# "Redundant" finite subcovering of a compact space.

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..

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.

• Every closed subspace of a compact space is compact even without the Hausdorff assumption. Commented Mar 10, 2015 at 21:16
• 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.) Commented Mar 10, 2015 at 21:50
• @user87690 Yeah, you are right I don't need Hausdorff Commented Mar 12, 2015 at 17:44

Hint: move from $\mathcal{U}$ to another cover.

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.