Open Set Less One Point is Open

Theorem

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

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

Let $\alpha \in U$.

Then $U \setminus \set \alpha$ is open in $M$.

Corollary

Let $S = \set {\alpha_1, \alpha_2, \ldots, \alpha_n} \subseteq U$ be a finite set of points in $U$.

Then $U \setminus S$ is open in $M$.

Proof

Let $x \in U \setminus \set \alpha$.

Let $\delta = \map d {x, \alpha}$.

Let $\map {B_\epsilon} x \subseteq U$ be an open $\epsilon$-ball of $x$ in $U$.

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

Then:

$\map {B_\epsilon} x \subseteq U \setminus \set \alpha$

The result follows.

$\blacksquare$