equivalence class of equinumerous sets is not a set
Recall that two sets are equinumerous iff there is a bijection between them.
Let be a non-empty set, and the class of all sets equinumerous to . Then is a proper class.
since is in . Since , pick an element , and let . Then is a proper class, for otherwise would be the “set” of all sets, which is impossible. For each in , the set is in one-to-one correspondence with , with the bijection given by if , and . Therefore contains for every in the proper class . Furthermore, since whenever , we have that is a proper class as a result. ∎
|Title||equivalence class of equinumerous sets is not a set|
|Date of creation||2013-03-22 18:50:34|
|Last modified on||2013-03-22 18:50:34|
|Last modified by||CWoo (3771)|