Let $X$ be a compact metric space and $F\subset C(X)$ be a compact subset. Show that $F$ is equicontinuous.
Proof: Let $f\in F$ be an arbitrary function. I want to show that $\forall \epsilon>0, \exists\,\delta>0$ such that, if $|x-y|<\delta$ then $|f(x)-f(y)|< \epsilon \ $ for all $f\in F$ and $\forall x,y\in X$.
I'm stuck in here. Should I take a sequence of functions?
How do I use the compactness of $X$ and compactness of $F$.
Best Answer
Since $F$ is compact, for given $\epsilon > 0$ there exist $f_i \in F$ such that $F \subseteq \cup_{i=1}^nB(f_i, \frac{\epsilon}{3})$. Choose $\delta_i>0$ by the uniform continuity of $f_i$ for $\frac{\epsilon}{3}$ and let $\delta$ be the minimum of $\delta_i$'s. Let $f \in F$ be arbitrary, then $f \in B(f_j, \frac{\epsilon}{3})$ for some $f_j$ thus $|f(x)-f_j(x)|< \frac{\epsilon}{3}$ for any $x$ , now $d(x_1,x_2)<\delta$ implies \begin{align} &|f(x_1)-f(x_2)| \leq |f(x_1)-f_j(x_1)|+|f_j(x_1)-f_j(x_2)|+|f_j(x_2)-f(x_2)|< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon. \end{align}