Synthetic Basis and Analytic Basis are Compatible

Theorem
Let $\left({S, \tau}\right)$ be a topological space.

Then $\mathcal B$ is an analytic basis for $\tau$ $\tau$ is the topology on $S$ generated by the synthetic basis $\mathcal B$.

Necessary Condition
Suppose that $\mathcal B$ is an analytic basis for $\tau$.

We proceed to check the axioms for $\mathcal B$ to be a synthetic basis on $S$.

Open set axiom $(O3)$ states that $S \in \tau$.

Therefore, by the definition of an analytic basis:
 * $\displaystyle \exists \mathcal S \subseteq \mathcal B: S = \bigcup \mathcal S$

By Equivalent Conditions for Cover by Collection of Subsets, $\mathcal B$ is a cover for $S$.

That is, axiom $(B1)$ for a synthetic basis is satisfied by $\mathcal B$.

Suppose that $A, B \in \mathcal B$.

By the definition of an analytic basis, $\mathcal B \subseteq \tau$.

Therefore $A, B \in \tau$.

Open set axiom $(O2)$ states that $A \cap B \in \tau$.

Therefore, by the definition of an analytic basis:
 * $\displaystyle \exists \mathcal A \subseteq \mathcal B: A \cap B = \bigcup \mathcal A$

That is, axiom $(B2)$ for a synthetic basis is satisfied by $\mathcal B$.

Therefore, $\mathcal B$ is a synthetic basis on $S$.

Let $\tau'$ be the topology on $S$ generated by the synthetic basis $\mathcal B$.

It follows from the definition of an analytic basis that $\tau \subseteq \tau'$.

Since the subset relation is transitive, we can apply open set axiom $(O1)$ for a topology to conclude that:
 * $\displaystyle \forall U \in \tau': \exists \mathcal A \subseteq \mathcal B \subseteq \tau: U = \bigcup \mathcal A \in \tau$

That is, $\tau' \subseteq \tau$.

By definition of set equality:
 * $\tau = \tau'$

Sufficient Condition
Suppose that $\mathcal B$ is a synthetic basis on $S$, and that $\tau$ is the topology on $S$ generated by $\mathcal B$.

Then:
 * $\displaystyle \mathcal B = \left\{{\bigcup \left\{{B}\right\}: \left\{{B}\right\} \subseteq \mathcal B}\right\} \subseteq \tau$

By definition, $\mathcal B$ is an analytic basis for $\tau$.