# Neighborhood in Metric Space has Subset Neighborhood

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## Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $a \in A$ be a point in $M$.

Let $N$ be a neighborhood of $a$ in $M$.

Then there exists a neighborhood $N'$ of $a$ such that:

- $(1): \quad N' \subseteq N$
- $(2): \quad N'$ is a neighborhood of each of its points.

## Proof

By definition of neighborhood:

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

By Open Ball is Neighborhood of all Points Inside, $N' = \map {B_\epsilon} a$ fulfils the conditions of the statement.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 5$