Intersection of Neighborhoods 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 $N, N'$ be neighborhoods of $a$ in $M$.

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

Proof
By definition of neighborhood:


 * $\exists \epsilon_1 \in \R_{>0}: B_{\epsilon_1} \left({a}\right) \subseteq N$

where $B_{\epsilon_1} \left({a}\right)$ is the open $\epsilon_1$-ball of $a$ in $M$.


 * $\exists \epsilon_2 \in \R_{>0}: B_{\epsilon_2} \left({a}\right) \subseteq N'$

where $B_{\epsilon_2} \left({a}\right)$ is the open $\epsilon_2$-ball of $a$ in $M$.

Thus by definition of intersection:


 * $B_\epsilon \left({a}\right) \subseteq N \cap N'$

where $\epsilon = \min \left\{ {\epsilon_1, \epsilon_2}\right\}$

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