Cardinality of Image of Mapping of Intersections is not greater than Weight of Space

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

Let $f:X \to \tau$ be a mapping such that
 * $\forall x \in X: \left({x \in f \left({x}\right) \land \forall U \in \tau: x \in U \implies f \left({x}\right) \subseteq U}\right)$.

Then the cardinality image $\left\vert{ \operatorname{Im} \left({f}\right) }\right\vert \leq w \left({T}\right)$.

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

By Image of Mapping of Intersections is Smallest Basis:
 * $\operatorname{Im} \left({f}\right) \subseteq \mathcal B$.

Thus by Subset implies Cardinal Inequality:
 * $\left\vert{ \operatorname{Im} \left({f}\right) }\right\vert \leq \left\vert{ \mathcal B }\right\vert = w \left({T}\right)$.