Superset of Neighborhood in Metric Space is Neighborhood

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $a \in A$ be a point in $M$. Let $N$ be a neighborhood of $a$ in $M$.

Let $N \subseteq N' \subseteq A$.

Then $N'$ is a neighborhood of $a$ in $M$.

Proof
By definition of neighborhood:


 * $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq N$

where $B_\epsilon \left({a}\right)$ is the open $\epsilon$-ball of $a$ in $M$.

By Subset Relation is Transitive:
 * $B_\epsilon \left({a}\right) \subseteq N'$

The result follows by definition of neighborhood of $a$.