From: |-|ercules on

Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings.
___________________________________________________________________________________________________


2 or 3 posters on sci.math have agreed with that statement.

BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers"
incase any of you want to feign comprehension disability.

Is it too much for you to fathom that I disagree whether no box containing the box numbers
that don't contain their own box numbers is self referential on some level? MAYBE, just MAYBE
there is another explanation than sets larger than infinity?

Do you believe there are numerous different digits (at finite positions) all along the expansion
of some reals that are not on the computable reals list?

What does that mean?

The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable"
which is a clear contradiction.

Herc
--
the nonexistence of a box that contains the numbers of all the boxes
that don't contain their own box number implies higher infinities.
- Cantor's Proof (the holy grail of paradise in mathematics)
From: hagman on
On 12 Jun., 12:29, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings.
> ___________________________________________________________________________________________________
>
> 2 or 3 posters on sci.math have agreed with that statement.
>
> BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers"
> incase any of you want to feign comprehension disability.
>
> Is it too much for you to fathom that I disagree whether no box containing the box numbers
> that don't contain their own box numbers is self referential on some level?  MAYBE, just MAYBE
> there is another explanation than sets larger than infinity?
>
> Do you believe there are numerous different digits (at finite positions) all along the expansion
> of some reals that are not on the computable reals list?
>
> What does that mean?
>
> The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable"
> which is a clear contradiction.
>
> Herc
> --
> the nonexistence of a box that contains the numbers of all the boxes
> that don't contain their own box number implies higher infinities.
> - Cantor's Proof (the holy grail of paradise in mathematics)

What's wrong with self-reference in a statement?
The set of all sets not containing themselves would be a self-
referential (aka. impredicative) definition and hence leads to a
contradiction (i.e. there cannot be an object satisfying the
definition).
OTOH, given a collection F of boxes, where each box has a number and
contains some numbers,
whether or not one of these boxes contains its own number is a
precisely defined predicate.
For a single box this can be checked because the number n of the box
can be determined by
assumption and for each number, including that n, it can be determined
by assumption, wehteer or not n is in the box.
Finally, it is by means of quantification also possible to talk about
whether or not the collection of F has the property that any or none
of its boxes contains its own number.
From: |-|ercules on
"hagman" <google(a)von-eitzen.de> wrote...
> On 12 Jun., 12:29, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>> Every possible digit sequence is computable to ALL (an INFINITE AMOUNT of) finite initial substrings.
>> ___________________________________________________________________________________________________
>>
>> 2 or 3 posters on sci.math have agreed with that statement.
>>
>> BTW: ALL (an INFINITE AMOUNT of) natural numbers are in "all natural numbers"
>> incase any of you want to feign comprehension disability.
>>
>> Is it too much for you to fathom that I disagree whether no box containing the box numbers
>> that don't contain their own box numbers is self referential on some level? MAYBE, just MAYBE
>> there is another explanation than sets larger than infinity?
>>
>> Do you believe there are numerous different digits (at finite positions) all along the expansion
>> of some reals that are not on the computable reals list?
>>
>> What does that mean?
>>
>> The only interpretation is "There is a finite substring between 2 digits (inclusive) that is not computable"
>> which is a clear contradiction.
>>
>> Herc
>> --
>> the nonexistence of a box that contains the numbers of all the boxes
>> that don't contain their own box number implies higher infinities.
>> - Cantor's Proof (the holy grail of paradise in mathematics)
>
> What's wrong with self-reference in a statement?
> The set of all sets not containing themselves would be a self-
> referential (aka. impredicative) definition and hence leads to a
> contradiction (i.e. there cannot be an object satisfying the
> definition).
> OTOH, given a collection F of boxes, where each box has a number and
> contains some numbers,
> whether or not one of these boxes contains its own number is a
> precisely defined predicate.
> For a single box this can be checked because the number n of the box
> can be determined by
> assumption and for each number, including that n, it can be determined
> by assumption, wehteer or not n is in the box.
> Finally, it is by means of quantification also possible to talk about
> whether or not the collection of F has the property that any or none
> of its boxes contains its own number.


What's wrong is the wacky conclusions mathematicians get every time
they self reference a collectic set and negate itself.

You only make a point with Russell's set by putting it in perspective. i.e. does it
mean that there is no set of all sets?

Why accept such a naively defined set at all? The smarter definition would be
"the set of all sets, that don't contain themselves, barring the set of all sets that
don't contain themselves". Obviously it can't contain itself and be viable, so
DECLARE THAT FACT. Collective sets come with an E&OE!

Herc
From: Colin on
On Jun 12, 7:21 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> [snip]

Why do you insist on flogging a dead horse? Even wikipedia writes

"Although the set of real numbers is uncountable, the set of
computable numbers is countable and thus almost all real numbers are
not computable. The computable numbers can be counted by assigning a
Gödel number to each Turing machine definition. This gives a function
from the naturals to the computable reals. Although the computable
numbers are an ordered field, the set of Gödel numbers corresponding
to computable numbers is not itself computably enumerable, because it
is not possible to effectively determine which Gödel numbers
correspond to Turing machines that produce computable reals. In order
to produce a computable real, a Turing machine must compute a total
function, but the corresponding decision problem is in Turing degree 0′
′. Thus Cantor's diagonal argument cannot be used to produce
uncountably many computable reals; at best, the reals formed from this
method will be uncomputable."

So, the computable reals are countable, and Cantor's diagonal argument
won't show they're uncountable. No one disputes this. Why do you keep
insisting that there are people who do dispute it and keep trying to
argue with them? You're arguing with people that don't exist.



From: |-|ercules on
"Colin" <colinpoakes(a)hotmail.com> wrote
>. Thus Cantor's diagonal argument cannot be used to produce
> uncountably many computable reals; at best, the reals formed from this
> method will be uncomputable."

OK, but you can make a superset including the computable reals
with the godel index.

It's not conceptually difficult to question Cantor's argument on a
hypothetical list. But there's little point going into detail, perhaps
if it was acknowledged modifying the diagonal results in a new
digit sequence that is not computable, then we could increase the
scope to work out the mechanics. But it wouldn't demonstrate
anything at this point, rather the opposite!

Herc