Please can you check my proof? I proved:
Let $X$ be a compact Hausdorff space and let $C(X)$ denote the space of continuous functions $f: X \to \mathbb R$ endowed with the sup norm $\|\cdot \|_\infty$. Then $S \subseteq C(X)$ is relatively compact if it is pointwise bounded and equicontinuous.
Here is my proof:
Let $S \subseteq C(X)$ be equicontinuous and pointwise bounded. Then $\overline{S}$ is also (uniformly) equicontinuous and pointwise bounded. The goal is to show that $\overline{S}$ is compact. By the general Heine-Borel theorem for metric spaces a subset of a metric space is compact if and only if it is complete and totally bounded. Furthermore, closed subsets of complete metric spaces are complete hence it is enough to show that $\overline{S}$ is totally bounded.
To this end let $\varepsilon > 0$. Since $\overline{S}$ is uniformly equicontinuous there exists $\delta > 0$ such that $|x-y|< \delta$ implies that $|f(x) -f(y)| < \varepsilon / 3$ for every $f \in \overline{S}$. Since $X$ is compact it may be covered using finitely many $\delta$ balls. Let $x_1, \dots, x_n$ denote the centres of these $\delta$ balls. Since $\overline{S}$ is pointwise bounded there are constants $K_i$ such that $\sup_{f \in \overline{S}}|f(x_i)|\le K_i$.
For $i \in \{1, \dots, n \}$ the interval $[-K_i,K_i]$ is totally bounded hence there are finitely many open balls $B_{i1}, \dots , B_{im_i}$ of length less than ${\varepsilon \over 3}$ that cover it. Define $A_{j_1, \dots, j_n} = \{f \in \overline{S} \mid f(x_i) \in B_{ij_k} \text{ for all } 1\le k\le n \}$. That is, for every $i$, $f(x_i)$ is in one of the balls covering $[-K_i,K_i]$. To finish the argument we claim that (1) the sets $A_{j_1, \dots, j_n}$ cover $\overline{S}$ and (2) every set $A_{j_1, \dots, j_n}$ is contained in an $\varepsilon$ ball:
(1) Let $f\in \overline{S}$ Then for every $i \in \{1, \dots, n\}$ we have $f(x_i) \in [-K_i,K_i]$. Hence for every $i$, $f(x_i) \in B_{ij_k}$ for some $j_k$. That is, $f \in A_{j_1, \dots , j_n}$.
(2) Let $f,g \in A_{j_1, \dots, j_n}$ and $x \in X$ be arbitrary. Then there is $x_i$ such that $x \in B(x_i,\delta)$. Then
$$ |f(x) – g(x)| \le |f(x) – f(x_i)| + |f(x_i) – g(x_i)| + |g(x_i) – f(x_i)| < \varepsilon$$
Best Answer
Your argument is correct. You actually proved that $A_{j_1, \dots, j_n}$ has diameter at most $\epsilon$ (which of course implies it's "contained in an $\varepsilon$-ball"). Personally I would stick with diameter formulation, because phrasing the result in terms of $\varepsilon$-ball makes the reader wonder what the center of that ball might be (which we don't know and don't care about; in fact $A_{j_1, \dots, j_n}$ may well be empty).