Characterization of Closed Ball in P-adic Numbers

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Theorem

Let $p$ be a prime number.

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

Let $\Z_p$ be the $p$-adic integers.

For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {{B_\epsilon}^-} a$ denote the closed ball of center $a$ of radius $\epsilon$.


Let $x, y \in \Q_p$.

Let $n \in \Z$.


The following statements are equivalent:

$(1)\quad x \in \map {B^{\,-}_{p^{-n}}} y$
$(2)\quad \norm{x -y}_p \le p^{-n}$
$(3)\quad \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$
$(4)\quad x - y \in p^n \Z_p$
$(5)\quad x + p^n \Z_p = y + p^n \Z_p$


Proof

From P-adic Numbers form Non-Archimedean Valued Field:

$\norm {\,\cdot\,}_p$ is a non-Archimedean norm.


Condition $(1)$ iff Condition $(2)$

This follows directly from the definition of a closed ball in the $p$-adic numbers.

$\Box$


Condition $(1)$ iff Condition $(3)$

By definition, $\map {B^{\,-}_{p^{-n}}} y$ is a closed ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$.

From Centers of Closed Balls in Non-Archimedean Division Rings:

$x \in \map {B^{\,-}_{p^{-n}}} y \leadsto \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$

From Center is Element of Closed Ball in P-adic Numbers:

$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y \leadsto x \in \map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$

$\Box$


Condition $(2)$ iff Condition $(4)$

\(\ds \norm{x - y}_p \le p^{-n}\) \(\leadstoandfrom\) \(\ds \norm{x - y}_p \le \norm{p^n}_p\) Definition of $p$-adic norm on integers
\(\ds \) \(\leadstoandfrom\) \(\ds \dfrac {\norm{x - y}_p} {\norm{p^n}_p} \le 1\) Dividing both sides of equation by $p^{-n}$
\(\ds \) \(\leadstoandfrom\) \(\ds \norm{p^{-n} \paren{x-y} }_p \le 1\) Norm of Quotient in Division Ring
\(\ds \) \(\leadstoandfrom\) \(\ds p^{-n} \paren{x-y} \in \Z_p\) Definition of $p$-adic integers
\(\ds \) \(\leadstoandfrom\) \(\ds x-y \in p^n\Z_p\)

$\Box$


Condition $(3)$ iff Condition $(5)$

From Closed Balls of P-adic Number,

$\map {B^{\,-}_{p^{-n}}} x = x + p^n \Z_p$

and

$\map {B^{\,-}_{p^{-n}}} y = y + p^n \Z_p$

Hence:

$\map {B^{\,-}_{p^{-n}}} x = \map {B^{\,-}_{p^{-n}}} y$ if and only if $x + p^n \Z_p = y + p^n \Z_p$

$\blacksquare$


Sources

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