Union of Relations Compatible with Operation is Compatible

Theorem

Let $\struct {S, \circ}$ be a closed algebraic structure.

Let $\FF$ be a family of relations on $S$.

Let each element of $\FF$ be compatible with $\circ$.

Let $\QQ = \bigcup \FF$.

Then $Q$ is a relation compatible with $\circ$.

Proof

Let $x, y, z \in S$.

Let $x \mathrel \QQ y$.

Then for some $\RR \in \FF$:

$x \mathrel \RR y$.

Since $\RR$ is a relation compatible with $\circ$:

$\tuple {x \circ z} \mathrel \RR \tuple {y \circ z}$

Since $\RR \in \FF$:

$\RR \subseteq \bigcup \FF = \QQ$

Thus

$\tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$

We have shown that:

$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$

A similar argument shows that:

$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {z \circ x} \mathrel \QQ \tuple {z \circ y}$

So $Q$ is a relation compatible with $\circ$.

$\blacksquare$