Neighborhood Space induced by Topological Space induced by Neighborhood Space

Theorem
Let $\left({S, \mathcal N}\right)$ be a neighborhood space.

Let $\left({S, \tau}\right)$ be the topological space induced by $\mathcal N$ on $S$.

Let $\left({S, \mathcal N'}\right)$ be the neighborhood space induced by $\tau$ on $S$.

Then $\mathcal N = \mathcal N'$.

Proof
Let $x \in S$.

Let $\mathcal N_x$ be the set of all neighborhoods of $x$.

Let $N \in \mathcal N_x$ be a neighborhood of $x$.

From Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is the superset of some open set $U$ in $\left({S, \mathcal N}\right)$.

By Neighborhood Space is Topological Space we have that $U$ is an open set of $\left({S, \tau}\right)$.

Thus, by definition, $N$ is a neighborhood of $x$ in the context of the topological space $\left({S, \tau}\right)$.

Thus by definition of the neighborhood space induced by $\tau$ on $S$, it follows that:
 * $N \in \mathcal N'_x$

where $\mathcal N'_x$ is the set of all neighborhoods of $x$ in $\left({S, \mathcal N'}\right)$.

Thus $\mathcal N_x \subseteq \mathcal N'_x$.

Now suppose that $N \in \mathcal N'_x$.

Then, by definition of neighborhood, $N$ is the superset of some open set $U$ in $\left({S, \tau}\right)$ such that $x \in U$.

Since $U \in \tau$ it follows that $U$ is an open setin $\left({S, \mathcal N}\right)$.

So by Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is a neighborhood of $x$ in $\left({S, \mathcal N}\right)$.

That is:
 * $N \in \mathcal N_x$

Thus it follows that $\mathcal N_x \subseteq \mathcal N'_x$

It follows by definition of set equality that $\mathcal N_x = \mathcal N'_x$.

As $x$ is arbitrary, it applies to all $x \in S$.

Hence the result.