Center is Element of Closed Ball

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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:

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


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 $(M1)$:

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

By assumption:

$\epsilon > 0$

Hence:

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

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

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

$\blacksquare$