Date: Jan 24, 2013 7:36 AM Author: Jesse F. Hughes Subject: Re: ZFC and God WM <mueckenh@rz.fh-augsburg.de> writes:

> On 24 Jan., 12:36, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

>> WM <mueck...@rz.fh-augsburg.de> writes:

>> > On 23 Jan., 19:04, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

>>

>> >> > Understand the Binary Tree. After you will have understood it, you

>> >> > will understand, why it is important.

>>

>> >> No, let's first settle the point at hand.

>>

>> > I do it by means of tools that I choose without any censorship from

>> > your side.

>>

>> I'm not being unreasonable here. You say ZF is inconsistent. I want

>> to see whether you can indeed show that.

>>

>> So, I'd like to know what inconsistency you can show in ZF. You have

>> already said (don't let me put words in your mouth! Correct me if I'm

>> wrong) that you can show ZF proves

>>

>> U_n {1,...,n} is not infinite. (*)

>

> It is not actually infinite. The cardinality is not larger than every

> n.

>>

>> Since we know it also proves the negation of (*), this would settle

>> your claim.

>>

>> Now, I'd just like to see the proof of (*). Nothing else. Just show

>> me that proof and we'll discuss it.

>

> Ok.

Well, what you present below is *not* a proof of (*). You've changed

goals against. Nonetheless, I'll read this, since this is the

argument you presented earlier in the thread.

>

> 1) Certainly you agree that in ZF we have the set T of all terminating

> decimal fractions t_i of the reals in the unit interval, i.e., finite

> sequences of digits, indexed by the FISs {1,...,n}.

>

> 2) Certainly you agree that the set T is countable.

>

> 3) Certainly you agree that the set can be diagonalized.

Yes, but let's be perfectly clear here. What we have now is this:

a function t: N -> R such that, for all i in N, t(i) is a real

number with terminating decimal representation. We'll write this as

t_i and we'll write t_i(j) for the j'th digit of t_i (in its

terminating representation, which is, of course, unique).

Let d in R be defined by

d(j) = 7 if t_j(j) != 7

d(j) = 6 if t_j(j) = 6.

Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in

N.

Is this what you mean up 'til now?

> 4) Certainly you agree that, since all t_i = (t_i1, t_i2, ..., t_in)

> have only a finite, though not limnited, number n of digits, the

> diagonalization for every t_i yields a finite d_i =/= t_ii.

> (The i on the left hand side cannot be larger than the i on the right

> hand side. In other words, "the list" is a square. Up to every i it

> has same number of lines and columns. )

No idea what you mean by the parenthetical remark.

I do agree that d_i is defined for every i in N. In particular, (d_i)

is an infinite sequence of digits. Is this what you're claiming, too?

> So everything here happens among FISs. And d cannot be longer than

> every t_i. Nevertheless d_i differs from every t_ii. So we see that ZF

> proves the uncountability of a countable set.

You've lost me. I don't know what you mean when you say, "everything

here happens among FISs." And I'm also puzzled by the meaning of the

next sentence.

Here are some obvious things.

d(j) is defined for every j in N.

d(j) != 0 and d(j) != 9 for any j in N.

Hence the number d does not have a terminating decimal

representation.

Every t_i has a terminating decimal representation.

Hence, every t_i has a shorter representation than d.

In other words,

(A i in N)(the shortest representation of t_i is shorter than

the shortest representation of d)

This looks like I do *not* agree with your claim that "d cannot be

longer than every t_i". It clearly is longer, and I've seen nothing

in your argument that actually serves to prove your obviously mistaken

claim.

So, what did you mean when you said "d cannot be longer than every

t_i" and why do you think this?

--

Jesse F. Hughes

"NONE of you noticed that Bush's rise has been coincident with my

arguments with you knuckleheads?"

-- James S. Harris, on how sci.math is a Republican tool