Synthetic Basis formed from Synthetic Sub-Basis

Theorem
Let $X$ be a set.

Let $\mathcal S$ be a synthetic sub-basis on $X$.

Define:
 * $\displaystyle \mathcal B = \left\{{\bigcap \mathcal A: \varnothing \neq \mathcal A \subseteq \left\{{X}\right\} \cup \mathcal S, \, \mathcal A \text{ is finite}}\right\}$

Then $\mathcal B$ is a synthetic basis on $X$.

Proof
Clearly, $\mathcal B \subseteq \mathcal P \left({X}\right)$.

Since $X \in \mathcal B$, it trivially follows that axiom B1 for a synthetic basis is satisfied.

Let $B_1, B_2 \in \mathcal B$. Then there exist non-empty and finite $\mathcal A_1, \mathcal A_2 \subseteq \left\{{X}\right\} \cup \mathcal S$ such that:
 * $\displaystyle B_1 = \bigcap \mathcal A_1$
 * $\displaystyle B_2 = \bigcap \mathcal A_2$

It follows that:
 * $\displaystyle B_1 \cap B_2 = \bigcap \left({\mathcal A_1 \cup \mathcal A_2}\right)$

Clearly, $\mathcal A_1 \cup \mathcal A_2$ is non-empty.

By Union Smallest, $\mathcal A_1 \cup \mathcal A_2 \subseteq \left\{{X}\right\} \cup \mathcal S$.

We have that $\mathcal A_1 \cup \mathcal A_2$ is finite.

Hence $B_1 \cap B_2 \in \mathcal B$, so it trivially follows that axiom B2 for a synthetic basis is satisfied.