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

Question

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.

Answer 1

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.)

Answer 2

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.