Necessity in Arzela-Ascoli theorem

Question

I am trying to prove necessity of boundedness and equicontinuity in Arzela-Ascoli and I don't know how to go about it. More precisely,I have:

Let $K$ be a compact metric space, and $A\subset C^0(K)$ a non-empty subset, Then A is pre-compact if and only if the following two conditions are satisfied: 1) A is bounded 2) A is equicontinuous

I am guessing that the first one is easy, with an argument like since $\bar A$ is compact in a metric space it is closed and bounded and since $A\subset \bar A$, $A$ is also bounded? For the second, I am not sure how to proceed: I am guessing that using the property that every sequence has a converging subsequence is good direction?

Thanks.

Answer 1

Your argument concerning the boundedness of $A$ is correct.

For equicontinuity, I would proceed as follows.

1. Let $\varepsilon > 0$.

2. Show that compactness of $\overline{A}$ implies that there exist $f_1,\dots,f_n \in A$ such that for every $f \in A$ we have $\|f - f_k\|_{\infty} < \frac{\varepsilon}{3}$ for some $k$.

3. Note that compactness of $K$ implies that the finite set $\{f_1,\dots,f_n\}$ is equicontinuous so we find $\delta > 0$ such that $|x - y| < \delta$ implies $|f_j(x) - f_j(y)| < \frac{\varepsilon}{3}$ for all $1 \leq j \leq n$.

4. For $f \in A$ take $k$ as in 2. and look at $$|f(x) - f(y)| \leq |f(x) - f_k(x)| + |f_k(x) - f_k(y) | + |f_k(y) - f(y)|$$ Now use 2. and 3.