Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Union of Closed Balls
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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:
- $\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{\paren {m - n} } }$
Proof
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:
- $\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} }$
$\blacksquare$