Countable Basis for P-adic Numbers

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.

Then:
 * $\mathcal B_p = \set{ \map q : q \in \Q, n \in \Z}$

is a countable basis for $\struct{\Q_p, \tau_p}$.

Proof
From Sequence of Powers of Number less than One, $\sequence{p^{-n}}$ is a real null sequence.

From Null Sequence induces Local Basis in Metric Space, for all $a \in \Q_p$ the set $\set { \map {B_{p^{-n}}} a : n \in \Z}$ is a local basis of $a$.

From Union of Local Bases is Basis, the set:
 * $\mathcal B’ = \displaystyle \bigcup_{a \in \Q_p} \set { \map {B_{p^{-n}}} a : n \in \Z} = \set { \map {B_{p^{-n}}} a : a \in Q_p, n \in \Z}$

is a basis for $\tau_p$.

Let $a \in \Q_p$ and $n \in \Z$.

By definition of the $p$-adic numbers, $\Q$ is everywhere dense in $\Q_p$.

Hence:
 * $\exists q \in \Q: q \in \map {B_{p^{-n}}} a$

From Characterization of Open Ball in P-adic Numbers,
 * $\map {B_{p^{-n}}} q = \map {B_{p^{-n}}} a$

Hence:
 * $\map {B_{p^{-n}}} a \in \mathcal B_p$

Since $q$ and $n$ were arbitrary, then:
 * $\mathcal B’ \subseteq \mathcal B_p$

It follows that $\mathcal B_p$ is a basis for $\tau_p$.

It remains to show that $\mathcal B_p$ is countable.

First note that $\mathcal B_p$ is a union of countable sets:
 * $\mathcal B_p = \displaystyle \bigcup_{q \in \Q} \set { \map {B_{p^{-n}}} q : n \in \Z}$