On Sep 11, 12:12 am, logic...@comcast.net wrote: > On Sep 10, 7:49 pm, William Hughes <wpihug...@hotmail.com> wrote:> On Sep 10, 10:37 pm, logic...@comcast.net wrote: > > [re] and z must be an element of B not in A. > > [wh] So if there is no element of B that is not in A > [wh] this fails. > > [re] It means F is not a bijection.
No, it doesn't. if F maps from A to B and F is injective and surjective then F is a bijection. Nothing about there being an element of B that is not in A.
> > [wh] What if A is a set such that there is a bijection from > [wh] A to a proper subset of A? > > [re] The point of the proof is to show such sets lead to > contradiction.
Using the assumption that such sets lead to a contradiction in order to prove that such sets lead to a contradiction is circular. Your assumption that z exists is equivalent to assuming that there is no set A such that there is a bijection from A to a proper subset of A.
