Set of Congruence Classes on Algebraic Structure forms Complete Lattice

Theorem
Let $\struct {S, \odot}$ be an algebraic structure.

Let $\map \RR \odot$ be the set of all congruence relations on $\struct {S, \odot}$.

Then $\struct {\map \RR \odot, \subseteq}$ is a complete lattice.

Proof
elements of $\map \RR \odot$ are subsets of $S \times S$.

First we consider the trivial relation $S \times S$ itself.

From Trivial Relation is Universally Congruent, $S \times S$ is a congruence relation on $\struct {S, \odot}$.

Let $\TT$ be a subset of $\map \RR \odot$.

Consider the intersection $\HH = \ds \bigcap \TT$.

Let $x_1, y_1, x_2, y_2 \in S$ be arbitrary, such that:
 * $x_1 \mathrel \HH y_1$ and $x_2 \mathrel \HH y_2$

We have:

Hence $\HH = \ds \bigcap \TT$ is a congruence relation on $\struct {S, \odot}$.

That is:
 * $\ds \bigcap \TT \in \struct {\map \RR \odot, \subseteq}$

The appropriate conditions are fulfilled, and from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
 * $\struct {\map \RR \odot, \subseteq}$ is a complete lattice.