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Let $M$ be compact and $\mathcal{U}$ an open covering of M such that each $p \in M$ is contained in at least two members of $\mathcal{U}$. Show that $\mathcal{U}$ reduces to a finite subcovering with the same property.

I've been struggling with the question for quite a while but did not get anywhere..

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3 Answers 3

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First, take a finite cover of your space $M.$

Note that since it is finite, the set of point $p \in M$ such that $p$ is only in one of your $\mathcal{U_i}$ will be a closed set. (You need to use the finiteness of your cover here)

Since this is closed, it will be compact (here I am assuming that your space is Hausdorff though). Then consider a cover of that set by elements of $\mathcal{U}$ that are not in your first cover.

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    $\begingroup$ Every closed subspace of a compact space is compact even without the Hausdorff assumption. $\endgroup$
    – user87690
    Commented Mar 10, 2015 at 21:16
  • $\begingroup$ Thanks for the help, guys! In fact, I misunderstood the question in the beginning (urgh). Just out of curiosity, suppose $U$ reduces to a minimal finite subcovering $U^{'}$ (in the sense that $U^{'}$ has no subcovering), is it necessary that $U^{'}$ is redundant? (My guess is no.) $\endgroup$ Commented Mar 10, 2015 at 21:50
  • $\begingroup$ @user87690 Yeah, you are right I don't need Hausdorff $\endgroup$ Commented Mar 12, 2015 at 17:44
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Hint: move from $\mathcal{U}$ to another cover.

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I would like to elaborate on part of Maxime Scott's solution given above, where it is told that the set $S$ of points which belong to only one member of the finite subcover has to be closed:

Call the finite subcover of $\mathcal{U}$ as $\mathcal{U}'$. The set of points belonging to at least two $U_i \in \mathcal{U}'$ is $$M\setminus S=\bigcup_{1\le i < j \le \lvert \mathcal{U}' \rvert} \{U_i \cap U_j : U_i \in \mathcal{U}'\}.$$ Since all $U_i\cap U_j$ are open, their union is open as well, so $M\setminus S$ is open. Hence $S$ is closed.

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