Membership Relation is Not Symmetric

Theorem
Let $\Bbb S$ be a set of sets in the context of pure set theory

Let $\mathcal R$ denote the membership relation on $\Bbb S$:
 * $\forall \tuple {a, b} \in \Bbb S \times \Bbb S: \tuple {a, b} \in \mathcal R \iff a \in b$

$\mathcal R$ is not in general a symmetric relation.

Proof
In the extreme pathological edge case:
 * $S = \set S$

it is seen that:
 * $S \in S$

and so:
 * $\forall x \in S: \tuple {a, b} \in \mathcal R \implies \tuple {b, a} \in \mathcal R$

demonstrating that $\mathcal R$ is symmetric in this specific case.

However, in this case $\set S$ is a set on which the Axiom of Foundation does not apply.

This is seen in Set is Not Element of Itself.

Hence this set is not supported by Zermelo-Fraenkel set theory.

Consider the set:


 * $T = \set {\O, \set \O}$

Then we immediately see that while:
 * $\O \in \set \O$

we have that:
 * $\set \O \notin \O$

and so $\mathcal R$ is seen to be not symmetric.