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 $f$.

Proof
By definition of weight there exists a basis $\mathcal B$ such that
 * $\left\vert {\mathcal B} \right\vert = w \left({T}\right)$

where
 * $w \left({T}\right)$ denotes the weight of $T$,
 * $\left\vert {\mathcal B} \right\vert$ denotes the cardinality of $\mathcal B$.

By assumption that weight is finite:
 * $\left\vert {\mathcal B} \right\vert$ is finite.

Then by Cardinality of Set is Finite iff Set is Finite:
 * $\mathcal B$ is finite.

Define a mapping $f: X \to \mathcal P \left({X}\right)$
 * $(1): \quad \forall x \in X: f \left({x}\right) = \bigcap \left\{{U \in \mathcal B: x \in U}\right\}$.

By definition of subset:
 * $\forall x \in X: \left\{{U \in \mathcal B: x \in U}\right\} \subseteq \mathcal B$.

By Subset of Finite Set is Finite
 * $\forall x \in X: \left\{{U \in \mathcal B: x \in U}\right\}$ is finite.

Then by General Intersection Property of Topological Space
 * $\forall x \in X: \bigcap \left\{{U \in \mathcal B: x \in U}\right\} \in \tau$.

So:
 * $f: X \to \tau$.

We will prove that
 * $(2): \quad \forall x \in X: \left({x \in f \left({x}\right) \land \forall U \in \tau: x \in U \implies f \left({x}\right) \subseteq U}\right)$.

Let $x \in X$.

By $(1)$:
 * $f \left({x}\right) = \bigcap \left\{{U \in \mathcal B: x \in U}\right\}$.

Thus by definition of intersection:
 * $x \in f \left({x}\right)$.

Let $U$ be an open subset of $X$.

Let $x \in U$.

By definition of basis
 * $\exists V \in \mathcal B: x \in V \subseteq U$.

Then
 * $V \in \left\{{U \in \mathcal B: x \in U}\right\}$.

Hence by Intersection is Subset:
 * $f \left({x}\right) \subseteq V$.

Thus by Subset Relation is Transitive
 * $f \left({x}\right) \subseteq U$.

This ends the proof of $(2)$.

We will prove that $\operatorname{Im} \left({f}\right)$ is a basis of $T$.

By $f: X \to \tau$ and definition of image
 * $\operatorname{Im} \left({f}\right) \subseteq \tau$.

Let $U$ be an open subset of $X$.

Let $x$ be a point $x \in X$ such that
 * $x \in U$.

By $(2)$:
 * $f \left({x}\right) \in \operatorname{Im} \left({f}\right) \land x \in f \left({x}\right) \subseteq U$.

By definition of basis this ends the proof of basis.

Thus the result.