Unions and Intersections of Open Sets are Open Let $(X,d)$ be a metric space.
Prove:


*

*the union of any open sets in $X$ is open in $X$

*the intersection of a finite number of open sets in $X$ is open in $X$


I could prove the first one but how do we prove the second one?
 A: Let $A_i$ be a finite collection of open sets in $X$ and let $A$ be their intersection, so that $A = \bigcap\limits_{i \in \mathbb N} A_i$. If the intersection is empty, then there is nothing to prove because the empty set is open (and closed) relative to any set. So assume the intersection is nonempty.
Define $r = \min\{r_1, \ldots, r_n\},$ where each $r_i$ corresponds to the open set $A_i$. Then for any $x \in A, B_r(x) \subseteq A$ is an open ball around $x$ which is fully contained in $A$, showing that $A$ is indeed open. 
$A_i$ must be a $\textit{finite}$ collection of open sets because we are only able to take the minimum of a $\textit{finite}$ number of radii. The minimum of an infinite set might not exist, and the infimum might be $0$. But open balls must have a $\textit{positive}$ radius, so we cannot construct an open ball here.
A: If the intersection is not empty, then take $x$ from this intersection. $x$ is also in every open set (say $U_1,...,U_k$) involved in the intersection, then there are positive $d_1,...,d_k$ with $B(x,d_i)\subseteq U_i$ for $i=1,...,k$. Now, take $d:=min \{d_i\}$ which is also a positive number. Then we have $B(x,d)$ is contained in the intersection, hence the result follows.
