Date: May 4, 2013 10:51 PM Author: ross.finlayson@gmail.com Subject: Re: Matheology § 258 On May 4, 1:48 pm, Dan <dan.ms.ch...@gmail.com> wrote:

> You can't prove it's not on the list , but you can't prove it is on

> the list either (if you could prove it , you would match the number

> 0.111.... to one of your finite definitions (eg. 0.1111 ) on the

> list ).

>

> > Therefore there are no actually infinite sequences at all - and there

> > is no Cantor-list, other than finitely defined lists.

>

> False . Either there are only finite-sequence lists with with a finite

> number of terms ,

> or there are infinite lists with infinite terms .

>

> 0

> 0 1

> 0 1 1

> 0 1 1 1

> 0 1 1 1 1

> ............

>

> You can have a formula for the list , f(a,b) , that gives the b'th

> digit of the a'th number of the list .

> You're saying b must be bounded (no infinite sequences of digits) ,

> but a can be infinite (an infinite amount of finite sequences) .

> This is inconsistent . Why should f(a,b) be a valid list but f(b,a) be

> an invalid list?

>

> In other words , how come this :

>

> 0

> 0 1

> 0 1 1

> 0 1 1 1

> 0 1 1 1 1

> ............

> ............

> ............

>

> is valid

> and this :

>

> 0 0 0 0 0 ...

> 1 1 1 1 ...

> 1 1 1 ...

> 1 1 ...

> 1 ...

>

> is not valid? What's possible for the columns of a list must be

> possible for the rows . The same constraints must apply both to the

> indexes of the real numbers in your list (vertical indices) , and to

> the indexes of your digits (horizontal indices) .

>

> More specifically : you have a list of (FINITE sequences of digits) .

> Your list an INFINITE number of (horizontal FINITE sequences of

> digits) :

>

> vertically infinite / horizontally finite .

> 0

> 0 1

> 0 1 1

> 0 1 1 1

> 0 1 1 1 1

> ............

> ............

> ............

>

> Why can't my list have a INFINITE number of (vertical FINITE

> sequences of digits) ?

>

> vertically finite / horizontally infinite .

> 0 0 0 0 0 ...

> 1 1 1 1 ...

> 1 1 1 ...

> 1 1 ...

> 1 ...

>

> The same indexing method for members of a list is used for digits of a

> member.

> What works for the columns must work for the rows . What works for the

> vertical must work for the horizontal .

> If you have infinite members you must allow infinite digits in a

> member . If you have infinite digits in any member, you must allow

> infinite members .

Almost seems like: making indices N x N, has that for the one image,

each element is finite yet unbounded, while for the other it is

infinite. This basically gets into considering a copy or instance of

N, then another, has in some manner the sputnik of quantification or

here correlation among the two, that for the same specification, one

has finite and unbounded elements, the other finite and unbounded and

as well: infinite elements.

Basically that an omega-sequence is constructed, here with the n-

sequence of unary digits being a natural integer, the omega-sequence

is in N where: the complement of the correlate/diagonal is an element

of the language of 1*, that the language admits infinite elements when

there are the two copies or instances of N, where in N by itself,

that's just the successor.

Why would this be? It seems simple to talk about as many copies of N

as as convenient to model free variables of unbounded automata, but

one might consider that these still come from some reserve of

distinguishable elements, of all of them. Then to have that two

copies of N so correlated/scaffolded see the admission of infinite

elements, that could only be from the language 1*, where with one or

more copies not so correlated there are only finite if unbounded

elements, might be seen as a development along the lines of the

Russell element of the Russell set as sputnik of quantification (of

regular sets), the omega-sequence of sputnik of correlation (of free

natural variables).

Regards,

Ross Finlayson