# Removing unnecessary sets of a refinement in a paracompact topological space.

Let $(X,\mathcal T)$ be a paracompact topological space and $\mathcal W$ be an open cover for $X$. Is there an open refinement $\mathcal U$ for $\mathcal W$, such that

$$\{U\in \mathcal U\mid (\forall V\in \mathcal U)(U\not\subset V) \}$$ is a cover for $X$?

$\not\subset$ means "not proper subset".

Edit:

$\{U\in \mathcal U\mid (\forall V\in \mathcal U)(U\not\subset V) \}$ must be an open cover in which all sets are maximal.

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Try \not\subset – Erick Wong Apr 11 '13 at 13:06
@ErickWong thanks. – user59671 Apr 11 '13 at 13:07

It’s true even if $X$ is only metacompact, meaning that every open cover has a point-finite open refinement. This follows from the following theorem, but first a definition: a family $\mathscr{F}$ of sets is irreducible if $F\nsubseteq\bigcup\left(\mathscr{F}\setminus\{F\}\right)$ for each $F\in\mathscr{F}$.
Theorem. Let $\mathscr{U}$ be a point-finite open cover of a space $X$. Then $\mathscr{U}$ has an irreducible subcover.
Proof. For each $x\in X$ let $\mathscr{U}(x)=\{U\in\mathscr{U}:x\in U\}$; this set is finite, so it has an irreducible subset $\mathscr{V}(x)$. Let $V(x)=\bigcup\mathscr{V}(x)$. Say that a set $D\subseteq X$ is $\mathscr{V}$-discrete if $V(x)\cap D=\{x\}$ for each $x\in D$. Let $$\mathfrak{D}=\{D\subseteq X:D\text{ is }\mathscr{V}\text{-discrete}\}\;;$$ then $\langle\mathfrak{D},\subseteq\rangle$ is a partial order. Let $\mathfrak C$ be a chain in this partial order, and let $D=\bigcup\mathfrak C$. If $x,y\in D$ with $x\ne y$, there is a $C\in\mathfrak C$ such that $x,y\in C$, so $y\notin V(x))$, and $D$ is $\mathscr{V}$-discrete. Thus, every chain in $\mathfrak D$ has an upper bound in $\mathfrak D$, and by Zorn’s lemma $\mathfrak D$ has a maximal element $D$. Let $\mathscr{V}=\bigcup_{x\in D}\mathscr{V}(x)$; then $\mathscr{V}$ is an irreducible subcover of $\mathscr{U}$. $\dashv$
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