Intersection of Rings of Sets

Theorem
Let $$\mathcal{R}_k$$ be a ring of sets, where $$k$$ is an element of an arbitrary set of indices.

Then their intersection $$\mathcal{R} = \bigcap_k \mathcal{R}_k$$ is itself a ring of sets.

Proof
Consider the set $$\mathcal{S} = \bigcup_k \mathcal{R}_k$$.

Now let $$S = \left\{{X \in Y: Y \in \mathcal{S}}\right\}$$.

This contains all the elements of all the sets contained in all the $$\mathcal{R}_k$$.

Now consider the power set $$\mathcal{P} \left({S}\right)$$ of $$S$$.

By Power Set is Algebra of Sets and the definition of algebra of sets, we have that $$\mathcal{P} \left({S}\right)$$ is a ring of sets.

Thus $$\left({\mathcal{P} \left({S}\right), *, \cap}\right)$$ is a ring (in the abstract algebraic sense of the term).

From the method of construction of $$\mathcal{P} \left({S}\right)$$, it is clear that all the $$\left({\mathcal{R}_k, *, \cap}\right)$$ are subrings of $$\mathcal{P} \left({S}\right)$$.

The result then follows from Intersection of Subrings.