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Let $K$ be a compact Hausdorff space and let $U,V\subset K$ be open. Let $\left(f_{i}\right)_{i\in I}$ be an increasing net of continuous non-negative functions such that $f_{i}\le 1$ and $f_{i}$ vanish outside of $U\cup V$. In my specific situation $\sup_{i\in I} f_{i}(x)=1$, for every $x$ in some dense subset of $U\cup V$, but I don't think it is relevant here.

Can we find two increasing nets $\left(g_{p}\right)_{p\in P}$ and $\left(h_{q}\right)_{q\in Q}$ of continuous non-negative functions such that $g_{p}$ and $h_{q}$ vanish outside of $U$ and $V$, respectively, and $g_{p}+h_{q}\le 1$, for every $p,q$, but for every $i$ there are $p,q$ such that $g_{p}+h_{q}\ge f_{i}$?

In the case when $U\cup V$ happens to be normal, since $U\backslash V$ and $V\backslash U$ are closed in $U\cup V$, there is $e:U\cup V\to [0,1]$ which equals $1$ on $U\backslash V$ and vanishes on $V\backslash U$. Define $e$ to be $0$ on $K\backslash\left(U\cup V\right)$. Then, taking $g_{i}=(1-e)f_{i}$ and $h_{i}=ef_{i}$ accomplishes the goal.

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  • $\begingroup$ Can't you just replace $U$ and $V$ with their closures? $\endgroup$
    – Nik Weaver
    Sep 9, 2022 at 11:58
  • $\begingroup$ @NikWeaver then the sets $\overline{U}\backslash V$ and $\overline{V}\backslash U$ are not necessarily disjoint $\endgroup$
    – erz
    Sep 9, 2022 at 17:44
  • $\begingroup$ Oh, I see. Yeah, it's not so obvious ... $\endgroup$
    – Nik Weaver
    Sep 10, 2022 at 13:46
  • $\begingroup$ You can replace $U\cup V$ by the cozero-set $O_f=\{x:f(x)\neq0\}$ of $f$; that is an $F_\sigma$-set and hence normal. Separate $O_f\cap(U\setminus V)$ and $O_f(V\setminus U)$ by the function $e$. $\endgroup$
    – KP Hart
    Sep 14, 2022 at 10:35
  • $\begingroup$ @KPHart what is $f$? $\endgroup$
    – erz
    Sep 14, 2022 at 18:16

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