# Russell Paradox and set theories

The Russell paradox arise in the Cantor set theory, but it can be avoided in the $ZF$ and in $NGB$ axiomatic set theory. Are there other axiomatic set theories in which this paradox can be avoided? Thanks.

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Aaargh. The Russel paradox does not arise in the set theory of Cantor. It was a paradox for the set theory of Frege. And no theory that we would call a set theory today allows for the paradox. – Michael Greinecker Mar 21 '12 at 16:10
Much like when defining a function you first check whether or not it is well defined, there is a list of usual paradoxes which one checks when defining axioms of set theory. Russell's one is probably the first. – Asaf Karagila Mar 21 '12 at 16:14
Quine's New Foundations avoids the paradox by forbidding comprehension over the poisonous predicates. – MJD Mar 21 '12 at 16:21
Every set theory built on standard logic tries to avoid the known difficulties. Whether it succeeds is another matter. The only exceptions are set theories based on paraconsistent logics, which limit the damage done by $A\land \lnot A$ in other ways. – André Nicolas Mar 21 '12 at 16:25
Most (if not all) modern set theories avoid Russell's paradox by limiting set construction principles. In particular, the Comprehension Scheme is restricted so that a definable class is a set only if it is a subclass of a set. – Arthur Fischer Mar 21 '12 at 16:27

As it is mentioned in comments above, the so called Russell paradox is a consequence of non-restricted Comprehension Schema according to which for any formula $\varphi(x)$, where $x$ is free, $\{x\mid\varphi(x)\}$ is a set. This paradox is actually a result of a logical truth ($R$ is a binary predicate): $$\neg\exists x\forall y(yRx\iff \neg yRy)\,.$$ In light of this, assuming non-restricted comprehension in a language in which you have at least one binary predicate you always get inconsistent theory. Thus if you want to build a set theory based on classical logic you must restrict the schema one way or another.