# Topology Defined by Neighborhood System

## Theorem

Let $S$ be a set.

Let $\family {\NN_x}_{x \mathop \in S}$ be an indexed family where $\NN_x$ is non-empty set of subsets of $S$.

Assume that

$(\text N 1): \quad \forall x \in S, U \in \NN_x: x \in U$
$(\text N 2): \quad \forall x \in S, U \in \NN_x, y \in U:\exists V \in \NN_y: V \subseteq U$
$(\text N 3): \quad \forall x \in S, U_1, U_2 \in \NN_x: \exists U \in \NN_x: U \subseteq U_1 \cap U_2$

Then $T = \struct {S, \tau}$ is a topological space where:

$\tau = \displaystyle \set {\bigcup \GG: \GG \subseteq \bigcup_{x \mathop \in S} \NN_x}$

Moreover, $\family {\NN_x}_{x \mathop \in S}$ is a neighborhood system of $T$.

## Proof

Define:

$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$

According to Topology Defined by Basis it should be proved that

$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$
$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$

Ad $(\text B 1)$:

Let $A_1, A_2 \in \BB$.

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:

$\exists x_1 \in S: A_1 \in \NN_{x_1}$

and

$\exists x_2 \in S: A_2 \in \NN_{x_2}$

By $(\text N 2)$:

$\exists V_1 \in \NN_x: V_1 \subseteq A_1$

and

$\exists V_2 \in \NN_x: V_2 \subseteq A_2$

By $(\text N 3)$:

$\exists U \in \NN_x: U \subseteq V_1 \cap V_2$

By definition of union:

$U \in \BB$
$V_1 \cap V_2 \subseteq A_1 \cap A_2$

Thus by Subset Relation is Transitive:

$U \subseteq A_1 \cap A_2$

Ad $(\text B 2)$:

Let $x \in S$.

By definition of empty set:

$\exists U: U \in \NN_x$

By $(\text N 1)$:

$x \in U$

By definition of union:

$U \in \BB$

Thus:

$\exists A \in \BB: x \in A$

By definition of $\tau$ and $\BB$:

$\tau = \set {\bigcup \GG: \GG \subseteq \BB}$

Thus by Topology Defined by Basis:

$T = \struct {S, \tau}$ is a topological space

It remains to prove that

$\family {\NN_x}_{x \mathop \in S}$ is a neighborhood system

Let $x \in S$.

Let $U \in \tau$.

By definition of $\tau$:

$\exists \GG \subseteq \BB: U = \bigcup \GG$

Let $x \in U$.

By definition of union:

$\exists V \in \GG: x \in V$

By definition of subset:

$V \in \BB$

By definition of union:

$\exists y \in S: V \in \NN_y$

By $(\text N 2)$:

$\exists W \in \NN_x: W \subseteq V$
$V \subseteq U$

Thus by Subset Relation is Transitive:

$\exists W \in \NN_x: W \subseteq U$

Thus by definition:

$\NN_x$ is a local basis.

$\blacksquare$