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


Re: Effect of gravitation in set theory
Posted:
Nov 24, 2011 8:23 PM


In article <a053f654a4af4c9e8fa35ebdcf774d84@14g2000yqo.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 23 Nov., 21:44, Virgil <vir...@ligriv.com> wrote: > > > > > I have proved that any infinite set of finite initial segments of N is > > a cover for N, even though WM claims otherwise, and WM has never been > > able to find any fault in that argument, > > I do not claim that your proof is false. > I claim that both proofs are correct.
You have not presented any proof, what you did present was an argument by magic.
> Hence the underlying theory is able to derive contradictory results.
Your "proof" is not a valid argument in ZF or ZFC or in any other standard set theory that does not limit sets to only finite sets. > > I proved that no S_k is necessary (and sufficient) to cober the actual > infinite set N. That implies that in no case more than one S_k is > necessary. And every S_k is finite.
While you did prove that no PARTICULAR S_k is required, but as the set of all but that one always is cover of N, your claim is of no importance.
In ZFC one can use the von Neumann model for N, in which case each member of N is also a finite initial segment of N, thus N is a cover of itself, so that any argument to the contrary sucks.
And WM's argument sucks anyway, for the reason shown above. > > How can you honestly claim that infinitely many insufficient thing are > sufficent?
Because it is so easy to see that they are! At least for those whose mathemticals vision is uncorrupted. > > However, my proof is correct with respect to mathematics.
But, as shown above, irrelevant. While no particular finite initial segment is necessary to cover N, the set of all such finite initial segements except one is always sufficient.
And it has also been shown that any infinite set of finite initial segments, even when missing you particular one, suffices to cover N.
Thus the truth of your claim in no way falsifies: "For a set of finite initial segments of N to cover N, it is necessary and sufficient that the set of finite initial segments be infinite."
And in no way interferes with N being (actually) infinite. 

