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 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