Neighborhood Space is Topological Space

Theorem
Let $\struct {S, \NN}$ be a neighborhood space.

Let $\tau = \set {N: N \in \NN}$ be the set of all open sets of $\struct {S, \NN}$.

Then $\struct {S, \tau}$ forms a topological space.

Proof
Each of the open set axioms is examined in turn:

$\text O 1$: Union of Open Sets
From Union of Open Sets of Neighborhood Space is Open, it follows that open set axiom $\text O 1$ is fulfilled.

$\text O 2$: Intersection of Open Sets
From Intersection of two Open Sets of Neighborhood Space is Open, it follows that open set axiom $\text O 2$ is fulfilled.

$\text O 3$: Underlying Set
From Whole Space is Open in Neighborhood Space, it follows that open set axiom $\text O 3$ is fulfilled.

All the open set axioms are fulfilled, and the result follows.