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In ZF set theory the axiom of regularity (also called axiom of foundation) says that:

In all nonempty sets x there is an element y such that x∩y=∅

As I been told that the intention of the axiom is to prohibit constructions as $x\in x$, but is this a possible strategy?

Given a binary predicate $R$, written in infix notation, there are unitary predicates $p$ such that:

$\qquad \exists y \forall x: x R y \leftrightarrow p(x)$.

But there also are some convincing counterexamples:

  • $ p(x) \leftrightarrow \neg(x R x)$
  • $ p(x) \leftrightarrow \neg(x R f(x))$, where $f$ is any surjection
  • $ p(x) \leftrightarrow$ all chains $x_1 R x, \: x_2 R x_1, \: x_3 R x_2, \ldots \:$ are finite.

I think there is a major mistake to try to avoid certain paradoxical loops. First, there are no guarantees that every possibility is prohibited. And second, it seems pathological to make constructions that avoid certain proof-techniques (it's like the attitude of a shyster).

So, what protection comes from this axiom? Historically, has it anything to do with the Russell-predicate $x\notin x$?

See also https://math.stackexchange.com/questions/944199/are-paradoxes-a-threat-against-mathematics

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Lehs
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    It is not the intent of Regularity to protect against paradoxes of the Russell type. That job is done (one hopes!) by the axioms that specify the tools for set construction, most specifically Separation and Replacement. – André Nicolas Sep 29 '14 at 04:36
  • @André Nicolas: is that the true history about this axiom of non Neumann? – Lehs Sep 29 '14 at 04:40
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    I think of Regularity as a technical tool to make the theory of ordinals more pleasant, and I believe that was also historically the main motivation. – André Nicolas Sep 29 '14 at 04:42
  • @André Nicolas: historically I think there was an other reason: https://en.wikipedia.org/wiki/Axiom_of_regularity#Regularity_and_Russell.27s_paradox – Lehs Sep 29 '14 at 04:53
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    It seems to me that the simplest, most natural way to avoid Russell-like paradoxes is a set theory in which no objects are assumed a prior to exist within that set theory. Then it would be impossible to actually prove the existence of any object or set, be it the Russell Set or any other object. If you want to do number theory, you don't really need a ZF-style axiom of infinity or to assume the existence of the empty set within your set theory.You can simply postulate the axioms for the natural numbers (e.g. Peano's axioms) outside of your set theory and proceed. – Dan Christensen Sep 29 '14 at 15:23

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No, the axiom of regularity doesn't guard us from Russell's paradox. To wit, $\sf ZFC-Reg$ is equiconsistent with $\sf ZFC$, and therefore it couldn't be that just $\sf Reg$ was the axiom which prevented Russell's paradox.

Asaf Karagila
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    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 14:33
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Russell's paradox results from considering the class $\mathbf{C} = \{x \colon x \not\in x\}$, assuming it's a set, and deriving a contradiction from this. (Namely, writing $\mathbf{C} = C$ since it's a set, we have $C \in C \Longleftrightarrow C \not\in C$.)

The axiom of regularity does show that $\mathbf{C} = \mathbf{V}$, where $\mathbf{V}$ is the universal class. But this does not in the least affect the validity of the argument that led to a contradiction. At most, it allows us to show that $\mathbf{V}$ is a proper class directly, without invoking the separation schema.

user179549
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