Topology Defined by Neighborhood System

Theorem
Let $S$ be a set.

Let $\left({\mathcal N_x}\right)_{x \mathop \in S}$ a indexed family where $\mathcal N_x$ is non-empty set of subsets of $S$.

Assume that
 * $(N1): \quad \forall x \in S, U \in \mathcal N_x: x \in U$
 * $(N2): \quad \forall x \in S, U \in \mathcal N_x, y \in U:\exists V \in \mathcal N_y: V \subseteq U$
 * $(N3): \quad \forall x \in S, U_1, U_2 \in \mathcal N_x: \exists U \in \mathcal N_x: U \subseteq U_1 \cap U_2$

Then $T = \left({S, \tau}\right)$ is a topological space where
 * $\tau = \displaystyle \left\{{\bigcup \mathcal G: \mathcal G \subseteq \bigcup_{x \mathop \in S} \mathcal N_x}\right\}$

Moreover, $\left({\mathcal N_x}\right)_{x \mathop \in S}$ is a neighborhood system of $T$.

Proof
Define $\mathcal B := \displaystyle \bigcup_{x \mathop \in S} \mathcal N_x$

According to Topology Defined by Basis it should be proved that
 * $(B1): \quad \forall A_1, A_2 \in \mathcal B: \forall x \in A_1 \cap A_2: \exists A \in \mathcal B: x \in A \subseteq A_1 \cap A_2$
 * $(B2): \quad \forall x \in S: \exists A \in \mathcal B: x \in A$

Ad. $(B1)$:

Let $A_1, A_2 \in \mathcal B$.

Let $x \in A_1 \cap A_2$.

By definition of intersection:
 * $x \in A_1 \land x \in A_2$

By definition of union:
 * $\exists x_1 \in S: A_1 \in \mathcal N_{x_1}$

and
 * $\exists x_2 \in S: A_2 \in \mathcal N_{x_2}$

By $(N2)$:
 * $\exists V_1 \in \mathcal N_x: V_1 \subseteq A_1$

and
 * $\exists V_2 \in \mathcal N_x: V_2 \subseteq A_2$

By $(N3)$:
 * $\exists U \in \mathcal N_x: U \subseteq V_1 \cap V_2$

By definition of union:
 * $U \in \mathcal B$

By Set Intersection Preserves Subsets:
 * $V_1 \cap V_2 \subseteq A_1 \cap A_2$

Thus by Subset Relation is Transitive:
 * $U \subseteq A_1 \cap A_2$

Ad $(B2)$:

Let $x \in S$.

By definition of empty set:
 * $\exists U: U \in \mathcal N_x$

By $(N1)$:
 * $x \in U$

By definition of union:
 * $U \in \mathcal B$

Thus:
 * $\exists A \in \mathcal B: x \in A$

By definition of $\tau$ and $\mathcal B$:
 * $\tau = \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}$

Thus by Topology Defined by Basis:
 * $T = \left({S,\tau}\right)$ is a topological space

It remains to prove that
 * $\left({\mathcal N_x}\right)_{x \mathop \in S}$ is a neighborhood system

Let $x \in S$.

Let $U \in \tau$.

By definition of $\tau$:
 * $\exists \mathcal G \subseteq \mathcal B: U = \bigcup \mathcal G$

Let $x \in U$.

By definition of union:
 * $\exists V \in \mathcal G: x \in V$

By definition of subset:
 * $V \in \mathcal B$

By definition of union:
 * $\exists y \in S: V \in \mathcal N_y$

By $(N2)$:
 * $\exists W \in \mathcal N_x: W \subseteq V$

By Set is Subset of Union/Set of Sets:
 * $V \subseteq U$

Thus by Subset Relation is Transitive:
 * $\exists W \in \mathcal N_x: W \subseteq U$

Thus by definition
 * $\mathcal N_x$ os a local basis.