# Definition of Compactness of a Topological Space: How can every open cover have a finite subcover

In Basic Topology, Armstrong states, "a topological space is compact if every open cover of X has a finite subcover." But, I'm mildly confused how this doesn't translate to the empty set being a finite subcover itself.

First, let X be a compact space and let U be an open cover of X. Thus, there exists a finite subcover of X, say $$\{x_1, ..., x_n\}$$. But then, this itself is an open cover of X, since a set of open sets whose union is the space is an open cover. Thus, there exists some element $$x_i$$ s.t. $$\bigcup\{x_1, \ldots, \widehat{x_i},\ldots x_n\}= X$$ i.e., a set without $$x_i$$ in it that's an open cover of X.

Why couldn't we repeat this until the set is empty and then we get $$\bigcup \emptyset = X$$ which is obviously a contradiction.

I obviously have some deep and fundamental misunderstanding of what open covers, subcovers, and compact spaces are, so I'd appreciate any insight.

• – MJD
Oct 3, 2021 at 21:41
• What does "a set without $x_i$ in it" mean? Each $x_j$ is a $subset$ of $X$. Oct 3, 2021 at 22:56

A "subcover" does not have to be a proper subcover. That is, the finite subcover of $$\{x_1,\dots,x_n\}$$ which is guaranteed to exist by compactness could just be $$\{x_1,\dots,x_n\}$$ itself, rather than some proper subset. You are correct that if you knew every open cover had a finite proper subcover, you could repeatedly get smaller and smaller subcovers to conclude the empty set is a cover and so the space would have to be empty. (Or more simply, you could reach that conclusion by just taking a proper subcover once of the specific open cover $$\{X\}$$, whose only proper subset is the empty set.)
Your error lies in the sentence “Thus, there exists some element $$x_i$$ s.t. $$\bigcup\{x_1, \ldots, \widehat{x_i},\ldots x_n\}= X$$”. That's not true. Since $$B=\{x_1,x_2,\ldots,x_n\}$$ is an open cover of $$X$$, it must have a finite subcover $$B'$$ of $$X$$, yes. But $$B$$ is already finite and therefore nothing prevents $$B'$$ from being $$B$$ itself.