Membership Relation is Not Symmetric

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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.

$\blacksquare$


Sources