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

Theorem

Let $p$ be a prime number.

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

Let $y \in \Q_p$

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

Let there exist $i \in \Z$:

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

Then:

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

Proof

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:

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

$\blacksquare$