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


Re: Effect of gravitation in set theory
Posted:
Nov 22, 2011 6:49 PM


In article <39c1f9b1edc04d049b9a1d16b81595ae@e15g2000vba.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 22 Nov., 12:15, Jürgen R. <jurg...@arcor.de> wrote: > > "WM" <mueck...@rz.fhaugsburg.de> schrieb im > > Newsbeitragnews:45a3e1ecb2ed4741b9f0ca0b9e1131cb@y42g2000yqh.googlegroup > > s.com... > > > > > On 22 Nov., 10:58, J rgen R. <jurg...@arcor.de> wrote: > > >> "WM" <mueck...@rz.fhaugsburg.de> schrieb im > > >> Newsbeitragnews:a144bf83bead4793aca24a14c148c478@b32g2000yqn.googlegr > > >> oups.com... > > > > >> > Wrong. 1 is not the first necessary summand. > > >> > 81 = 40 + 41. > > > > >> Nor is it the second necessary summand, nor > > >> the third. Guess what  that's why I called it > > >> unnecessary! > > > > > Because you have no idea of linear sets > > > > Indeed, I don't. What is a linear set?
http://www.ams.org/journals/tran/194205202/S00029947194200066052/S0 0029947194200066052.pdf Definition 1.2. A linear set is a subset of an ndimensional vector space closed under vector subtraction and multiplication by arbitrary scalars. > > > > I suppose there are also nonlinear sets of > > natural numbers. Never heard of those either. > > Look up linear ordering and inclusion monotonic sequences of sets. > > > > > and do not understand that, > > > when going through the numbers one by one, 40 is the least element of > > > the set of natural numbers that are necessary to give sum 81. > > > > Well, in that case, since we have 80 = 28 + 27 + 26, we > > have proven the surprising fact that 40 is less than 27. > > You have not understood your own example? You wrote: > > Let us prove, by your logic, that the number 81 (I suppose you agree > that 81 exists) is not the sum of smaller integers: > a) 1 is not a necessary summand since 81 = 2 + 79 > > 1 to 39 obviously are not necessary, because 40 and 41 will do. But if > 40 is dropped, then it is impossible to have 81 as a sum of smaller > numbers. Therefore 40 is the smallest natural that is necessary to get > the sum 81. > > You tried to use induction but obviously did not understand that 39 is > not necessary does not imply 40 is not necessary. > > And now I am fed up with that silly example. You are invited to show > that my proof fails, but not by a wrong analogy, but by mathematics. > > I defined: The finite initial segment S_n = (1, ..., n) of N is > necessary in a union of finite initial segments that cover N, if > there is a natural k in S_n that is not in any S_m with m > n.
That is a corrupt definition. Any standard definition of "necessary" would say that S_k is necessary if there is no cover of N by a set of finite initial segments that does not include S_j as a member.
> > I proved by induction: S_1 is not necessary because of S_2. > S_n is not necessary because S_(n+1) will do. > > Hence, there is not more than one S_k necessary to cover N.
Nonsequitur!
You have not proved (and cannot because it is false) that NO SET of finite initial segments can cover N, But {S_n : n inN} is a set of finite initial segment that clearly covers N
> In > particular it is easy to see that not two or more S_k are necessary.
What is easy for WM to see also happens to be quite false.
One can easily show that no finite set of finite initial segments can succeed in covering N, but those less willfully blind than WM can also easily see that EVERY infinite set of finite initial segments necessarily covers N. At least in ZF.
What goes on in the matheology of Wolkenmuekenheim does not affect ZF. 

