Open Set Less One Point is Open/Corollary

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

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

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

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

Proof
Follows directly from the above and Intersection of Open Subsets.

Let:
 * $U_1 = U \setminus \left\{{\alpha_1}\right\}, U_2 = U \setminus \left\{{\alpha_2}\right\}, \ldots, U_n = U \setminus \left\{{\alpha_n}\right\}$

From the above, $U_1, U_2, \ldots, U_n$ are all open in $M$.

From Intersection of Open Subsets, $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is open.