Can one escape from the "mirror-image" of Russell's Paradox? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:30:49Z http://mathoverflow.net/feeds/question/21880 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21880/can-one-escape-from-the-mirror-image-of-russells-paradox Can one escape from the "mirror-image" of Russell's Paradox? Garabed Gulbenkian 2010-04-19T19:52:20Z 2010-04-21T02:31:05Z <p>Let s denote the "virtual set" (x:x is an element of x). Does there exist a set theory T (based on classical logic and not so far proved inconsistent) such that: (1) T allows the existence of all the sets in ZF as well as many infinite self-membered sets. (2) T allows s to exist as an actual set and provides an answer to the question "Is s an element of s?" without engendering any paradox.</p> http://mathoverflow.net/questions/21880/can-one-escape-from-the-mirror-image-of-russells-paradox/22004#22004 Answer by Carl Mummert for Can one escape from the "mirror-image" of Russell's Paradox? Carl Mummert 2010-04-21T01:52:16Z 2010-04-21T02:31:05Z <p>Yes, there is such a set theory. It is ZF minus foundation plus Aczel's "anti-foundation axiom". </p> <p>One reference for this system is a book-length set of lecture notes by Aczel, <a href="http://standish.stanford.edu/pdf/00000056.pdf" rel="nofollow">http://standish.stanford.edu/pdf/00000056.pdf</a></p> <p>In this theory, the collection of all sets that contain themselves is nonempty (in fact, it is a proper class) and contains elements that are themselves infinite. </p> <p><s>If you wanted the collection of all sets that contain themselves to itself be a set, you could try to restrict the anti-foundation axiom to a some particular definable set of graphs; it is usually stated for the class of all graphs.</s></p>