Finite Intersection of Open Sets of Metric Space is Open

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

Let $$U_1, U_2, \ldots, U_n$$ be open in $$M$$.

Then $$\bigcap_{i=1}^n U_i$$ is open in $$M$$.

That is, the intersection of a finite number of open subsets is open.

Proof
Let $$x \in \bigcap_{i=1}^n U_i$$.

For each $$i \in \left[{1 \,. \, . \, n}\right]$$, we have $$x \in U_i$$.

Thus $$\exists \epsilon_i > 0: N_{\epsilon_i} \left({x}\right) \subseteq U_i$$.

Let $$\epsilon = \min_{i=1}^n \left\{{\epsilon_i}\right\}$$.

Then $$N_{\epsilon} \left({x}\right) \subseteq N_{\epsilon_i} \left({x}\right) \subseteq U_i$$ for all $$i \in \left[{1 \,. \, . \, n}\right]$$.

So $$N_{\epsilon} \left({x}\right) \subseteq \bigcap_{i=1}^n U_i$$.

The result follows.

Warning
This result breaks down when we consider an infinite number of subsets.

For example, the open interval $$\left({-\frac 1 n \, . \, . \, \frac 1 n}\right)\subseteq \reals$$ is open in $$\reals$$ for all $$n \in \mathbb{N}$$.

But $$\bigcap_{i=1}^\infty \left({-\frac 1 n \, . \, . \, \frac 1 n}\right) = \left\{{0}\right\}$$ which is not open in $$\reals$$.