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.

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, $$\bigcap_{i=1}^n U_i$$ is open.