Date: Jan 22, 2013 9:49 AM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:> On 22 Jan., 13:00, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>> "Jesse F. Hughes" <je...@phiwumbda.org> writes:>>>>>>>>>>>> > WM <mueck...@rz.fh-augsburg.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.>> So you know about something of d that is not in the union of its> finite initial segments? Yes, the God of matheology may provide the> answer, or, if not the answer, may at least press his followers to> believe that.Well, it's not a "union" in the usual sense, but let's let it pass.Here's the problem.  Let d(n) be the n'th digit of d, and similarlyb_k(n) the n'th digit of b_k.  Let d_k be the finite initial segmentof d consisting of the first k digits.  We can prove, as you say, that  (An)(d(n) != b_n(n)).From this, you claim that it follows that  (Ek)(An)(d_k != b_n)                                  (*)I simply do not see how that follows.  It certainly *is* true that  (Ak)(An < k)(d_k(n) != b_n(n)).But can you give me an argument that (*) is true?>>>> >> 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 understanding your Gods commands - at least I cannot understand why> you believe in that crap. But in mathematics, we have the following> case:>> A) $\mathbb{N} = \bigcup_{n=1}^\infty$ {n}> and also the union of all FISONs yields> B) $\mathbb{N} = \bigcup_{n=1}^\infty$ {1, 2, ..., n}> because it cannot be less than A.> Do you agree that the actually infinite path {1, 2, ...} does not> differ from B, such that "behind" every natural nothing can happen? In> particular the inclusion of {1, 2, ...} does not change anything in B?> Therefore  {1, 2, ...} cannot differ from the union B?I'm not at all sure what your question means, but I agree (of course)that the identity stated in (B) is correct.  The set N is the union ofall the sets {1,2,...,n} (using the convention that N's least elementis 1, of course).> By the way, every FISON, which stands for finite initial segment of> the naturals, is finite. This does not change by unioning as many as> are available. The union never gets larger than every FISON. Can you> understand that?Of course that claim is simply nonsense, but let's focus on the matterat hand.  Are you claiming that the "reason" that some d_k is missingfrom the list of b_n's is that  The union of any set of FISONs is a FISON?If so, then we can look at that latter claim, but if not, let's notget distracted.  -- Jesse F. Hughes"And I'm one of my own biggest skeptics as I had *YEARS* of wrongideas, and attempts that failed.  Worse, for some of them it took*MONTHS* before I figured out where I screwed up." -- James Harris