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.