I'm having trouble following a step on page 54 of Peter Meyer-Nieberg's Banach Lattices, proposition 2.1.5. The setting: $K$ is a compact, hausdorf, quasi-stonian (Quasi-extremely disconnected) space. Definition of quasi-stonian: The closure of every open $F_\sigma$ set is open. (I don't think that part is necessary for the step I'm having trouble with)
The goal is to show that $C(K)$ (The set of continuous real valued functions on $K$ is $\sigma$-Dedekind complete (meaning every order bounded sequence has an infinum and supremum)
So he starts with a sequence $(f_n)\in C(K)$ such that $0\le f_n\le 1$ (I assume that is sufficient for any order bounded sequence? Since that's the only one he proves it for)
He constructs the following two functions on $K$: $$f(s)=\inf \{f_n (s):n\in \mathbb N\} $$ $$\hat f(s)=\sup \{\inf \{f(t):t\in U\}:U\text { open}, s\in U\}$$ So, I see what's going on here, he's taking every open set s is in, taking the inf on those open sets, then taking the sup of those numbers.
He then goes to prove the continuity of $\hat f$. Where he loses me is that he asserts that for $a\in \mathbb R$, $\hat f ^{-1}((a,\infty))$ is open by construction.
That phrase makes it seem like it should be obvious, but I've been stuck for a week. I've tried starting with something like: Let $s\in \hat f ^{-1}((a,\infty))$ Let $\mathcal B$ be a basis of open sets on $K$ and $B\in \mathcal B$ an arbitrary basis set such that $s\in B$, then try to show that $\forall x\in B$, $f(x)>a$, since that would get that anything that shares a basis element with $s$, the whole basic open set is in there, and I'd be done. Given that $B$ would be one of the $U's$ we are taking the infinums over, it seemed like I might have something...but I couldn't get anywhere.
Any thoughts/help/clues/answers?
Let $s \in \hat{f}^{-1}((a,\infty))$. Then $\hat{f}(s) > a$, so by definition of the supremum, there exists an open set $U$ such that $s \in U$ and $\inf\{ f(t) : t \in U\} > a$. For any $s' \in U$, we have $$\hat{f}(s') \ge \inf\{ f(t) : t \in U\} > a.$$ Thus $U \subset \hat{f}^{-1}((a,\infty))$, which means that $\hat{f}^{-1}((a,\infty))$ is a neighborhood of $s$.