# Relating a finite number of sets in a base to a finite number of open sets

As part of the proof that the product of two compact spaces is itself compact, I'm seeing this lemma:

Let $$\mathfrak B$$ be a base for the open sets of a topological space $$Z$$. If, for each covering $$\{B_{\beta}\}_{\beta \in J}$$ of $$Z$$ by members of $$\mathfrak B$$, there is a finite subcovering, then Z is compact.

The proof (rather too long to post here), seems to be saying:

Because every open set in $$Z$$ can be described as a union of sets in $$\mathfrak B$$, no covering of $$Z$$ can have more members than the equivalent covering in $$\mathfrak B$$.

If there were more sets in an arbitrary covering $$\{A\}$$ of $$Z$$ than there are in an equivalent covering in $$\mathfrak B$$, then $$\{A\}$$ would include sets that are not equal to a union of sets in $$\mathfrak B$$.

Is this correct?

let $$\mathcal{U}$$ be an arbitrary open cover of $$Z$$. Let $$\mathcal{B}$$ be a defined by $$\mathcal{B}= \{O \in \mathfrak{B}: \exists U \in \mathcal{U}: B \subseteq U\}$$ which is an open cover of $$Z$$ by members of $$\mathfrak{B}$$.
(intermezzo: why is it a cover: let $$z \in Z$$. Then for some $$U_z \in \mathcal{U}$$, $$z \in U_z$$, because $$\mathcal{U}$$ is a cover. As $$U_z$$ is open and $$\mathfrak{B}$$ is a base, we know there is some $$B_z \in \mathfrak{B}$$ such that $$z \in B_z \subseteq U_z$$. But note that we've just shown that $$B_z \in \mathcal{B}$$ and it covers $$z$$..)
So by assumption, finitely many $$B_1, B_2, \ldots, B_n$$ from $$\mathcal{B}$$ cover $$Z$$ too and for each of these finitely many (so no axiom of choice needed) $$B_i$$ we pick a $$U_i \in \mathcal{U}$$ such that $$B_i \subseteq U_i$$, which is possible by definition of $$\mathcal{B}$$. Then certainly the $$U_1, U_2, \ldots U_n$$ form a finite subcover of $$\mathcal{U}$$. So $$Z$$ is compact as we started with an arbitary open cover and found a finite subcover.