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 $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a subset of the power set of $X$.

Then $\mathcal G$ is a synthetic sub-basis on $X$.

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

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

It is claimed that:
 * $\left({1}\right): \quad$ $\mathcal G \subseteq \tau \left({\mathcal G}\right)$.
 * $\left({2}\right): \quad$ For any topology $\mathcal T$ on $X$, the implication $\mathcal G \subseteq \mathcal T \implies \tau \left({\mathcal G}\right) \subseteq \mathcal T$ holds.

Since $\mathcal G \subseteq \mathcal B \subseteq \tau \left({\mathcal G}\right)$, it follows that $\mathcal G \subseteq \tau \left({\mathcal G}\right)$ since the subset relation is transitive.

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

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 $\tau \left({\mathcal G}\right) \subseteq \mathcal T$.