# Intersection of two Open Sets of Neighborhood Space is Open

## Theorem

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

Let $U$ and $V$ be open sets of $\struct {S, \NN}$.

Then $U \cap V$ is an open set of $\struct {S, \NN}$.

### Corollary

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

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

Let $\ds \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\struct {S, \NN}$.

Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\struct {S, \NN}$.

## Proof

Let $U$ and $V$ be open sets of $\struct {S, \NN}$.

Let $x \in S$ such that $x \in U \cap V$.

Then $x \in U$ and $x \in V$, both of which are neighborhoods of $x$ by definition of open set.

By neighborhood space axiom $(\text N 4)$ it follows that $U \cap V$ is a neighborhood of $x$.

As $x$ is arbitrary, it follows that the above is true for all $x \in U \cap V$.

It follows by definition that $U \cap V$ is an open set of $\struct {S, \NN}$.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): $\S 3.3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$