I have some questions about the two versions of Arzelà-Ascoli Theorem given in Folland's book "Real Analysis"
Version 1:
Given a compact Hausdorff space $X$, let $F$ be an equicontinuous, pointwise bounded subset of $C(X)$, the set of complex-valued continuous functions on $X$. The, $F$ is totally bounded in the uniform metric and the closure of $F$ in $C(X)$ is compact.
(Q1) The uniform metric is not well-defined for all $f \in C(x)$, so what does it mean to say that $F$ is totally bounded in the uniform metric? Also, how do we take the closure with respect to this metric?
Version 2:
If $X$ is a $\sigma$-compact locally compact Hausdorff space, $\dots$
Isn't $\sigma$-compact space always locally compact? Why is it assumed that the space is both $\sigma-$compact and locally compact?
The uniform metric in $C(X)$ is defined by $d(f,g)=\sup_{x\in X}\bigl|f(x)-g(x)\bigr|$. This makes sense for any two functions $f,g\in C(X)$, since $X$ is compact, which implies that $f-g$ is bounded.
On the other hand $\mathbb Q$ (with the usual topology) is $\sigma$-compact, since it's the union of all of its singletons, but not locally compact.