# Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1

## Theorem

Let $p$ be a prime number.

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

Let $n, m \in Z$, such that $n < m$.

Then:

$\forall y \in \Q_p: \norm{y}_p \le p^{-n}$ if and only if there exists $i \in \Z$:
$(1)\quad0 \le i \le p^\paren{m-n}-1$
$(2)\quad\norm{y - i p^n}_p \le p^{-m}$

## Proof

### Necessary Condition

Let $y \in \Q_p$.

Let $\norm{y}_p \le p^{-n}$.

Now:

 $\displaystyle \norm{y}_p$ $\le$ $\displaystyle p^{-n}$ $\, \displaystyle \leadsto \,$ $\displaystyle p^n \norm{y}_p$ $\le$ $\displaystyle 1$ Multiply both sides by $p^n$ $\, \displaystyle \leadsto \,$ $\displaystyle \norm{p^{-n} }_p \norm{y}_p$ $\le$ $\displaystyle 1$ Definition of $p$-adic norm $\, \displaystyle \leadsto \,$ $\displaystyle \norm{p^{-n} y}_p$ $\le$ $\displaystyle 1$ Norm axiom (N2) : (Mulitplicativity) $\, \displaystyle \leadsto \,$ $\displaystyle \map {B_1^-} {p^{-n}y}$ $=$ $\displaystyle \map {B_1^-} 0$ Characterization of Closed Ball in P-adic Numbers
$\exists \mathop k \in \Z : \norm{p^{-n} y - k}_p \le p^\paren{n-m}$
$\exists \mathop 0 \le i \le p^\paren{m-n}-1 : p^\paren{m-n} \divides k - i$

By definition of the $p$-adic norm: $\norm{k - i}_p \le p^\paren{n-m}$

It follows that:

 $\displaystyle \norm{p^{-n} y - i}_p$ $\le$ $\displaystyle \max \set{\norm{p^{-n} y - k}_p, \norm{i - k}_p}$ Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric $\displaystyle$ $\le$ $\displaystyle p^\paren{n-m}$ $\, \displaystyle \leadsto \,$ $\displaystyle \norm{p^{-n} }_p \norm{y - i p^n}_p$ $\le$ $\displaystyle p^\paren{n-m}$ Norm axiom (N2) : (Mulitplicativity) $\, \displaystyle \leadsto \,$ $\displaystyle p^n \norm{y - i p^n}_p$ $\le$ $\displaystyle p^\paren{n-m}$ Definition of $p$-adic norm $\, \displaystyle \leadsto \,$ $\displaystyle \norm{y - i p^n}_p$ $\le$ $\displaystyle p^{-m}$ Divide both sides by $p^{-n}$

$\Box$

### Sufficient Condition

Let $y\in \Q_p$.

Let there exist $i \in \Z$:

$(1)\quad0 \le i \le p^\paren{m-n}-1$
$(2)\quad\norm{y - i p^n}_p \le p^{-m}$

Now:

 $\displaystyle \norm{ y }_p$ $=$ $\displaystyle \norm{ y - i p^n + i p^n}_p$ $\displaystyle$ $\le$ $\displaystyle \max \set {\norm{ y - i p^n}_p, \norm{i p^n}_p}$ Norm axiom $(N4)$ : (Ultrametric Inequality)

By assumption:

$\norm{ y - i p^n} \le p^{-m} \le p^{-n}$

and:

 $\displaystyle \norm{i p^n}_p$ $=$ $\displaystyle \norm{i}_p \norm{p^n}_p$ Norm axiom $(N2)$ : (Multiplicativity) $\displaystyle$ $\le$ $\displaystyle 1 \cdot p^{-n}$ As $i \in \Z \subseteq \Z_p$ $\displaystyle$ $=$ $\displaystyle p^{-n}$

Hence:

$\max \set {\norm{ y - i p^n}_p, \norm{i p^n}_p} \le p^{-n}$.

So:

$\norm{ y }_p \le p^{-n}$.

$\blacksquare$