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.