Sphere is Disjoint Union of Open Balls 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.

Let $a \in \Q_p$.

For all $\epsilon \in \R_{>0}$:

let $\map {S_\epsilon} a$ denote the sphere of $a$ of radius $\epsilon$.
let $\map {B_\epsilon} a$ denote the open ball of $a$ of radius $\epsilon$.


Then:

$\ds \forall n \in Z: \map {S_{p^{-n} } } a = \bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n}$


Proof

For all $\epsilon \in \R_{>0}$:

let $\map {B^-_\epsilon} a$ denote the closed ball of $a$ of radius $\epsilon$.


Let $n \in \Z$.

Then:

\(\ds \map {S_{p^{-n} } } a\) \(=\) \(\ds \map {B^-_{p^{-n} } } a \setminus \map {B_{p^{-n} } } a\) Sphere is Set Difference of Closed and Open Ball in P-adic Numbers
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop = 0}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} } \setminus \map {B_{p^{-n} } } a\) Closed Ball is Disjoint Union of Open Balls in P-adic Numbers
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} \cup \map {B_{p^{-n} } } {a + 0 \cdot p^n} } \setminus \map {B_{p^{-n} } } a\) Union is Associative and commutative
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} \cup \map {B_{p^{-n} } } a } \setminus \map {B_{p^{-n} } } a\) $a + 0 \cdot p^n = a$
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop = 1}^{p-1} \map {B_{p^{-n} } } {a + i p^n} } \setminus \map {B_{p^{-n} } } a\) Set Difference with Union is Set Difference
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop = 1}^{p - 1} \paren {\map {B_{p^{-n} } } {a + i p^n} \setminus \map {B_{p^{-n} } } a }\) Set Difference is Right Distributive over Union


From Closed Ball is Disjoint Union of Open Balls in P-adic Numbers:

$\set {\map {B_{p^{-n} } } {a + i p^n}: i = 0, \dots, p - 1}$ is a set of pairwise disjoint open balls.


Continuing from above:

\(\ds \map {S_{p^{-n} } } a\) \(=\) \(\ds \bigcup_{i \mathop = 1}^{p - 1} \paren {\map {B_{p^{-n} } } {a + i p^n} \setminus \map {B_{p^{-n} } } a}\)
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n}\) Set Difference with Disjoint Set

$\blacksquare$


Also see

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