Set of Ideals forms Complete Lattice

Theorem
Let $$\left({R, +, \circ}\right)$$ be a group, and let $$\mathbb R$$ be the set of all ideals of $$R$$.

Then $$\left({\mathbb R, \subseteq}\right)$$ is a complete lattice.

Proof
Let $$\varnothing \subset \mathbb S \subseteq \mathbb R$$.

By Intersection of Ideals:
 * $$\bigcap \mathbb S$$ is the largest ideal of $$R$$ contained in each of the elements of $$\mathbb S$$.


 * The intersection of the set of all ideals of $$R$$ containing $$\bigcup \mathbb S$$ is the smallest ideal of $$R$$ containing $$\bigcup \mathbb S$$.

Thus:
 * Not only is $$\bigcap \mathbb S$$ a lower bound of $$\mathbb S$$, but also the largest, and therefore an infimum.


 * The supremum of $$\mathbb S$$ is the join of the set of all ideals of $$\mathbb S$$.

From Sum of Ideals is an Ideal, this supremum is:
 * $$\sum_{S \in \mathbb S} \left({S, +, \circ}\right)$$

i.e.:
 * $$\left({S_1, +, \circ}\right) + \left({S_2, +, \circ}\right) + \cdots + \left({S_n, +, \circ}\right)$$

where addition of ideals is as defined in subset product.

Therefore $$\left({\mathbb R, \subseteq}\right)$$ is a complete lattice.