Synthetic Basis and Analytic Basis are Compatible

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

Then $\mathcal B$ is an analytic basis for $\vartheta$ iff $\vartheta$ is the topology on $X$ generated by the synthetic basis $\mathcal B$.

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

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

Axiom $\left({3}\right)$ for a topology states that $X \in \vartheta$.

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

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

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, $\mathcal B \subseteq \vartheta$; therefore, $A, B \in \vartheta$.

Axiom $\left({2}\right)$ for a topology states that $A \cap B \in \vartheta$.

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 $X$.

Let $\vartheta'$ be the topology on $X$ generated by the synthetic basis $\mathcal B$.

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

Since the subset relation is transitive, we can apply axiom $\left({1}\right)$ for a topology to conclude that:
 * $\displaystyle \forall U \in \vartheta': \exists \mathcal A \subseteq \mathcal B \subseteq \vartheta: U = \bigcup \mathcal A \in \vartheta$

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

Hence, by Equality of Sets, $\vartheta = \vartheta'$.

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

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

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