Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis

Proof
Let $X$ be a set.

Let $\mathcal S \subseteq \mathcal P \left({X}\right)$ be a synthetic sub-basis on $X$.

Let $\mathcal B$ be the synthetic basis on $X$ generated by the synthetic sub-basis $\mathcal S$.

Let $\tau \left({\mathcal S}\right)$ be the topology on $X$ generated by the synthetic basis $\mathcal B$.

We now show that:
 * $\left({1}\right): \quad$ $\mathcal S \subseteq \tau \left({\mathcal S}\right)$.
 * $\left({2}\right): \quad$ For any topology $\mathcal T$ on $X$, the implication $\mathcal S \subseteq \mathcal T \implies \tau \left({\mathcal S}\right) \subseteq \mathcal T$ holds.

We have that:
 * $\mathcal S \subseteq \mathcal B \subseteq \tau \left({\mathcal S}\right)$

Hence by transitivity of $\subseteq$:
 * $\mathcal S \subseteq \tau \left({\mathcal S}\right)$

Suppose that $\mathcal T$ is a topology on $X$ such that $\mathcal S \subseteq \mathcal T$.

From General Intersection Property of Topological Space:
 * $\mathcal B \subseteq \mathcal T$

By open set axiom $\left({1}\right)$:
 * $\tau \left({\mathcal S}\right) \subseteq \mathcal T$

It follows that $\tau \left({\mathcal S}\right)$ is the unique topology on $X$ satisfying conditions $\left({1}\right)$ and $\left({2}\right)$.