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In set theory, if $R$ is a relation, it is considered a set such that if $aRb$, then $(a,b)\in R$. However, if $\in$ (the "element of" symbol) is considered to be a relation in the same regard, then I can prove that this is inconsistent with ZF.

Theorem: If $\in$ is a relation, then there exists $U$ in ZF such that $\forall x(x\in U)$.

Proof: Let $\in$ be a relation. Let this relation also be denoted $S$. By the axiom of the power set, for any set $x$ there exists a family $y$ such that $x\in y$. By the definition of a relation, $x\in y \implies (x,y)\in S$. Using the Kuratowski's definition of an ordered pair $$(x,y):=\big\{\{x\},\{x,y\}\big\}.$$ Therefore, $x\in y \implies \big\{\{x\},\{x,y\}\big\}\in S$. By the axiom of the union, there exists the arbitrary union $T=\cup S$. So $\big\{\{x\},\{x,y\}\big\}\in S\implies\{x\}\in T$. Once again, there must exist the arbitrary union $U=\cup T$. Hence, $\{x\}\in T\implies x\in U$. Therefore, if $\in$ is a relation, there exists $U$ such that $\forall x(x\in U)$ in ZF. $\square$

Of course, this immediately becomes inconsistent, as then Russel's Paradox arises.

Does this mean that ZF is inconsistent, or that $\in$ is not considered to be a relation in the same regard as other relations? If the latter, how is $\in$ then defined?

Graviton
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    It's a class. Just like equality is a relation, but it's not a set. Or how the power set function is a function but not a set. Etc. – Asaf Karagila Jul 16 '21 at 07:40
  • As a follow up @Asaf, how does one make sense of a class? Is a class just anything that can be constructed using normal comprehension? – Math Simp Jul 16 '21 at 12:45
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    @MathSimp: Depending on the exact interpretation of the term, it is usually means a definable collection, where definable means with parameters. Every set corresponds to a class in the natural way, but of course not every class corresponds to a set. – Asaf Karagila Jul 16 '21 at 12:46
  • It is a binary "relation symbol" (see the syntax of first-order logic) of ZFC. – Poscat Sep 25 '22 at 07:16

1 Answers1

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$\in$ is special. It is undefined. Its behaviour is detailed in the ZF axioms, and that's it.

It does behave exactly like a (class) relation. And it is easy to define such a relation that coincides with $\in$. But the symbol $\in$ itself isn't a relation.

Arthur
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  • Is there a formal way of asserting that $\in$ is special? Just upon first glance, that seems to me a very fragile way of avoiding a paradox, to just claim that $\in$ is functionally equivalent to the notion of a relation, yet is classified differently. – Graviton Jul 16 '21 at 07:49
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    @Graviton Basically every axiomatisation of everything ever needs undefined terms and / or symbols. It is the starting point from where everything in the theory builds. There are not enough words and symbols to define everything, so this is a trade-off one must accept. ZF set theory takes symbols and definitions from logic (such as $\forall$ and $\lnot$ and $=$), adds a single new symbol $\in$, describes through its axioms how $\in$ behaves, and everything builds from there. – Arthur Jul 16 '21 at 07:54