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.
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?