Synthetic Basis formed from Synthetic Sub-Basis

Theorem
Let $$A$$ be a set.

Let $$\mathcal S$$ be a synthetic sub-basis for $$A$$.

Then the collection of all finite intersections of sets from $$\left\{{A}\right\} \cup \mathcal S$$ forms a synthetic basis for $$A$$.

Proof
From the definition, $$\mathcal S \subseteq \mathcal P \left({A}\right)$$, where $$\mathcal P \left({A}\right)$$ is the power set of $$A$$.

Let $$\mathcal S = \left\{{S_1, S_2, \ldots}\right\}$$.

Let $$\mathcal B$$ be the collection of all finite intersections of sets from $$\left\{{A}\right\} \cup \mathcal S$$.

It follows that each of $$A, S_1, S_2, \ldots$$ are themselves finite intersections of sets from $$\left\{{A}\right\} \cup \mathcal S$$, as they are intersections of themselves with themselves (or with $$A$$, if you like).

Thus $$\forall S \in \left\{{A}\right\} \cup \mathcal S: S \in \mathcal B$$.

We need to show that $$\mathcal B$$ is a synthetic basis for $$A$$.


 * B1: As $$A \in \mathcal B$$, it follows trivially that $$A$$ is the union of sets from $$\mathcal B$$.
 * B2: All elements of $$\mathcal B$$ are formed as the intersection of a finite number of sets from $$\mathcal S$$. Thus the intersection of any two of these is bound to be another element of $$\mathcal B$$.