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 $\RR$ denote the membership relation on $\Bbb S$:
- $\forall \tuple {a, b} \in \Bbb S \times \Bbb S: \tuple {a, b} \in \RR \iff a \in b$
$\RR$ 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:
- $\forall x \in S: \tuple {a, b} \in \RR \implies \tuple {b, a} \in \RR$
demonstrating that $\RR$ 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 $\RR$ is seen to be not symmetric.
$\blacksquare$
Sources
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Exercise $1.5.1$