Synthetic Sub-Basis and Analytic Sub-Basis are Compatible

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

Let $\mathcal S \subseteq \tau$.

Then $\mathcal S$ is an analytic sub-basis for $\tau$ iff $\tau$ is the topology on $X$ generated by the synthetic sub-basis $\mathcal S$.

Necessary Condition
Follows directly from the definitions of the generated topology and an analytic sub-basis.

Sufficient Condition
Let $\tau'$ be the topology on $X$ generated by the synthetic sub-basis $\mathcal S$.

By the definitions of the generated topology and an analytic sub-basis, we have $\tau \subseteq \tau'$.

By the definition of the generated topology, it follows that $\tau' \subseteq \tau$.

By Equality of Sets, we have $\tau = \tau'$.