In article <b67bf90c-b332-4edc-a58c-720877e7c879@g39g2000pro.googlegroups.com>, Graham Cooper <grahamcooper7@gmail.com> wrote:
> On Aug 19, 9:52 am, SPQR <S...@roman.gov> wrote: > > In article > > <2dd21dc3-c759-46e3-a3ed-42493bf77...@q12g2000prf.googlegroups.com>, > > Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > > On Aug 18, 2:05 am, at1with0 <at1wi...@gmail.com> wrote: > > > > Just to reiterate/elaborate. > > > > You can prove that there is no function from N onto R which implies R > > > > is uncountable. You do not need a diagonal trick to prove it because > > > > you can prove that no function exists from a set onto its powerset in > > > > general. This is easy to prove in the finite case but the same proof > > > > works for the infinite case. The only thing left to show is that R is > > > > equipollent to the powerset of N. > > > > > HOLD IT RIGHT THERE! > > > > > Jim Burns and now you have INCORRECTLY STATED that if a proof works on > > > any example it works for an infinite set of examples. > > > > What they may have stated is that if it works on EVERY example, which it > > does, then it works on the whole set of examples however large that set > > of examples may be! > > > Right! This is an incorrect deduction! > > > Herc
