Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis

Theorem
The definitions of the generated topology are equivalent.

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 $\mathcal B$.

It is claimed 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.

Since $\mathcal S \subseteq \mathcal B \subseteq \tau \left({\mathcal S}\right)$, it follows that $\mathcal S \subseteq \tau \left({\mathcal S}\right)$ because $\subseteq$ is a transitive relation.

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

By axiom $\left({3}\right)$ for a topology and from General Intersection Property of Topological Space, it follows that $\mathcal B \subseteq \mathcal T$.

By axiom $\left({1}\right)$ for a topology, it follows that $\tau \left({\mathcal S}\right) \subseteq \mathcal T$.

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