Neighborhoods in Standard Discrete Metric Space

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Theorem

Let $M = \struct {A, d}$ be a metric space where $d$ is the standard discrete metric.

Let $a \in A$.


Then $\set a$ is a neighborhood of $a$ which forms a basis for the system of neighborhoods of $a$.


Proof

By definition of the standard discrete metric:

$\map d {x, y} = \begin {cases}

0 & : x = y \\ 1 & : x \ne y \end {cases}$

Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.

Then:

\(\ds \map {B_\epsilon} a\) \(=\) \(\ds \set {x \in A: \map d {x, a} < \epsilon}\) Definition of Open $\epsilon$-Ball of $a$
\(\ds \leadsto \ \ \) \(\ds \map {B_\epsilon} a\) \(=\) \(\ds \set a\)

So by definition $\set a$ is a neighborhood of $a$.


Let $\NN_a$ be the system of neighborhoods of $a$.

Let $N \in \NN_a$ be a neighborhood of $a$.

Then:

\(\ds a\) \(\in\) \(\ds N\) Point in Metric Space is Element of its Neighborhood
\(\ds \leadsto \ \ \) \(\ds \set a\) \(\subseteq\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds \map {B_\epsilon} a\) \(\subseteq\) \(\ds N\) for all $\epsilon \in \R_{>0}$ such that $\epsilon < 1$

Hence the result by definition of basis of a system of neighborhoods.

$\blacksquare$


Sources

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