Minimal Ring Generated by System of Sets

Theorem
Let $$\mathcal S$$ be a non-empty system of sets.

Then there is a unique ring of sets $$\mathcal R \left({\mathcal S}\right)$$ which:
 * contains $$\mathcal S$$;
 * is contained by every ring of sets which also contains $$\mathcal S$$.

This ring of sets $$\mathcal R \left({\mathcal S}\right)$$ is called the minimal ring generated by $$\mathcal S$$.

Uniqueness
Suppose there were two such rings of sets $$\mathcal R \left({\mathcal S}\right)$$ and $$\mathcal R \left({\mathcal S}\right)'$$.

Then by definition their intersection $$\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)'$$ would also contain $$\mathcal S$$.

By Intersection of Rings of Sets, $$\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)'$$ is also a ring of sets.

From Intersection Subset, $$\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)' \subseteq \mathcal R \left({\mathcal S}\right)$$ and $$\mathcal R \left({\mathcal S}\right) \cap \mathcal R \left({\mathcal S}\right)' \subseteq \mathcal R \left({\mathcal S}\right)'$$.

Hence either $$\mathcal R \left({\mathcal S}\right)$$ or $$\mathcal R \left({\mathcal S}\right)'$$ can not be minimal.

So if $$\mathcal R \left({\mathcal S}\right)$$ exists, it has to be unique.

Existence
Consider the union:
 * $$X = \bigcap_{A \in \mathcal S} A$$

of all sets in $$\mathcal S$$.

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

From Power Set is Algebra of Sets and by definition of algebra of sets, $$\mathcal P \left({X}\right)$$ is a ring of sets containing $$\mathcal S$$.

Now let $$\Sigma$$ be the set of all rings of sets contained in $$\mathcal P \left({X}\right)$$ which also contain $$\mathcal S$$.

Then consider the intersection:
 * $$\mathcal R \left({\mathcal S}\right) = \bigcap_{\mathcal T \in \Sigma} \mathcal T$$

of all these rings of sets.

It is clear that $$\mathcal R \left({\mathcal S}\right)$$ has the properties specified.

In particular, as all $$\mathcal T \in \Sigma$$ contain $$\mathcal S$$, then so does $$\mathcal R \left({\mathcal S}\right)$$.

Finally, note that if $$\mathcal R'$$ is a ring of sets containing $$\mathcal S$$, then:
 * $$\mathcal R^* = \mathcal R' \cap \mathcal P \left({X}\right)$$

is a ring of sets in $$\Sigma$$.

Hence $$\mathcal R \left({\mathcal S}\right) \subseteq \mathcal R^* \subseteq \mathcal R'$$ as we needed to show.