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

Corollary

Let $M$ and $U$ be as above.

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

Then $U - S$ is open in $M$.

Proof

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

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

Let $N_\epsilon \left({x}\right) \subseteq U$ be an $\epsilon$-neighborhood of $x$ in $U$.

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

Then $N_\epsilon \left({x}\right) \subseteq U - \left\{{\alpha}\right\}$.

The result follows.

$\blacksquare$

Proof of Corollary

Follows directly from the above and Intersection of Open Subsets.

Let $U_1 = U - \left\{{\alpha_1}\right\}, U_1 = U - \left\{{\alpha_2}\right\}, \ldots, U_n = U - \left\{{\alpha_n}\right\}$.

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

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

$\blacksquare$