# Center is Element of Closed Ball

## Theorem

Let $M = \left({A, d}\right)$ 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:

$a \in \map { {B_\epsilon}^-} a$

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

$\map d {a,a} = 0$

By assumption:

$\epsilon > 0$

Hence:

$\map d {a,a} < \epsilon$

By definition of the closed $\epsilon$-ball of $a$ $\map {{B_\epsilon}^-} a$ in $M$:

$a \in \map {{B_\epsilon}^-} a$

$\blacksquare$