Closed Ball is Closed/Metric Space

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.

Then $\map {B_\epsilon^-} x$ is a closed set of $M$.

Proof
We show that the complement $A \setminus B_\epsilon^- \left({x}\right)$ is open in $M$.

Let $a \in A \setminus \map {B_\epsilon^-} x$.

Then by definition of closed ball:
 * $\map d {x, a} > \epsilon$

Put:
 * $\delta := \map d {x, a} - \epsilon > 0$

Then:
 * $\map d {x, a} - \delta = \epsilon$

Let $b \in \map {B_\delta} a$.

Then:

and so:
 * $b \notin \map {B_\epsilon^-} x$

Then:
 * $\map {B_\delta} a \subseteq A \setminus \map {B_\epsilon^-} x$

so $A \setminus \map {B_\epsilon^-} x$ is open in $M$.

Hence, by definition of closed set:
 * $\map {B_\epsilon^-} x$ is closed in $M$.