Closed Ball is Closed

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $x \in A$.

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

Let $B_\epsilon^- \left({x}\right)$ be the closed $\epsilon$-ball of $x$ in $M$.

Then $B_\epsilon^- \left({x}\right)$ is a closed set of $M$.

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

Let $a \in A \setminus B_\epsilon^- \left({x}\right)$.

Then by definition of closed ball:
 * $d \left({x, a}\right) > \epsilon$

Put:
 * $\delta := d \left({x, a}\right) - \epsilon > 0$

Then:
 * $d \left({x, a}\right) - \delta = \epsilon$

Let $b \in B_\delta \left({a}\right)$.

Then:

and so:
 * $b \notin B_\epsilon^- \left({x}\right)$

Then:
 * $B_\delta \left({a}\right) \subseteq A \setminus B_\epsilon^- \left({x}\right)$

so $A \setminus B_\epsilon^- \left({x}\right)$ is open in $M$.

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