I am trying to show the following:
Let $\mathcal{U} = \{f: S \rightarrow \mathbb{R} \ \ \text{s.t.} \ f \ \text{is bounded}\}$ and assume that $S$ is totally bounded. Denote by $A$ the finite $\epsilon$-net of $S$, so that for $\epsilon > 0$ $S = \bigcup_{a \in A} B_\epsilon(a)$, where $B_\epsilon(a) = \{s \in S \ \text{s.t.} \ \sup_{g \in \mathcal{U}}|g(s) - g(a)| < \epsilon \}$.
I am trying to show that $E = \{g: A \rightarrow \mathbb{R} \ \text{s.t.} \ g(a) = f(a) \ \forall a \in A, \ \text{for some} \ f \in \mathcal{U}\}$ equipped with the sup norm $||\cdot||_{\infty}$ is totally bounded.
Intuitively, I understand that I need to find a finite set $D \subset E$ such that for any $y \in E$ there exists an $x \in D$ such that $\max_{a \in A} |y(a) - x(a)| < \epsilon$. I can see that the set $D$ takes the form $D = \bigcup_{\alpha \in A} \{x^{\alpha} \}$ where $x^\alpha : A \rightarrow \mathbb{R}$ and that I need to put an additional restriction on the function $x^\alpha$ so that $D$ is finite and $||y - x^\alpha||_{\infty} < \epsilon$, however I haven't had any luck so far with finding the right one.
This is false. $E$ is not even bounded (hence not totally bounded). Note that $f \in \mathcal U$ implies that $nf \in \mathcal U$ for every positive integer $n$. Hence $g \in E$ implies $ng \in E$ for every positive integer $n$.