Intersection of Relations Compatible with Operation is Compatible

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

Let $\mathscr F$ be a indexed family of relations on $S$.

Suppose that each element of $\mathscr F$ is compatible with $\circ$.

Let $\QQ = \bigcap \mathscr F$ be the intersection of $\mathscr F$.

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

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

Suppose that $x \mathrel \QQ y$.

Then for each $\RR \in \mathscr F$:


 * $x \mathrel \RR y$

Then since $\RR$ is a relation compatible with $\circ$:


 * $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$

Since this holds for each $\RR \in \mathscr F$:


 * $\paren {x \circ z} \mathrel \QQ \paren {y \circ z}$

We have shown that:


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

A similar argument shows that:


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

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