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 $\text N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.
By neighborhood space axiom $\text N 2$, $x \in U$.
$\Box$
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 $\text N 3$, $N$ is then a neighborhood of $x$.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.7$