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How does axiom of regularity ensures that Russell's Paradox doesn't occurs. Given the axiom of regularity states: If A is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A.

Anonymous
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    Are you sure that's the axiom of regularity? Or are you not working with ZF? In ZF a set can only contain other sets. – Vsotvep Jun 18 '19 at 12:42
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    And in any case you can never avoid a contradiction by adding additional axioms. If you have a proof of a contradiction from a certain set of axioms, then this proof will continue to prove a contradiction no matter what else you add to your axiom system. – hmakholm left over Monica Jun 18 '19 at 14:29

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It doesn't.

You need the other axioms of $\mathsf{ZF}$ to define how to construct sets, and thus to see that the object in Russell's paradox is a proper class.

Note that the domain of discourse is $V$, the class of all sets. Any variable that is used in a first-order formula in the language of set theory refers to elements of $V$, and thus to sets. This means that proper classes and sets can only contain sets.

If $R=\{x\mid x\notin x\}$ is a set, this would lead to a contradiction, since then both $R\in R$ and $R\notin R$ are provable. However, if we make our axioms such that $R$ is a proper class, then $R$ cannot contain itself, since $R$ is not an element in our universe.

It is the axiom (scheme) of unrestricted comprehension of naive set theory that makes it possible to define the set $\{x\mid x\notin x\}$. In $\mathsf{ZF}$ such a thing is not possible, since we're using the weaker axiom (scheme) of separation, that says you can only define subsets of sets with first-order formulas:

Unrestrictied comprehension: $\forall\bar v\,\exists y\forall x(x\in y\leftrightarrow \phi(x,\bar v))$

Separation: $\forall\bar v\forall z\exists y\forall x(x\in y\leftrightarrow (\phi(x,\bar v)\land x\in z))$

The axiom of regularity says that any non-empty set contains an element that is disjoint from it.

Regularity: $\forall x(x\neq\varnothing\to \exists y\in x( y\cap x=\varnothing)$

It makes sure that a set cannot contain itself, since if $x\in x$, then $\{x\}$ does not contain a $y$ such that $y\cap x=\varnothing$. Without it, most of mathematics is still perfectly provable, and Russell's paradox doesn't work since we use the (weak enough) axiom (scheme) of separation.

Using the axiom of regularity, we can prove that $R=V$, since $x\notin x$ is true for any set.

Vsotvep
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