Virgil
Posts:
8,833
Registered:
1/6/11


Re: Effect of gravitation in set theory
Posted:
Nov 22, 2011 7:01 PM


In article <2c592489f9454bc4b0a3de1800674567@o9g2000vbc.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 22 Nov., 18:24, William Hughes <wpihug...@gmail.com> wrote: > > On Nov 22, 1:15 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > On 22 Nov., 13:45, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Nov 22, 8:00 am, WM <mueck...@rz.fhaugsburg.de> wrote: > > > > > > <snip> > > > > > > > ... 40 is the smallest natural that is necessary to get > > > > > the sum 81. > > > > > > 81=50+31. Must be some strange use of "necessary" that is only > > > > valid in Wolkenmuekenheim. > > > > <snip evasion> > > > > In Wolkenmuekenheim, "40 is necessary" does not mean that > > 40 has to be used. Strange place Wolkenmuekenheim. > > That is correct. But as I already said, JR's example is completely > missing the point. 1 + 80 = 81 and 40 + 41 = 81. Both pairs achieve > the same. > But the set S_3 = {1, 2, 3} is of no use whatsoever, to get N.
But the infiniteness of set of finite initial segments is both sufficient necessary to have a set of them covering N.
> I can show and I have shown that there are no two or more S_k doing > anything that would not be done by one S_k.
Then you have again "shown" something false because {S_k: k in N} must cover every k in N. > > Everybody who handwavingly claims that infinitely many S_k would be > necessary (or sufficient) is caught in matheology and unable to > understand proofs by induction.
So WM now claims that ther are some members of N not in any member of {S_k: k in N}. I challenge WM to find any such member of N \ {S_k: k in N}. 

