"WM" <mueckenh@rz.fh-augsburg.de> wrote in message news:ee23e72e-206f-48c9-ac34-ad7b2046f299@c9g2000yqm.googlegroups.com... On 27 Mai, 23:15, Virgil <virg...@nowhere.com> wrote:
> > > > Of course there can be different models for N and there can be > > > > different names for the elements of N. But that does not matter. The > > > > natural numbers do not enter mathematics because someone "defines" > > > > them, names them, or makes models of them, but because the natural > > > > numbers are simply existing and mathematics is built upon them. > > > > But according to WM, no such thing as N can exist. > > > So WM wold throw out the naturals on which so much is built. > > > The question is not whether the complete set of all naturals exists. > > That is a question on which WM differs from the mainstream. > But WM cannot grant its existence in a given argument and then deny it > in the same argument and expect anyone to accept that sort of argument.
Grant the existence of two natural numbers m and n such that m/n = sqrt (2). Then falsify it. Grant the existence of a largest natural. Then falsify it. Grant the existence of a largest prime number. Then falsify it. Grant the existence of all natural numbers. Then falsify it.
All these proofs are proofs by contradiction.
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How do you falsify the existence of the set of all Natural numbers in ZF ? ZF includes an axiom of infinity, which pretty much directly guarantees that there is an infinite set of all finite ordinals. Assuming ZF is consistent, this is also known to be independent of the other ZF axioms. So unless you are arguing that ZF is inconsistent, I think you are going to have a lot of problems showing that there cannot be a set of all natural numbers.
