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Definitions/Background: A monotone normality operator on a topological space $X$ is a function $\mathcal G$ such that

  • for any two closed disjoint subsets $E,F$ of $X,$ $\mathcal G(E,F)$ is an open subset of $X$ such that $E\subseteq\mathcal G(E,F)\subseteq\overline{\mathcal G(E,F)}\subseteq X\setminus F,$ where the bar denotes the closure operator [call this the normality condition], and
  • for any closed subsets $E,F,E',F'$ of $X$ with $E\subseteq E',$ $F'\subseteq F,$ $E\cap F=\emptyset,$ and $E'\cap F'=\emptyset,$ we have $\mathcal G(E,F)\subseteq \mathcal G(E',F')$ [call this the monotone condition].

It's worth noting that if $\mathcal G$ is a monotone normality operator on $X,$ then the function $\mathcal H$ given by $\mathcal H(E,F):=\mathcal G(E,F)\setminus\overline{\mathcal G(F,E)}$ is also a monotone normal operator on $X,$ and $\mathcal H(E,F),\mathcal H(F,E)$ are disjoint for all disjoint closed subsets $E,F$ of $X$. Hence, when convenient, we may without loss of generality require that a monotone normality operator has this property [call it the symmetric disjointness property].

A topological space $X$ is said to be monotonically normal if $X$ is a $T_1$ space and there is a function $\mu$ such that

  • for any $x\in X$ and any open neighborhood $U$ of $x$ in $X,$ we have $x\in\mu(x,U)\subseteq U$, and
  • for any $x,y\in X$ and any respective open neighborhoods $U,V$ of $x,y$ in $X,$ if $\mu(x,U),\mu(y,V)$ are non-disjoint, then $x\in V$ or $y\in U$.

I am trying to show that a $T_1$ space $X$ is monotonically normal if and only if it has a monotone normality operator.

If $\mathcal G$ is a monotone normality operator on $X$ (WLOG with the symmetric disjointness property, then for any $x\in X$ and any open neighborhood $U$ of $x$ in $X,$ we may put $$\mu(x,U):=\mathcal G\bigl(\{x\},X\setminus U\bigr),$$ and readily show that $X$ is monotonically normal.

For the other direction, I'm having some trouble. Given $\mu$ as in the definition of monotonically normal, I defined $$\mathcal G(E,F):=\left(\bigcup_{x\in E}\mu(x,X\setminus F)\right)\setminus\overline{\left(\bigcup_{y\in F}\mu(y,X\setminus E)\right)}.$$ I have been able to show that this satisfies the normality condition, but I've been unable thus far to prove that it satisfies the monotone condition.

It may be that I've chosen a bad definition for $\mathcal G$ (it may not necessarily satisfy the monotone condition), or perhaps there's some property of $\mu$ that I can use to finish the proof.

If $\mu$ had a monotonicity condition of its own--say that $\mu(x,U)\subseteq\mu(x,V)$ whenever $U,V$ are open neighborhoods of $x$ with $U\subseteq V,$ then that would certainly do the trick, but I can't see why $\mu$ should have such a property, nor can I see how to use $\mu$ to define some $\nu$ with all the properties of $\mu$ and also such a monotonicity condition (like we used $\mathcal G$ to define $\mathcal H$).

As a side note (though I'm not sure it will help), in trying to understand more about $\mu,$ I've determined that $\overline{\mu(x,U)}\subseteq U$ whenever $U$ is an open neighborhood of $x$ in $X$, from which we can readily show that a monotonically normal space is completely regular (using a Urysohn's Lemma type argument with $\mu$ to choose the set at each stage) and Hausdorff.

Has anyone got any clues to get me unstuck?

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HINT: Define

$$\mathscr{G}(E,F)=\bigcup\{\mu(x,U): x\in E\text{ and }U\cap F=\varnothing\}\;.$$

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    $\begingroup$ (+1) The normality condition is easy enough, and it's clear that $\mathscr G(E,F)\subseteq\mathscr G(E',F)$ whenever $E,E',F$ are closed with $E\subseteq E'$ and $E'\cap F=\emptyset$. I can only get that $\mathscr G(E',F')\subseteq\mathscr G(E',F)$ when $E',F',F$ are closed with $F'\subseteq F$ and $E'\cap F'=\emptyset,$ which is the reverse of the inclusion that I need. Perhaps I'm missing something, but even if this doesn't quite work, I think I can adapt it to suit my needs. I'll think on it and get back to you. $\endgroup$ – Cameron Buie Sep 25 '13 at 3:54
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    $\begingroup$ @Cameron: If $E\subseteq E'$, $F'\subseteq F$, and $x\in E$, any open nbhd of $x$ that misses $F$ also misses $F'$, so $\mathscr{G}(E,F)\subseteq\mathscr{G}(E',F')$: every $\mu(x,U)$ that goes into building $\mathscr{G}(E,F)$ also goes into building $\mathscr{G}(E',F')$. $\endgroup$ – Brian M. Scott Sep 25 '13 at 4:01
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    $\begingroup$ Oh, for Pete's sake. I went from $F'\subseteq F$ to $X\setminus F'\subseteq X\setminus F$. No wonder it wasn't working! $\endgroup$ – Cameron Buie Sep 25 '13 at 4:03
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    $\begingroup$ @Cameron: Reminds me of the time squared $2$ and got $16$. $\endgroup$ – Brian M. Scott Sep 25 '13 at 4:05

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