Currently I'm reading baby Rudin and I've come across Arzelà-Ascoli Theorem (Theorem 7.25 in Rudin). I've read online that equicontinuity is exactly the additional property that we need to get something on $C(K)$ (with $K$ compact metric space) similar to Heine-Borel Theorem on euclidean spaces. To be more precise, it seems like we can say the following: Let $E$ be a subset of $C(K)$, $E$ is compact if and only if $E$ is bounded (uniformly), closed (uniformly) and equicontinuous. Now by Theorem 7.25, I think I can see why $E$ bounded, closed and equicontinuous implies $E$ compact. The problem is that I'm not able to prove the converse, namely is it true that $E$ compact implies $E$ equicontinuous? If that's not the case, is there a good example?
Arzelà-Ascoli and Compactness
arzela-ascolicompactnessreal-analysis
Best Answer
Let $E$ be compact. Then given $\epsilon>0$ , there exists a finite $\frac{\epsilon}{3}$ net $A=\{f_{1},...,f_{n}\}$ . Now $f_{i}$'s are uniformly continuous (by continuity on a compact metric space)
There exists $\delta_{i}$ s.t $|f_{i}(x)-f_{i}(y)|\leq \frac{\epsilon}{3}$ for all $x,y$ s.t $d(x,y)<\delta_{i}$ . So take the minimum of all the finitely many $\delta_{i}$ to get that $|f_{i}(x)-f_{i}(y)|<\frac{\epsilon}{3}\,,\forall x,y$ s.t. $d(x,y)<\delta$.
Now for any $f\in E$ . There exists $f_{i}$ such that $\sup_{x\in X}|f(x)-f_{i}(x)|<\frac{\epsilon}{3}$ by construction of the epsilon net. Then If $d(x,y)<\delta$ then,
$$|f(x)-f(y)|\leq |f(x)-f_{i}(x)| + |f_{i}(x)-f_{i}(y)|+|f_{i}(y)-f(y)|<3\cdot\frac{\epsilon}{3}$$ .
And this proves that $E$ is equicontinuous.
To sum it up, Arzela-Ascoli theorem says that a set $E\subset C(K)$ is compact if and only if it is closed, bounded and equicontinuous.