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Regarding Russell’s paradox, its dual is also interesting: Consider the set D := { s | s ∈ s }, the set of sets that do contain themselves. Does D contain itsel
by layer8 7mo ago
Regarding Russell’s paradox, its dual is also interesting: Consider the set D := { s | s ∈ s }, the set of sets that do contain themselves. Does D contain itself? It might or it might not, neither causes a contradiction. Tnis shows that you don’t need an antinomy for a set comprehension to be ill-defined.
- xanderlewis 7mo agoWhy is it ill-defined? As you said, there's no contradiction. Also, in the usual ZF set theory, it's empty.
- Koshkin 7mo ago... and, as such, it doesn't contain itself!
- layer8 7mo agoIt’s ill-defined in the sense that it doesn’t uniquely define the set. There are at least two different sets that D could be (one containing it and one not containing it), hence the expression doesn’t denote a well-defined set. (*) The axioms of ZF do not allow to form that expression, so the set doesn’t exist in ZF. (*) This is from a universist view. In a pluralist view, one wouldn’t say that the fact of the matter of whether D contains itself or not is independent from naive set theory, and that there are set universes where it is the case and others where it isn’t. But I would hold that naive set theory starts from a universist view.
- jesuslop 7mo agoI think "Foundation" axiom F forbids your recursive set, and there are models of both core set theory satisfying either F or ¬F, so F is independent of core set theory (core -> not including F or ¬F). F is normally assumed in set theory, but Aczel has worked with "ill" founded (¬F) set theory models. Just as with the axiom of choice. No religion wars, just people pushed to be explicit with assumptions.