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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?

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I think the following proof might be more helpfully formulated (not in terms of "size", or "more than"):

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.

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