# Subset in Neighborhood Space is Neighborhood iff it contains Open Set

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## Theorem

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

Let $x \in S$ be a point of $S$.

Let $N \subseteq S$ be a subset of $S$.

Then $N$ is a neighborhood of $x$ if and only if there exists an open set $U$ of $\struct {S, \NN}$ such that $x \in U \subseteq N$.

## Proof

### Necessary Condition

Let $N$ be a neighborhood of $x$.

Then by neighborhood space axiom $N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.

By neighborhood space axiom $N 2$, $x \in U$.

$\Box$

### Sufficient Condition

Let $N$ be such that:

- $\exists U \in \NN: x \in U \subseteq N$

By definition of open set, $U$ is a neighborhood of $x$.

By neighborhood space axiom $N 3$, $N$ is then a neighborhood of $x$.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): $\S 3.3$: Neighborhoods and Neighborhood Spaces: Lemma $3.7$