Let $X$ be a compact topological space and $\mathcal F\subseteq C(X,\mathbb C)$.
By the ordinary Arzelà-Ascoli theorem, we know that
- If $\mathcal F$ is relatively compact in the uniform topology, then $\overline{\mathcal F}$ is equicontinuous.
- If $\mathcal F$ is equicontinuous and pointwise bounded, then $\mathcal F$ is relatively compact.
Question: How precisely does this result generalize when $X$ is not compact?
My guess is that we should be able to show something like the following: Assume that $X$ is an arbitrary topological space.
- If $\mathcal F$ is relatively sequentially compact in the topology of compact convergence, then $\mathcal F$ is locally$^1$ uniformly equicontinuous.
- If $X$ is $\sigma$-compact and $\mathcal F$ is equicontinuous and pointwise bounded, then $\mathcal F$ is relatively compact in the topology of compact convergence.
A few remarks: We may note that "equicontinuity" and "uniform equicontinuity" of a subset of $C(K)$ coincide on every compact topological space $K$. Moreover, if $X$ is $\sigma$-compact, then the topology of compact convergence on $\mathbb C^X$ is metrizable and we know that on every metric space "relative compactness" and "relative sequential compactness".
Regarding 3.: Assume $\mathcal F$ is relatively compact in the topology of compact convergence. Let $K\subseteq X$ be compact and $$\mathcal F_K:=\left\{\left.f\right|_K:f\in\mathcal F\right\}.$$ Since $\mathcal F$ is relatively compact in the topology of compact convergence, $\mathcal F_K$ should be relative compact in the uniform topology. Thus, (1.) yields that $\mathcal F_K$ is uniformly equicontinuous.
Is this proof correct? Can we show more, for example, when $(X,d)$ is a $\sigma$-compact metric space?
Regarding the latter: Let $\varepsilon>0$. Since $\mathcal F_K$ is uniformly equicontinuous for every compact subset $K\subseteq X$, we would find a nonincreasing $(\delta_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\forall n\in\mathbb N:\forall f\in\mathcal F:\forall x,y\in X_n:d(x,y)<\delta_n\Rightarrow\left|f(x)-f(y)\right|<\varepsilon\tag2,$$ where $(X_n)_{n\in\mathbb N}$ is a nondecreasing sequence of compact subsets of $X$ with $X=\bigcup_{n\in\mathbb N}X_n$. But are we able to deduce something stronger than the already shown "local uniform equicontinuity" from this?
Regarding 4.: Assume $\mathcal F$ is equicontinuous and pointwise bounded. Then $\mathcal F_K$ is clearly equicontinuous and pointwise bounded as well, for every compact $K\subseteq X$. Assume $X$ is $\sigma$-compact and hence there are compact $X_n\subseteq X$ for $n\in\mathbb N$ such that $(X_n)_{n\in\mathbb N}$ is nondecreasing and $X=\bigcup_{n\in\mathbb N}X_n$. If $(f_n)_{n\in\mathbb N}\subseteq\mathcal F$, then we can apply (2.) inductively to see that there is a $g:X\to\mathbb C$ with $$\left.g\right|_{X_n}\in C(X_n,\mathbb C)$$ and $$\sup_{X_n}\left|f_{n_k}-g\right|\xrightarrow{k\to\infty}0\tag1$$ for some increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ for all $n\in\mathbb N$. Can we conclude $g\in C(X,\mathbb C)$?
$^1$ I'm not sure if this terminology is consistent in the literature, but with "locally" I mean that $\left\{\left.f\right|_K:f\in\mathcal F\right\}$ is (uniformly) equicontinuous for every compact $K\subseteq X$.