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

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

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

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

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

where $B_{\epsilon_i} \left({x}\right)$ is the open $\epsilon_i$-ball of $x$.

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

Then:
 * $B_\epsilon \left({x}\right) \subseteq B_{\epsilon_i} \left({x}\right) \subseteq U_i$

for all $i \in \left[{1 \,.\,.\, n}\right]$.

So:
 * $\displaystyle B_\epsilon \left({x}\right) \subseteq \bigcap_{i \mathop = 1}^n U_i$

The result follows.

Also see

 * Infinite Intersection of Open Sets of Metric Space may not be Open for how this result breaks down for an infinite number of open sets.