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May 11, 2020 at 6:01 answer added user76284 timeline score: 2
May 11, 2020 at 5:43 comment added user76284 Small note: As the current version of the Wikipedia article indicates, an empty set axiom is not needed if one includes $\bot$ as a positive formula.
Sep 14, 2017 at 8:33 comment added Mike Shulman To avoid Curry's paradox, paraconsistentists are usually forced to discard some instances of "contraction" ($P\to (P\to Q)$ yields $P\to Q$) as well -- in which case it seems to me you are more than halfway to linear logic instead, in which case (as Andreas mentioned) a naive comprehension rule can actually be consistent (as opposed to paraconsistent).
Sep 14, 2017 at 8:31 comment added Mike Shulman For the record, Russell's paradox doesn't depend on excluded middle; it works just as well in intuitionistic logic. And personally, I find the usual paraconsistent approach unimpressive because mere paraconsistency of a logic (i.e. ability to sustain the simultaneous truth of $P$ and $\neg P$ without deducing an arbitrary $Q$) is insufficient to avoid set-theoretic paradoxes, since Curry's paradox doesn't involve negation at all but goes directly to an arbitrary $Q$.
Aug 6, 2013 at 4:20 comment added Noah Schweber @Thomas: exactly. As Andreas mentioned, precisely how much you have to get rid of is sensitive, and I certainly don't know what the "closest-to-classical" logic is that can avoid Russell's Paradox in naive set theory. That said, the article "Paraconsistent Set Theory" by Graham Priest (chapter 8 in "Logic, Mathematics, Philosophy: Vintage Enthusiasms Essays in honour of John L. Bell"; see link.springer.com/content/pdf/10.1007%2F978-94-007-0214-1_8.pdf) looks like a good introduction to this area.
Jul 30, 2013 at 6:42 comment added Thomas Benjamin @noah: when you speak of non-classical logics, do you include systems of logic where excluded middle doesn't hold? Russell's paradox (seemingly) obtains because one assumes excluded middle. The same (possibly) implies the means for extricating ones self from Curry's paradox as well...
Sep 16, 2012 at 1:27 comment added Andreas Blass @Zhen Lin: What you wrote depends sensitively on the details of the propositional deduction rules. Specifically, the obvious deduction of $\bot$ won't work in a system like Girard's linear logic. The problem is that $\phi\to(\phi\to\psi)$ doesn't give you $\phi\to\psi$ in such systems. (I vaguely recall that an attempt was made to use linear logic to circumvent Russell's paradox, but the resulting set theory was terribly weak; there may have been better attempts since then but I don't recall seeing any.)
Aug 30, 2012 at 11:36 comment added Zhen Lin @Noah: Russell's paradox still occurs with naïve comprehension even in very weak systems: we are basically given $R \notin R \leftrightarrow R \in R$, and to deduce $\bot$ we just need the deduction rules for $\land$ and $\to$.
Aug 29, 2012 at 23:27 comment added Goldstern When you write "This seems to me arguably in the spirit of restricting naive comprehension because comprehension is still the main set construction principle, and in particular there is no need for powerset or replacement.", you seem to imply that replacement and power set are NOT instances of naive comprehension. If so, please explain what you mean by "naive comprehension". If not, then "ZF minus Foundation and Extensionality" is a natural weakening of comprehension, guided by "limitation of size". Except that I think that any "natural" set theory needs some version of extensionality.
Aug 29, 2012 at 19:34 answer added Joel David Hamkins timeline score: 5
Aug 29, 2012 at 18:16 comment added Noah Schweber Probably not a fix you intend to consider, but you can also escape Russell's paradox by keeping full comprehension, and instead adopting a non-classical logic as your deductive framework. I tend to find this approach not very interesting, but it certainly can yield a non-trivial set theory which is very different from ZFC, NF, positive set theory, etc.
Aug 29, 2012 at 18:03 history asked user26062 CC BY-SA 3.0