Topological Space induced by Neighborhood Space induced by Topological Space

Theorem
Let $S$ be a set.

Let $\tau$ be a topology on $S$, thus forming the topological space $\left({S, \tau}\right)$.

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

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

Then $\tau = \tau'$.

Proof
Let $U \in \tau$ be an open set of $\left({S, \tau}\right)$.

By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.

By definition, $U$ is an open set of $\left({S, \mathcal N}\right)$.

Thus by definition of neighborhood space induced by $\tau$ on $S$:
 * $U \in \mathcal N$

Then, by definition of the topological space induced by $\mathcal N$ on $S$:
 * $U \in \tau'$.

Thus $\tau \subseteq \tau'$.

Now let $U \in \tau'$ be an open set of $\left({S, \tau'}\right)$.

Then, by definition, in the neighborhood space $\left({S, \mathcal N}\right)$, $U$ is an open set of $\left({S, \mathcal N}\right)$.

That is, $U$ is a neighborhood in $\left({S, \mathcal N}\right)$ of each of its points.

But the neighborhoods of $\left({S, \mathcal N}\right)$ are the neighborhoods of $\left({S, \tau}\right)$.

Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\left({S, \tau}\right)$.

That is:
 * $U \in \tau$.

It follows by definition of set equality that $\tau = \tau'$.