It's well known it's impossible to solve the Russel paradox using only the axioms of the Cantor set theory. The Quine's New foundation $(NF)$ is able to prevent this paradox and others like for example, the Burali - Forti. Are there paradoxes impossible to solve inside the axioms of $NF$ that are solved inside $NFU$? Thanks.

  • 1
    $\begingroup$ I do not understand what you mean by "it's impossible to solve the Russell paradox using only the axioms of the Cantor set theory." On the contrary, the paradox is very successfully solved there. You need to clarify what you mean. $\endgroup$ – Andrés E. Caicedo Sep 24 '13 at 15:08
  • $\begingroup$ In the 'naive' set theory I know it's impossible to solve the antinomy. It needs to extend the theory using the ZF axioms or the NGB. In the NF (or NFU), I know the Russel's antinomy is avoided without any other extension $\endgroup$ – Riccardo.Alestra Sep 25 '13 at 8:49
  • $\begingroup$ I think you misunderstand the relation of ZF/NBG/NF to naive set theory. None are extensions of NST; they are all NST with bits removed--they are subtheories of NST. Since there are no known inconsistencies in NF, there's nothing interesting to say about whether NFU avoids them. $\endgroup$ – Malice Vidrine Jul 2 '16 at 8:00

All the "classical" paradoxes with naive set theory fail to go through in NF; however, the consistency of NF even relative to ZFC is open (well, Holmes has a claimed consistency proof but I don't think it's been vetted yet).

Interestingly, it is known that NFU and its various familiar extensions are consistent, relative to very weak theories:

  • NFU itself is relatively consistent with PA.

  • NFU + Infinity + Choice is relatively consistent with a particular weak set theory called "MacLane set theory."

Moreover these consistency proofs are not very hard. So there's an interesting jump in strength going from NFU to NF.

Another manifestation of this jump in strength is the remarkable fact that NF disproves the axiom of choice while choice is consistent with ZFU! There doesn't seem to be a snappy reason for this; Specker's proof is quite complicated (although see here for an outline of the proof). The disproof of choice obviously isn't a paradox in the sense of a proof of inconsistency, but it is a very odd result, especially given that a rejection of choice isn't implicit in the motivation for NF.

This is the closest thing to a paradox which is impossible to solve inside the axioms of NF that is solved inside NFU; it's not really a "paradox" in the modern sense, but it is a proof of an arguably highly counterintuitive result (the negation of choice). And this argument dies a horrible death if we try to run it in NFU.

So I would say the following:

While we are confident that NFU is paradox-free (since we are confident, I hope!, in the consistency of Peano arithmetic), the evidence in favor of the consistency of NF is much weaker; there is a possible proof of consistency relative to a fragment of ZFC, which is still (I think) unvetted, and there are fundamental weirdnesses about NF which do not apply to NFU which suggest (at least to me) that we shouldn't expect NF to be consistent just because NFU is.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.