Center is Element of Closed Ball
Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon \in \R_{>0}$ be a positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $M$.
Then:
- $a \in \map {{B_\epsilon}^-} a$
Normed Division Ring
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $\struct{R, \norm {\,\cdot\,} }$.
Then:
- $a \in \map { {B_\epsilon}^-} a$
P-adic Numbers
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
Then:
- $a \in \map { {B_\epsilon}^-} a$
Proof
By metric axiom $(\text M 1)$:
- $\map d {a, a} = 0$
By assumption:
- $\epsilon > 0$
Hence:
- $\map d {a, a} < \epsilon$
By definition of the closed $\epsilon$-ball of $a$ $\map {{B_\epsilon}^-} a$ in $M$:
- $a \in \map { {B_\epsilon}^-} a$
$\blacksquare$