Neighborhood Space induced by Topological Space induced by Neighborhood Space

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

Let $\struct {S, \tau}$ be the topological space induced by $\NN$ on $S$.

Let $\struct {S, \NN'}$ be the neighborhood space induced by $\tau$ on $S$.

Then $\NN = \NN'$.

Proof
Let $x \in S$.

Let $\NN_x$ be the set of all neighborhoods of $x$.

Let $N \in \NN_x$ 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 $\struct {S, \NN}$.

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

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

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

where $\NN'_x$ is the set of all neighborhoods of $x$ in $\struct {S, \NN'}$.

Thus $\NN_x \subseteq \NN'_x$.

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

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

Since $U \in \tau$ it follows that $U$ is an open setin $\struct {S, \NN}$.

So by Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is a neighborhood of $x$ in $\struct {S, \NN}$.

That is:
 * $N \in \NN_x$

Thus it follows that $\NN_x \subseteq \NN'_x$

It follows by definition of set equality that $\NN_x = \NN'_x$.

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

Hence the result.