Let $\langle X,\mathscr{O}\rangle$ be a topological space. We say that disjoint $A,B\in 2^X$ are functionally separated iff there exists a continuous function $f\colon X\to [0,1]$ such that:

  • if $x\in A$, then $f(x)=0$
  • if $x\in B$, then $f(x)=1$

We define $\langle X,\mathscr{O}\rangle$ to be completely regular iff for a closed set $C\in 2^X$ and a point $x\notin X$, $F$ and $\{x\}$ are functionally separated. Let $\langle\mathrm{r}\mathscr{O},+,\cdot,-,\emptyset,X\rangle$ be the complete boolean algebra of regular open subsets of completely regular space $\langle X,\mathscr{O}\rangle$. In this algebra define: \[ A\ll B\iff \mbox{$A$ and $-B$ are functionally separated.} \] What I am trying to prove are the following two properties of $\ll$:

  1. if $A\ll C$ and $B\ll D$, then $A\cdot B\ll C\cdot D$
  2. if $A\ll C$, then $\exists_{B\in 2^X\setminus\{\emptyset\}}\,A\ll B\ll C$

Concerning 1. suppose $f\colon X\to[0,1]$ and $g\colon X\to[0,1]$ are such that:

  • $x\in A\rightarrow f(x)=0$ and $x\in B\rightarrow g(x)=0$
  • $x\in -C\rightarrow f(x)=1$ and $x\in -D\rightarrow g(x)=1$

Now put $h(x):=\max\{f(x),g(x)\}$. Thus we have:

  • $x\in A\cdot B\rightarrow x\in A\wedge x\in B\rightarrow f(x)=0 \wedge g(x)=0\rightarrow h(x)=0$
  • $x\in -C+-D\rightarrow x\in\mathrm{Int}\,\mathrm{Cl}\, (-C\cup-D)\rightarrow\: ???$.

And I got stuck at the second dot above. Could you please give my any hint how to proceed?

Concerning 2. I would appreciate any suggestion.


2 Answers 2


For 1 note that $x ∈ \operatorname{Cl}(-C) ∪ \operatorname{Cl}(-D)$.

For 2 consider $B := f^{-1}[[0, 1/2)]$.

  • $\begingroup$ Could you please develop slightly 2nd point? I am not sure if I understand it properly. $\endgroup$ Nov 10, 2015 at 8:17
  • $\begingroup$ The idea is that particular values of $f$ on the sets doesn't matter, the poit is there is space between them. We have $f(A) = 0$, $f(-B) ≥ 1/2$, so for $g := \min(f, 1/2)$ we have $g(A) = 0$, $g(B) = 1/2$, … $\endgroup$
    – user87690
    Nov 10, 2015 at 9:52

If $x\in -(C\cdot D)$, $$\begin{split}\Rightarrow & x \in \text{Cl}(-C)\text{ or }x\in \text{Cl}(-D) \\ \Rightarrow & f(x) =1\text{ or }g(x)=1 \\ \Rightarrow &h(x) = 1 \end{split}$$ as $f, g$ has range $[0,1]$).

  • $\begingroup$ If $x\in-(C\cdot D)$, then $x\in(-C+-D)$ from which you cannot infer that $x\in-C$ or $x\in-D$. To be more precise, if $x\in-(C\cdot D)$, then $x\in\mathrm{Cl}(-C)$ or $x\in\mathrm{Cl}(-D)$, yet the sets I consider are not closed in general. Or am I missing something? $\endgroup$ Nov 8, 2015 at 12:39
  • $\begingroup$ @MadHatter : For $Cl$, does that stand for (topological) closure? $\endgroup$
    – user99914
    Nov 8, 2015 at 12:45
  • $\begingroup$ Yes, it is the standard topological closure. $\endgroup$ Nov 8, 2015 at 12:46
  • 1
    $\begingroup$ Um.... Then the argument is the same. if $f =1$ on $A$, then $f = 1$ also on $\overline A$. @MadHatter $\endgroup$
    – user99914
    Nov 8, 2015 at 12:50

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