\begingroup$ I am really sorry. I missed an important point. It was given that $X$ is compact. I am updating the original post. $\endgroup$ – Andrew Ray Dec 8 '13 at 7:40
  • $\begingroup$ The version of the contrapositive you stated is incorrect. Rather, for every $\delta>0$ there is an $a$ with $d(a,f(a))<\delta$. But note that changing $\delta$ may change $a$. Some work is needed from here to see that there is indeed a fixed point. $\endgroup$ – Andrés E. Caicedo Dec 8 '13 at 7:42
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Your intuition to show the contrapositive is correct! Here is my idea:

Note that compactness implies sequential compactness since we are working in a metric space. Assume that for every $\delta > 0$, there is some $a\in X$ for which $d(a,f(a)) < \delta$. In particular, this means that, for all $n\in N$ where $N$ is the set of natural numbers, there is some $a_n$ for which $d(a_n, f(a_n)) < \frac{1}{n}$. Since the metric space is sequentially compact, every sequence has a convergent subsequence, and in particular, we can choose $a_{n_k}$ so that $a_{n_k}\rightarrow a \in X$. At the same time we note that $$d(a,f(a)) = \lim_{k\rightarrow \infty} d(a_{n_k},f(a_{n_k})) \leq \lim_{k\rightarrow\infty} \frac{1}{n_k} = 0$$ and so $a$ is a fixed point.

Can you think of a way to prove this without using sequential compactness?

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  • $\begingroup$ Your previous to last $=$ would be best as a $\le$. $\endgroup$ – Andrés E. Caicedo Dec 8 '13 at 7:53
  • $\begingroup$ Good catch! Sorry, it's late here. $\endgroup$ – user106555 Dec 8 '13 at 7:55
  • $\begingroup$ Thanks a lot for the help. It is indeed a pretty solution. $\endgroup$ – Andrew Ray Dec 8 '13 at 7:55
  • $\begingroup$ No problem. Are you allowed to assume the equivalence of compactness and sequential compactness in your class? $\endgroup$ – user106555 Dec 8 '13 at 7:56
  • $\begingroup$ Yes I am. I have seen that proof in class :) $\endgroup$ – Andrew Ray Dec 8 '13 at 7:57
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Do you know the theorem that the image of a compact set under a continuous function is compact? Consider the function $x \to d(x, f(x))$.

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