Smallest Normal Subgroup containing Set

Theorem
Let $$S \subseteq G$$ where $$G$$ is a group.

Then there exists a unique smallest normal subgroup of $$G$$ which contains $$S$$.

Proof
Let $$\mathbb{S}$$ be the set of all normal subgroups of $$G$$ that contain $$S$$.

$$\mathbb{S} \ne \varnothing$$, since $$S \subseteq G \triangleleft G$$.

Let $$N = \bigcap H: H \in \mathbb{S}$$, that is, the intersection of all elements of $$\mathbb{S}$$.

By Intersection of Normal Subgroups is Normal, $$N \triangleleft G$$, and by the definition of intersection, $$S \subseteq N$$.

By the method of construction, $$N$$ is the smallest such subgroup.

By the definition of "smallest", $$N$$ is unique.