Closed Balls Centered on P-adic Number is Countable/Open Balls
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Theorem
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Then the set of all open balls centered on $a$ is the countable set:
- $\BB = \set {\map {B_{p^{-n} } } a : n \in \Z}$
Proof
Let $\epsilon \in \R_{\ge 0}$.
Lemma
- $\exists n \in \Z : p^{-\paren {n + 1} } < \epsilon \le p^{-n}$
From Open Ball contains Smaller Open Ball:
- $\map {B_\epsilon} a \subseteq \map {B_{p^{-n} } } a$
From Open Ball contains Strictly Smaller Closed Ball:
- $\map {B^-_{p^{-\paren {n + 1} } } } a \subseteq \map {B_\epsilon} a$
From Open Ball in P-adic Numbers is Closed Ball
- $\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$
Hence:
- $\map {B_{p^{-n} } } a \subseteq \map {B_\epsilon} a$
By definition of set equality:
- $\map {B_\epsilon} a = \map {B_{p^{-n} } } a$
Since $\epsilon \in \R_{>0}$ was arbitrary then:
- $\forall \epsilon \in \R_{>0} : \exists n \in \Z : \map {B_\epsilon} a = \map {B_{p^{-n} } } a$
Hence the set of all open balls centered on $a$ is:
- $\BB = \set {\map {B_{p^{-n} } } a : n \in \Z}$
From Surjection from Countably Infinite Set iff Countable, it follows that $\BB$ is a countable set.
$\blacksquare$