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

Put $\delta = \dfrac{\epsilon - d \left({x, a}\right)}{2}$.

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

Then $b \notin B_\epsilon^- \left({x}\right)$, as:

Then $B_\delta \left({a}\right) \subseteq A \setminus B_\epsilon^- \left({x}\right)$, so $A \setminus B_\epsilon^- \left({x}\right)$ is open.

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