Finite Weight Space has Basis equal to Image of Mapping of Intersections

Theorem
Let $T = \left({X, \tau}\right)$ be a topological space with finite weight.

Then there exist a basis $\mathcal B$ of $T$ and a mapping $f:X \to \tau$ such that
 * $\mathcal B = \operatorname{Im} \left({f}\right)$ and
 * $\forall x \in X: \left({x \in f \left({x}\right) \land \forall U \in \tau: x \in U \implies f \left({x}\right) \subseteq U}\right)$.

where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$.