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 $\mathcal B$.