Image of Mapping of Intersections is Smallest Basis

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 image $\operatorname{Im} \left({f}\right)$ is subset of every basis of $T$.

Proof
Let $\mathcal B$ be a basis.

Let $V \in \operatorname{Im} \left({f}\right)$.

Then by definition of image there exists a point $b \in X$ such that
 * $V = f \left({b}\right)$.

Then $V$ is open because $\operatorname{Im} \left({f}\right) \subseteq \tau$.

By assumption of mapping $f$:
 * $b \in V$.

Then by definition of basis there exists a subset $U \in \mathcal B$ such that
 * $b \in U \subseteq V$.

By definition of basis:
 * $\mathcal B \subseteq \tau$.

Then by definition of subset:
 * $U \in \tau$.

Then by assumption of mapping $f$:
 * $f \left({b}\right) \subseteq U$.

Thus by definition of set equality:
 * $V = U \in \mathcal B$.