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