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I've encountered a homework question I don't know how to solve, and most people I asked even tend to go so far as to say that the main statement in the question is false. I'm not really convinced of the latter (the question is from Warner's Foundations of Differentiable Manifolds and Lie Groups), so I was wondering whether anyone can shed some light on the issue. Any help would be greatly appreciated :-)

First we define $\{ U_\alpha \}$ to be an open cover of a manifold $M$. The question is now to prove that there exists a refinement $\{ V_\alpha \}$ of $\{ U_\alpha \}$, such that $\overline{ V_\alpha } \subset U_\alpha $ for all $\alpha$ (the closure of every $V_\alpha$ must be a subset of every $U_\alpha$).

This is actually all the information that is given... we've been considering some possible 'catches' to the question, but none of them resulted in something useful. I hope somebody can help out!

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When you say "the closure of every $V_\alpha$ must be a subset of every $U_\alpha$", you mean "the closure of every $V_\alpha$ must be a subset of the corresponding $U_\alpha$"? –  joriki Sep 20 '12 at 23:09
That's actually one of the 'catches' we encountered... the book doesn't say more than this, but it pays great attention to detail, so I assume that this is what they mean. –  JorenB Sep 20 '12 at 23:18
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2 Answers

up vote 3 down vote accepted

If your definition of manifold requires $M$ to be second countable and Hausdorff, or even just paracompact and Hausdorff, then $M$ is metrizable and hence both metacompact and normal. Let $\mathscr{U}=\{U_\alpha:\alpha\in A\}$; metacompactness implies that $\mathscr{U}$ has a point-finite open refinement $\mathscr{R}$, meaning that each point of $M$ is in only finitely many members of $\mathscr{R}$. For each $R\in\mathscr{R}$ there is an $\alpha(R)\in A$ such that $R\subseteq U_{\alpha(R)}$. For each $\alpha\in A$ let $W_\alpha=\bigcup\{R\in\mathscr{R}:\alpha(R)=\alpha\}$, and let $\mathscr{W}=\{W_\alpha:\alpha\in A\}$; then $\mathscr{W}$ is a point-finite open refinement of $\mathscr{U}$, and $W_\alpha\subseteq U_\alpha$ for each $\alpha\in A$.

It now suffices to find an open cover $\mathscr{V}=\{V_\alpha:\alpha\in A\}$ of $M$ such that $\operatorname{cl}V_\alpha\subseteq W_\alpha$ for each $\alpha\in A$. Such a cover is called a shrinking of $\mathscr{V}$, and it’s a standard result that in a normal space every point-finite open cover is shrinkable. Applying this result, we get the desired $\mathscr{V}$.

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The result follows from Zorn's lemma. Consider the family $\mathbb F$ consisting of functions from the index set (say) $I$ into the topology $\tau$ so that $f(\alpha)=U_\alpha$ or $\overline f(\alpha)\subset U_\alpha$ and $\bigcup_{\alpha\in I} f(\alpha)=M$. Let us order $\mathbb F$ by defining $f\leq g$ if $f(\alpha)=g(\alpha)$ for all $\alpha \in I$ so that $f(\alpha)\ne U_\alpha$. It is easy to see that the partial order satisfies the hypothesis of Zorn's lemma and that a maximal element gives you the desire refinement.

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If $f_0$ is a maximal element of $\mathbb{F}$, can we be sure $\{f_0(\alpha)\}$ covers $M$? –  Julian Rosen Sep 20 '12 at 23:41
@PinkElephants I forgot to add the condition that for all $f\in\mathbb F$ $\bigcup_{\alpha\in I} f(\alpha)=M$. –  azarel Sep 21 '12 at 0:28
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