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 $$\mathcal{B}$$.