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 $4.8: \ N \, 5$