Point in Metric Space has Countable Neighborhood Basis

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $a \in A$.


Then there exists a basis for the neighborhood system of $a$ which is countable.


Proof

Consider the countable set:

$\BB_a := \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$

That is, let $\BB_a$ be the set of all open $\epsilon$-balls of $a$ such that $\epsilon$ is of the form $\dfrac 1 n$ for (strictly) positive integral $n$.

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

Then by definition of neighborhood:

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

From Between two Real Numbers exists Rational Number:

$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$

Let $\epsilon$ be expressed in canonical form as:

$\epsilon = \dfrac p q$

where $p$ and $q$ are coprime integers and $q > 0$.

Then:

$\epsilon := \dfrac 1 q \le \dfrac p q$

Thus $0 < \epsilon < \epsilon'$.

So:

\(\ds x\) \(\in\) \(\ds \map {B_\epsilon} a\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, a}\) \(<\) \(\ds \epsilon\) Definition of Open Ball of Metric Space
\(\ds \leadsto \ \ \) \(\ds \map d {x, a}\) \(<\) \(\ds \epsilon'\) as $\epsilon < \epsilon'$
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \map {B_{\epsilon'} } a\) Definition of Open Ball of Metric Space
\(\ds \leadsto \ \ \) \(\ds \map {B_\epsilon} a\) \(\subseteq\) \(\ds \map {B_{\epsilon'} } a\) Definition of Subset
\(\ds \leadsto \ \ \) \(\ds \map {B_\epsilon} a\) \(\subseteq\) \(\ds N\) Subset Relation is Transitive

From Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} a$ is a neighborhood of $a$.

But as $\epsilon = \dfrac 1 q$ where $q \in \Z_{>0}$ it follows that:

$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}$

Hence the result.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Point_in_Metric_Space_has_Countable_Neighborhood_Basis&oldid=507257"