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

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

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

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

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

The result follows.

Alternatively, note that a metric space is a topological space.

The result follows from the definition of topology.

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

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

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