> On Saturday, 2 August 2014 22:54:24 UTC+2, Martin Shobe wrote: > >> Less unambiguously, you can show that every finite initial segment of N >> is not sufficient. > > And *if there is more in |N* than every finite initial segment, > then the limit shows that even this "more" is not sufficient.
In N there is every IFS+N - same way that there is in every proper segment in R there is another whole R.
> But who would confess his belief that |N contains more than every finite natural number (tantamount to every finite segment)? He would become ridiculous in mathematics because he cannot answer what this "more" should be.
More is just N because N = any IFS + N.
>> Therefore you must believe in something unmathematical. >> >> What's unmathematical about thinking that N isn't a finite initial >> segment of N. > > You are trying to cheat again and again.
And you are not trying BUT YOU ARE AN ASSHOLE.
> |N is not a finite initial segments but all finite initials segments or numbers.
N = ALL IFN's IN N PLUS N.
> What else?
As many as you want as long they are below the number of 2^N (power set).
>>> It follows from the proof that every natural numbers fails. Enough for a mathematician. >> > >> Go ahead and prove that "the rationals cannot be enumerated by the >> naturals" follows from "The number of unit intervals, each one >> containing infinitely many rationals without index =< n, increases >> infinitely". > > No problem. The fact already that you are trying to cheat would make every objective reader suspicious.
YOU ARE AN ASSHOLE.
>>> For that "proof" you have to assume that every is tantamount with all. This, however, is a very naive way of thinking that infinite sets can >> be exhausted like finite sets. > >> >> It's still proven. >> > It is proven to be very naive.
Because you are not.
> A proof is a convincing argument.
And it needs to be formally RIGHT.
> Your argument will not convince anybody > without matheological indoctrination > when being contrasted with mine.