Open Set Less One Point is Open

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

Let $U \subseteq M$ be an open set of $M$.

Let $\alpha \in U$.

Then $U \setminus \left\{{\alpha}\right\}$ is open in $M$.

Proof
Let $x \in U \setminus \left\{{\alpha}\right\}$.

Let $\delta = d \left({x, \alpha}\right)$.

Let $B_\epsilon \left({x}\right) \subseteq U$ be an open $\epsilon$-ball of $x$ in $U$.

Let $\zeta = \min \left\{{\epsilon, \delta}\right\}$.

Then $B_\epsilon \left({x}\right) \subseteq U \setminus \left\{{\alpha}\right\}$.

The result follows.