Intersection of Neighborhoods in Metric Space is Neighborhood

Theorem

Let $M = \struct {A, d}$ 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:

$\exists \epsilon_1 \in \R_{>0}: \map {B_{\epsilon_1} } a \subseteq N$

where $\map {B_{\epsilon_1} } a$ is the open $\epsilon_1$-ball of $a$ in $M$.

$\exists \epsilon_2 \in \R_{>0}: \map {B_{\epsilon_2} } a \subseteq N'$

where $\map {B_{\epsilon_2} } a$ is the open $\epsilon_2$-ball of $a$ in $M$.

Thus by definition of intersection:

$\map {B_\epsilon} a \subseteq N \cap N'$

where $\epsilon = \min \set {\epsilon_1, \epsilon_2}$

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

$\blacksquare$

Also see

• Intersection of Neighborhoods in Topological Space is Neighborhood

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 4$