3 ms·
If that set is well-defined, it obviously contains itself. Under ZFC, no set can contain itself, so that set is not well-defined. http://en.wikipedia.org/wiki/
by philh 17y ago
If that set is well-defined, it obviously contains itself. Under ZFC, no set can contain itself, so that set is not well-defined.
http://en.wikipedia.org/wiki/Axiom_of_regularity http://en.wikipedia.org/wiki/Axiom_of_regularity
But there may well be set theories where it is well-defined, hence does contain itself.
- gambling8nt 17y agoAs noted above, the collection of all sets cannot be a set in any set theory with specification, regardless of regularity: Call the collection of all sets S. We specify T by T=\{x| x \in S \wedge \neg x \in x\} T is then the subclass of S of sets that do not contain themselves. Thus with specifiction (provable from replacement, or as an axiom by itself), it is contradictory for S to be a set. If you reject the law of excluded middle, you can have a intuitionist set theory where S is neither a set nor not a set; alternatively, you can have a set theory without specification one constructed based on a type theory might meet this requirement.