Why is Russell's set non-well-founded?

Question

It is usually said that the Russell set (the set $R$ of all sets that are not members of themselves) is non-well-founded, but I honestly do not understand why.

If an object is non-well-founded, it should somehow violate the Axiom of Regularity below (which is sometimes called Foundation Axiom, such as in https://plato.stanford.edu/entries/nonwellfounded-set-theory/#2.2):

Axiom of Regularity: For every non-empty set $x$, there is some $y \in x$ such that $y\cap x=\varnothing$.

But I cannot see how exactly the Russell set violates the axiom. Actually, it seems to respect it: since the empty set does not contain itself, it is a member of the Russell set $R$, and so there is a $y \in R$ such that $y\cap R=\varnothing$, namely, the empty set $\varnothing$. So what do people mean when they say that the Russell set is non-well-founded? Can someone guide me through the strategy to prove this?

Observation: I've read the discussion in Non WellFounded Set theories and Russell's Paradox and it was very helpful, but I still don't understand this particular point. Noah, in his answer, said this: "Let $𝑋$ be the set of all well-founded sets. Note that by regularity this is equal to the set of all sets not containing themselves." This must mean that the axiom of regularity is equivalent to the statement "a set cannot be a member of itself," but I don't understand why. I'm sure I'm missing something here, but I don't know what.

Thanks a lot!

Answer 1

Let's break down why the Russell Class causes logical issues.

Something important to notice is that The Russell Class being a set leading to a contradiction, is a Pure Theory of Set Theory.

That is, you don't need any axioms. So, the following holds in All Set Theories based on first order predicate calculus with the ∈ relation added to it.

Theorem: $\lnot\exists y\forall x(x\in y \iff \lnot (x\in x))$

Proof:

By contradiction Assume such a y exists ( is a set)

So for some set $y$, $\forall x(x\in y \iff \lnot (x\in x))$

So, we let $x = y$

$y\in y \iff \lnot (y\in y)$

A contradiction.

In Particular, we could consider

ZF - Foundation + ($\lnot $( Foundation))

as a collection of Axioms for a Set Theory.

In there, we have a set that is not well-founded. But, The Russell Class is still not a set.

Notice that without any axioms ∈ is a "dummy" relation and could mean anything. So not only is the above theorem true for ∈ it holds for any relation.

We could interpret elementhood as $=$ Then the above theorem would be: No set is equal to every set that is not equal to themselves.

Why bring this up? To point out how the contradiction, at it's core is deeper than sets containing certain kinds of sets, or well-foundedness.

The same contradiction holds, regardless of how we interpret $\in$. So, the theorem in full glory, is really about all relations. and The Russell Class is just a particular example.