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For lack of a better term and for this purpose only, let us call a refinement $\mathcal{V}$ of a cover $\mathcal{U}$ faithful if $\mathcal{V}$ can be written as $\mathcal{V}=\{V_U:U\in\mathcal{U}\}$ where $V_U\subseteq U$ for each $U\in\mathcal{U}$.

Does every open cover of a paracompact space have a faithful open refinement that is locally finite?

In other words (and maybe clearer): if, for every open cover $\mathcal{U}$ of a topological space, we demand the existence of such an open and locally finite cover $\{V_U:U\in\mathcal{U}\}$ that $V_U\subseteq U$ for each $U\in\mathcal{U}$, do we get an equivalent or strictly stronger notion than paracompactness?

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    So by faithful you really want that the refinement to be in bijection with the original covering, right?2012-03-29
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    @Asaf: Not necessarily in bijection: $V_{U_1}=V_{U_2}$ for distinct $U_1,U_2$ is allowed.2012-03-29
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    Do remind me, then, how does that differ from the definition of a paracompact space?2012-03-29
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    @Asaf: I don't immediately see that they are the same, since in the definition of refinement we require only that each refining set lies inside some set of the original cover.2012-03-29
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    So what you're actually looking for is either a confirmation that we can find a refinement that for *every* set in the open cover there is a refined set which is a subset of it; or a counterexample for that: some open cover that if we require every set to have a subset then we will lose the locally finiteness property?2012-03-29
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    @Asaf: I rephrased the question. Hopefully it's clearer now.2012-03-29

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I usually call this a precise refinement, but that may just be me.

The answer is yes. Let $\mathscr{U}$ be an open cover of the paracompact space $X$. Then $\mathscr{U}$ has a locally finite open refinement $\mathscr{R}$. For each $R\in\mathscr{R}$ fix $U(R)\in\mathscr{U}$ such that $R\subseteq U(R)$. For each $U\in\mathscr{U}$ let $V_U=\bigcup\{R\in\mathscr{R}:U(R)=U\}$, and let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. Clearly $\mathscr{V}$ is a precise open refinement of $\mathscr{U}$, so it only remains to show that $\mathscr{V}$ is locally finite.

Fix $x\in X$. $\mathscr{R}$ is locally finite, so there is an open nbhd $G$ of $x$ that meets only finitely many $R\in\mathbb{R}$, say $R_1,\dots,R_n$. Then $G\cap V_{U(R_k)}\ne\varnothing$ for $k=1,\dots,n$, but that’s it: if $G\cap V_U\ne\varnothing$, there must be an $R\in\mathscr{R}$ such that $G\cap R\ne\varnothing$ and $U(R)=U$, in which case $V_U=V_{U(R_k)}$ for some $k$.

Note that it is possible that some $V_U$ are empty.

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    Thanks a lot! And _precise_ is indeed a better adjective for it.2012-03-29