Meno_ wrote:But even here it is not settled, since contradiction and non contradiction are in an Absolute Russell set , self inclusive sets?

You raise two interesting points.

First, there is nothing wrong with self-inclusive sets.

Why can't we have a set \(x\) such that \(x \in x\)? There is no reason we can't, and in fact this is perfectly consistent with the other axioms of set theory. But for intuitive reasons -- namely, \(x \in x\) violates our intuitions about sets -- we don't want to allow that. So we simply

declare an axiom that says we can't have \(x \in x\).

Of course then we might still have \(x \in y\) and \(y \in x\), or even longer chains such as \(x \in y\), \(y \in z\), and \(z \in x\). So there is a clever axiom that outlaws all of these circular chains of inclusion. It's called the

axiom of foundation or sometimes the axiom of regularity.

What happens if we don't include foundation in our axioms? Then we get the study of

non well-founded set theory. It's obscure but it's studied and even applied in some disciplines.

So first point, there is nothing inherently wrong with self-membership. It's only outlawed in standard set theory because it doesn't fit our intuition about what sets should be.

The second point is that browser has a misunderstanding about contradictions. Just because you have some proof that ends in a contradiction, it doesn't mean math is broken or that everything is true. It just means that you have to throw out the assumption that led to the contradiction.

For example in Euclid's famous proof of the infinitude of primes, we start by assuming that we have a finite list of all the primes, then we show that this leads to a contradiction. We haven't broken math or proved everything is true. All we've done is shown that the assumption that there are finitely many primes is false. Nothing else.

Browser keeps saying that because we have some proof that leads to a contradiction we can use that contradiction to show that math is inconsistent. But that's wrong. All we're showing in the Russell proof is that the class, or collection, of things that are not members of themselves can not possibly be a set. That's all we've shown. There are no implications beyond that fact.

Meno_ wrote:The feeling I have is, that redefinition does nothing but reassert the primacy of naive logic.

Interesting word choice. Naive set theory is the essentially Frege's failed set theory in which sets can be formed out of unrestricted predicates, such as the "set of all things that are not members of themselves." Russell showed that this idea leads to a contradiction. So naive set theory fails. Sets can't be thought of as simply collections of things satisfying some predicate. Rather, a set is something that conforms to our axioms, which are chosen carefully to avoid contradictions.

Note: "Naive Set Theory" is also the name of a standard undergrad set theory text by Paul Halmos. It's NOT actually about naive set theory; it's about axiomatic set theory. No idea why Halmos chose that inaccurate title but it's a great book, highly recommended for people interested in set theory. Very readable.

https://en.wikipedia.org/wiki/Naive_Set_Theory_(book)

Meno_ wrote: Can that allowance withstand other succeeding arguments? Or, is it other systems of classes , when weighed in, Change the balance ?

Russell's paradox shows that the collection of all things that are not members of themselves can not possibly be a set. In ZFC (the standard axiom system for math) there is no such thing as a proper class, so in ZFC we simply say the Russell set doesn't exist. But there are other systems of set theory that formalize proper classes, and then the Russell class has official standing as a proper class: a well-defined collection that's "too big" to be a set.