
Re: ZFC and God
Posted:
Jan 22, 2013 9:49 AM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 22 Jan., 13:00, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> "Jesse F. Hughes" <je...@phiwumbda.org> writes: >> >> >> >> >> >> > WM <mueck...@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. > > 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 similarly b_k(n) the n'th digit of b_k. Let d_k be the finite initial segment of 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 of all the sets {1,2,...,n} (using the convention that N's least element is 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 matter at hand. Are you claiming that the "reason" that some d_k is missing from 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 not get distracted.
 Jesse F. Hughes "And I'm one of my own biggest skeptics as I had *YEARS* of wrong ideas, and attempts that failed. Worse, for some of them it took *MONTHS* before I figured out where I screwed up."  James Harris

