Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $p$ be a prime number.

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

Let $a \in \Q_p$.

For all $\epsilon \in \R_{>0}$, let $\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$.

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


Then:

$(1) \quad \map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n} } - 1} \map {B^-_{p^{-m} } } {a + i p^n}$
$(2) \quad \set {\map {B^-_{p^{-m} } } {a + i p^n} : i = 0, \dots, p^{\paren {m - n} } - 1}$ is a set of pairwise disjoint closed balls


Proof

Condition $(1)$

Lemma

$\forall y \in \Q_p: \norm y_p \le p^{-n}$ if and only if there exists $i \in \Z$ such that:
$(1)\quad 0 \le i \le p^{\paren {m - n}} - 1$
$(2)\quad \norm {y - i p^n}_p \le p^{-m}$

$\Box$


Let $0 \le i \le p^{\paren{m - n}} - 1$.

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

By definition of a closed ball:

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

From Lemma:

$\norm {x - a}_p \le p^{-n}$

By definition of a closed ball:

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

Since $x$ was arbitrary:

$\map {B^-_{p^{-m} } } {a + i p^{-n} } \subseteq \map {B^-_{p^{-n} } } a$

Since $i$ was arbitrary:

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


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

By definition of a closed ball:

$\norm {x - a}_p \le p^{-n}$

From Lemma:

$\exists i \in \N : 0 \le i \le p^{\paren {m - n}} - 1: \norm {x - a - i p^{-n} } \le p^{-m}$

By definition of a closed ball:

$\exists i \in \N : 0 \le i \le p^{\paren {m - n}} - 1 : x \in \map {B^-_{p^{-m} } } {a + i p^{-n} }$

Hence:

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


It follows that:

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

$\Box$


Condition $(2)$

Let $0 \le i, j \le p^{\paren {m - n}} - 1$.

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

From Characterization of Open Ball in P-adic Numbers:

$\norm {\paren {x -a} - i p^n}_p \le p^{-m}$

and:

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


We have that P-adic Norm satisfies Non-Archimedean Norm Axioms.

Then:

\(\ds \norm {i p^n - j p^n}_p\) \(\le\) \(\ds p^{-m}\) Corollary to P-adic Metric on P-adic Numbers is Non-Archimedean Metric
\(\ds \leadsto \ \ \) \(\ds \norm {p^n}_p \norm {i - j}_p\) \(\le\) \(\ds p^{-m}\) Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity
\(\ds \leadsto \ \ \) \(\ds p^{-n} \norm {i - j}_p\) \(\le\) \(\ds p^{-m}\) Definition of $p$-adic norm
\(\ds \leadsto \ \ \) \(\ds \norm {i - j}_p\) \(\le\) \(\ds p^{n - m}\) multiplying both sides by $p^n$.
\(\ds \leadsto \ \ \) \(\ds p^{\paren {m - n} }\) \(\divides\) \(\ds \paren {i - j}\) Definition of $p$-adic norm
\(\ds \leadsto \ \ \) \(\ds j\) \(\equiv\) \(\ds i \mod p^{\paren {m - n} }\) Definition of Congruence Modulo $p$
\(\ds \leadsto \ \ \) \(\ds i\) \(=\) \(\ds j\) Integer is Congruent to Integer less than Modulus
\(\ds \leadsto \ \ \) \(\ds \map {B^-_{p^{-m} } } {a + i p^n}\) \(=\) \(\ds \map {B^-_{p^{-m} } } {a + j p^n}\)

The result follows.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Closed_Ball_is_Disjoint_Union_of_Smaller_Closed_Balls_in_P-adic_Numbers&oldid=576428"