Relation Intersection Inverse is Greatest Symmetric Subset of Relation

Theorem
Let $\mathcal R$ be a relation on a set $S$.

Let $\mathcal P \left({\mathcal R}\right)$ be the power set of $\mathcal R$.

By definition, $\mathcal P \left({\mathcal R}\right)$ can be considered as the set of all relations on $S$ that are subsets of $\mathcal R$.

Then the greatest element of $\mathcal P \left({\mathcal R}\right)$ that is symmetric is:


 * $\mathcal R \cap \mathcal R^{-1}$

Proof
By Intersection of Relation with Inverse is Symmetric Relation:
 * $\mathcal R \cap \mathcal R^{-1}$ is a symmetric relation.

Suppose for some $\mathcal S \in \mathcal P \left ({\mathcal R} \right )$ that $S$ is symmetric and not equal to $\mathcal R \cap \mathcal R^{-1}$.

We will show that it is a proper subset of $\mathcal R \cap \mathcal R^{-1}$.

Suppose $(x, y) \in \mathcal S$.

Then $(x, y) \in \mathcal R$.

By $\mathcal S$ being symmetric $(y, x) \in \mathcal S$.

So $(y, x) \in \mathcal R$.

By definition of inverse relation $(x, y) \in \mathcal R^{-1}$.

By definition of intersection $(x, y) \in \mathcal R \cap \mathcal R^{-1}$.

Thus:


 * $(x, y) \in \mathcal S \implies (x, y) \in \mathcal R \cap \mathcal R^{-1}$

By definition of subset:


 * $\mathcal S \subseteq \mathcal R \cap \mathcal R^{-1}$

Finally, as we assumed $\mathcal S \neq \mathcal R \cap \mathcal R^{-1}$:


 * $\mathcal S \subset \mathcal R \cap \mathcal R^{-1}$

Hence the result.