Continuous function on a compact metric space with no fixed point

Question

Let $(X,d)$ be a compact metric space, and suppose $f:X \to X$ is a continuous map with no fixed point. Show that there is an $\epsilon > 0$ so that $d(f(x),x) \ge \epsilon \ \forall x \in X$.

I considered a function $g:X \to \mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?

Answer 1

The function $g : X \to \mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $\epsilon > 0$ let $I_{\varepsilon} = \{r\in \mathbb{R} \mid r > \varepsilon\}$, which is an open set in $\mathbb{R}$. Then $g^{-1}(I_{\varepsilon})$ is open in $X$, and the family of sets $\{g^{-1}(I_{\varepsilon}) \mid \varepsilon > 0 \}$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = \bigcup_{i=1}^{n} g^{-1}(I_{\varepsilon_i})$. Let $\varepsilon_{min} = \min\{\varepsilon_i,\dots,\varepsilon_n\}$. Then for every $x\in X$ we have $x \in g^{-1}(I_{\varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) \in I_{\varepsilon_i}$, so $d(f(x),x) > \varepsilon_i \geq \varepsilon_{min}$, and we are done.

Answer 2

For metric spaces, compactness and sequential compactness are the same. Assume that $\forall \epsilon > 0$, $\exists x \in X$ such that $d(f(x),x)< \epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.