Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis

Proof
Let $X$ be a set.

Let $\SS \subseteq \powerset X$ be a synthetic sub-basis on $X$.

Let $\BB$ be the synthetic basis on $X$ generated by the synthetic sub-basis $\SS$.

Let $\map \tau \SS$ be the topology on $X$ generated by the synthetic basis $\BB$.

We now show that:
 * $(1): \quad$ $\SS \subseteq \map \tau \SS$
 * $(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds.

We have that:
 * $\SS \subseteq \BB \subseteq \map \tau \SS$

Hence by transitivity of $\subseteq$:
 * $\SS \subseteq \map \tau \SS$

Suppose that $\TT$ is a topology on $X$ such that $\SS \subseteq \TT$.

From General Intersection Property of Topological Space:
 * $\BB \subseteq \TT$

By :
 * $\map \tau \SS \subseteq \TT$

It follows that $\map \tau \SS$ is the unique topology on $X$ satisfying conditions $(1)$ and $(2)$.