Open Set Less One Point is Open/Corollary

Theorem

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

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

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

Follows directly from Open Set Less One Point is Open and Finite Intersection of Open Sets of Metric Space is Open.

Let:

$U_1 = U \setminus \set {\alpha_1}, U_2 = U \setminus \set {\alpha_2}, \ldots, U_n = U \setminus \set {\alpha_n}$

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

From Finite Intersection of Open Sets of Metric Space is Open $\ds \bigcap_{i \mathop = 1}^n U_i$ is open.

$\blacksquare$