## The intersection of a nonempty collection of subrings is a subring

Prove that the intersection of any nonempty collection of subrings of a ring is also a subring.

Let $R$ be a ring and let $A$ be a nonempty set of subrings of $R$.

Now $\bigcap A \subseteq R$ is a subgroup, so it suffices to show that $\bigcap A$ is closed under multiplication. To that end, let $x,y \in \bigcap A$. Then $x,y \in S$ for all $S \in A$, and $xy \in S$ for all $S \in A$. Thus $xy \in \bigcap A$, and by definition $\bigcap A \subseteq R$ is a subring.