Synthetic Basis and Analytic Basis are Compatible

Theorem
Let $\vartheta$ be the topology on a set $A$ arising from a (synthetic) basis $\mathcal B$ for $A$.

Then $\mathcal B$ is an analytic basis for $\vartheta$.

Similarly, let $T = \left({A, \vartheta}\right)$ be a topological space.

Let $\mathcal B$ be an analytic basis for $\vartheta$.

Then $\mathcal B$ is a synthetic basis for $A$.

Proof
Follows directly from the definitions for analytic basis and synthetic basis.