Closed Balls Centered on P-adic Number is Countable/Open Balls

From ProofWiki
Jump to navigation Jump to search

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:

$\mathcal B = \set{\map {B_{p^{-n}}} a : n \in \Z}$

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 $\epsilon \in \R_{\gt 0}$ was arbitrary then:

$\forall \epsilon \in \R_{\gt 0} : \exists n \in \Z : \map {B_\epsilon} a = \map {B_{p^{-n} } } a$

Hence the set of all open balls centered on $a$ is:

$\mathcal B = \set{\map {B_{p^{-n}}} a : n \in \Z}$

From Surjection from Countably Infinite Set iff Countable, it follows that $\mathcal B$ is a countable set.

$\blacksquare$