Finite Intersection of Open Sets of Neighborhood Space is Open

Corollary to Intersection of two Open Sets of Neighborhood Space is Open
Let $\left({S, \mathcal N}\right)$ be a neighborhood space.

Let $n \in \N_{>0}$ be a natural number.

Let $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\left({S, \mathcal N}\right)$.

Then $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$.

Proof
Proof by induction:

Let $U_1, U_2, \ldots$ be open sets of $\left({S, \mathcal N}\right)$.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$.

$P \left({1}\right)$ is true, as this just says:
 * $U_1$ is an open set of $\left({S, \mathcal N}\right)$.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $U_1 \cap U_2$ is an open set of $\left({S, \mathcal N}\right)$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$.

Then we need to show:
 * $\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i$ is an open set of $\left({S, \mathcal N}\right)$.

Induction Step
This is our induction step:

We have that:
 * $\displaystyle \bigcap_{i \mathop = 1}^{k+1} U_i = \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$

From the induction hypothesis:
 * $\displaystyle \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\left({S, \mathcal N}\right)$.

From the basis for the induction:
 * $\displaystyle \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ is an open set of $\left({S, \mathcal N}\right)$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $ \displaystyle\forall n \in \N_{>0}: \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\left({S, \mathcal N}\right)$.