
Re: ZFC and God
Posted:
Jan 22, 2013 7:00 AM


"Jesse F. Hughes" <jesse@phiwumbda.org> writes:
> WM <mueckenh@rz.fhaugsburg.de> writes: > >> On 21 Jan., 19:07, Zuhair <zaljo...@gmail.com> wrote: >> >>> Doesn't that say that mathematics following ZFC is only grounded in >>> Mythology driven principles! >>> >>> Doesn't that mean that ZFC based mathematics is too imaginary that >>> even if consistent still it is based and rooted in fantasy that cannot >>> really meet reality! >> >> ZFC is not consistent unless inconsistencies are defined to be no >> inconsistencies, distinctions need not be distinguishable, >> incomletenesses need not be incomplete, and so on. >> >> Consider, for instance, all terminating binary fractions b_n >> 0.0 >> 0.1 >> 0.00 >> 0.01 >> 0.10 >> 0.11 >> 0.000 >> where some numbers are represented twice (in fact each one appears >> infinitely often). Constructing the diagonal d we find that d differs >> from *every* b_n *at a finite place*. >> >> Since the above list is complete, which is possible because all >> terminating fractions, as a subset of all fractions, are countable, it >> is impossible that the diagonal differs from all entries b_n at a >> finite place. If this was possible, the list would have a gap, namely >> a finite initial segment of d. That means, the diagonal up to every >> bit can be found in the list. And after every finite place there is >> nothing that could distinguish two numbers. >> >> Therefore the diagonal does not increase the cardinal number of the >> listed entries b_n. > > This is your proof that ZF is inconsistent, is it? >> >> The diagonal may be infinitely long. But what does that mean? > > It means that d is not a terminating fraction, you moron, so d is > not part of the set of all terminal fractions and hence you haven't > shown that *this* enumeration is not surjective, much less that the > set of terminal fractions is uncountable.
I guess I overlooked the important bit of your argument. It was this:
it is impossible that the diagonal differs from all entries b_n at a finite place. If this was possible, the list would have a gap, namely a finite initial segment of d. That means, the diagonal up to every bit can be found in the list. And after every finite place there is nothing that could distinguish two numbers.
But there's nothing to support your claim that, if d is not in the list, then there is a finite initial segment of d not in the list.
>> Every given number of bits is surpassed. But the same holds for the >> entries of the list. The only difference could be a bit of the >> diagonal that has no finite index. But such bits are not part of >> mathematics and of Cantor's argument. > > You are incapable of very basic mathematical reasoning. You are a > shame to your school.  Jesse F. Hughes "It's easy folks. Just talk about my approach to your favorite mathematician. If they can't be interested in it, they've demonstrated a lack of mathematical skill."  James Harris

