# Open and Closed Balls in P-adic Numbers are Totally Bounded

## Theorem

Let $p$ be a prime number.

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

Let $a \in \Q_p$.

Let $n \in \Z$.

Then the open ball $\map {B_{p^{-n} } } a$ and closed ball $\map {B^-_{p^{-n} } } a$ are totally bounded subspaces.

## Proof

We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$.

From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.

Let $d$ denote the subspace metric induced by the norm $\norm {\,\cdot\,}_p$ on $\map {B^-_{p^{-n} } } a$.

That is, $d: \map {B^-_{p^{-n} } } a \times \map {B^-_{p^{-n} } } a \to \R_{>0}$ is the metric defined by:

$\forall x, y \in \map {B^-_{p^{-n} } } a: \map d {x,y} = \norm {x - y}_p$

By the definition of totally bounded it needs to be shown that:

for every $\epsilon \in \R_{>0}$ there exist finitely many points $x_0, \dots, x_k \in \map {B^-_{p^{-n} } } a$ such that:
$\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {x_i - x}_p \le \epsilon$
for all $x \in \map {B^-_{p^{-n} } } a$.
$\exists N \in \N: \forall k > N: p^{-k} \le \epsilon$.

Let $m = \max \set {n + 1, N}$.

Then $m > n$.

$\map {B^-_{p^{-n} } } a = \displaystyle \bigcup_{i \mathop = 0}^{p^\paren {m - n} - 1} \map {B^-_{p^{-m} } } {a + i p^n}$

Let $x \in \map {B^-_{p^{-n} } } a$.

Let $k = p^\paren {m - n} - 1$.

Then there exists $0 \le j \le k$ such that:

$x \in \map {B^-_{p^{-m} } } {a + j p^n}$

That is:

$\norm {x - a + j p^n}_p \le p^{-m} \le \epsilon$

Hence:

$\displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \norm{x - a + j p^n}_p \le \epsilon$

Since $x \in \map {B^-_{p^{-n} } } a$ was arbitrary, then:

$\forall x \in \map {B^-_{p^{-n} } } a : \displaystyle \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \epsilon$

The result follows.

$\blacksquare$