I am learning set theory and I am curious if we could have unrestricted comprehension while blocking Russell's paradox using the axiom of regularity/foundation. To my very limited knowledge, axiom of regularity is not needed to block paradoxes since if ZF without regularity is inconsistent, then adding regularity would not make it consistent (and the restricted separation schema has already done the work). It seems to me that axiom of regularity also blocks the existence of a universal set and since it says no sets can be a member of itself, it should also block the construction of the Russell sets (since $\forall x (x\not\in x)$ is a description of the universal set under the Axiom of Regularity...?)


2 Answers 2


Adding an axiom can never make an inconsistent system consistent. Axioms can't block paradoxes. You should think of this the other way round: the axioms of regularity and replacement are designed to strengthen ZF in useful ways while avoiding inconsistency.

  • $\begingroup$ I see. Thanks a lot (below I have a question regarding the logical relationship between axiom of regularity and principle of vicious circle). Just to clarify the effect of adding new axioms: does this also have to do with the fact that the comprehension axiom is a schema rather than a first-order sentence? I am asking because on a naive picture, a first-order theory is a set of sentences. If all sentences are non-schematic, and the theory is finite, then it is equivalent to a theory of one axiom which is the conjunction of all these axioms. $\endgroup$ Mar 4, 2018 at 15:29
  • $\begingroup$ So restricting an axiom seems equivalent to adding new conjucnts, therefore adding new axioms. And here I suppose here we avoid Russell's paradox by restricting comprehension axiom...? $\endgroup$ Mar 4, 2018 at 15:30
  • $\begingroup$ Yes a finitely axiomatizable theory can be axiomatized with a single axiom. Conceptually, an axiom schema is an infinite conjunction. Restricting a schema reduces the number of conjuncts in that conceptual infinite conjunction. In the case of the restricted comprehension axiom, it removes the conjuncts that lead to Russell's paradox. $\endgroup$
    – Rob Arthan
    Mar 4, 2018 at 18:45

Rob wrote correctly, you cannot revive a dead person by giving him an extra leg to stand on.

But here is a direct proof: Consider the Russell class, $\{x\mid x\notin x\}$. By comprehension this is a set, and by regularity it is in fact the set of all sets. But now it must be a member of itself, and this is a contradiction to regularity.

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    $\begingroup$ Or, just take $\{x:x=x\}$ :P $\endgroup$ Mar 4, 2018 at 8:56
  • $\begingroup$ Yeah, or that. :P $\endgroup$
    – Asaf Karagila
    Mar 4, 2018 at 8:56
  • $\begingroup$ I see. So is it right to think that the axiom of regularity is not a set-formation restriction: it doesn't say that you can't form a set that is a member of itself. It only says that all sets are not member of themselves. In particular, you can construct a set that is both a member and not a member of itself (hence the Russell set). I guess I was confused between the axiom of regularity and the principle of vicious circle (or the iterative conception of sets where you can only form a set whose elements are already sets formed in previous stages). $\endgroup$ Mar 4, 2018 at 15:16
  • $\begingroup$ Is it right to think that these two are not equivalent (though the latter presumably implies the former?) $\endgroup$ Mar 4, 2018 at 15:16

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