CANTOR DISPROOF <<<<<<<<<<<<<<<< [Math]

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From: |-|ercules on 27 Jun 2010 08:44

"Sylvia Else" <sylvia(a)not.here.invalid> wrote
>
> You purport to extrapolate from all finite permutations to all infinite
> sequences.

The difference is I'm using finite prefix permutations. What would be good
is an axiom or theorem to assert

All finite prefix sequences + no constraint on the suffix -> all infinite sequences

Herc

From: Jim Burns on 27 Jun 2010 09:57

George Greene wrote:
> On Jun 26, 6:05 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>> It is perplexing if outputs of all computer programs are listed,
>> how do you find a program to compute the diagonal digit
>> at the position that is contradictory?
>
> You just write the program that says output(n) = 9 - L(n,n)
> AND YOU'RE DONE, VOILA,
> WHOOT, THERE IT IS!
> It is a VERY SIMPLE program.
> But there is no "contradictory" position.
> The number being computed simply IS NOT ON the list.

Consider the list L (which is in no way an attempt
to list all the reals),
.4999...
.0900...
.00900...
...
that is,
L(n) = { .499... , n = 1
{ 9/10^n , otherwise

Your program produces
output = .5000...

However, .5000... = .4999...
so the proposed anti-diagonal /is/ on this list.

The problem is not with Cantor's argument,
it is with your 9 - L(n,n). Use a better
digit-selecting function and it goes away.

Jim Burns

From: Jim Burns on 27 Jun 2010 10:23

Sylvia Else wrote:
> On 27/06/2010 4:34 PM, |-|ercules wrote:
>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>> If you think my induction only works on finite prefixes
>>>> and not over entire infinite expansions
>>>> then YOU prove that assertion.
>>>
>>> Consider the following proposition:
>>>
>>> For any *finite* list of infinite digit sequences,
>>> one can use the anti-diagonal method to produce
>>> a sequence that is not in the list.
>>>
>>> Do you have any difficulty with that?
>>
>> No. This is precisely my point.
>>
>> 123
>> 456
>> 789
>>
>> Diag = 159
>> Anti-Diag = 260
>>
>> 260 is a NEW DIGIT SEQUENCE.
>
> OK, so you accept that it is true for any finite list.
> That is, that for any finite list there is a sequence
> that is not in the list, or to put it another way,
> all finite lists omit at least one sequence.
> Note that the requirement that the length of the list
> be finite doesn't impose any maximum on the length.

Please pardon my interruption.

I would like to point out that you don't need to extrapolate
to the infinite case here, because every real on the list is at
a finite position.

If we assume that the anti-diagonal is on the infinite list,
then it must be on the list for some finite position n*.
If we then truncate the list after n*, we have a
finite list which (by the same argument) /we have agreed/
cannot have the anti-diagonal anywhere, including position n*.
Except that it does. Contradiction.

And so the anti-diagonal cannot be on the infinite list, either.

Jim Burns

> By analogy with your argument that extrapolates from a list that
> contains all finite permutations to a list that contains all infinite
> sequences, I'll argue that by extroplating from the fact that all finite
> lists omit at least one sequence one can conclude that an infinite list
> omits at least one sequence.
>
> The latter of course contradicts your thesis, but either extrapolating
> from the finite to the infinite is valid, or it isn't. Without some
> demonstration that the circumstances are materially different, you can't
> argue that the extrapolation is valid in one case, and invalid in the
> other.

From: |-|ercules on 27 Jun 2010 16:55

"Jim Burns" <burns.87(a)osu.edu> wrote ...
> Sylvia Else wrote:
>> On 27/06/2010 4:34 PM, |-|ercules wrote:
>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>>> If you think my induction only works on finite prefixes
>>>>> and not over entire infinite expansions
>>>>> then YOU prove that assertion.
>>>>
>>>> Consider the following proposition:
>>>>
>>>> For any *finite* list of infinite digit sequences,
>>>> one can use the anti-diagonal method to produce
>>>> a sequence that is not in the list.
>>>>
>>>> Do you have any difficulty with that?
>>>
>>> No. This is precisely my point.
>>>
>>> 123
>>> 456
>>> 789
>>>
>>> Diag = 159
>>> Anti-Diag = 260
>>>
>>> 260 is a NEW DIGIT SEQUENCE.
>>
>> OK, so you accept that it is true for any finite list.
>> That is, that for any finite list there is a sequence
>> that is not in the list, or to put it another way,
>> all finite lists omit at least one sequence.
>> Note that the requirement that the length of the list
>> be finite doesn't impose any maximum on the length.
>
> Please pardon my interruption.
>
> I would like to point out that you don't need to extrapolate
> to the infinite case here, because every real on the list is at
> a finite position.
>
> If we assume that the anti-diagonal is on the infinite list,
> then it must be on the list for some finite position n*.
> If we then truncate the list after n*, we have a
> finite list which (by the same argument) /we have agreed/
> cannot have the anti-diagonal anywhere, including position n*.
> Except that it does. Contradiction.
>
> And so the anti-diagonal cannot be on the infinite list, either.
>
> Jim Burns
>

Isn't Sylvia making a NON correct proof to illustrate a point about MY proof?
Works on all finite cases --/--> Works on infinite case

There is another "mid-way" example where you form a finite diagonal on the infinite list,
more akin to my induction of prefixes on an infinite string, which would not work in your case.

Herc

From: Sylvia Else on 27 Jun 2010 20:14

On 28/06/2010 6:55 AM, |-|ercules wrote:
> "Jim Burns" <burns.87(a)osu.edu> wrote ...
>> Sylvia Else wrote:
>>> On 27/06/2010 4:34 PM, |-|ercules wrote:
>>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>>>> If you think my induction only works on finite prefixes
>>>>>> and not over entire infinite expansions
>>>>>> then YOU prove that assertion.
>>>>>
>>>>> Consider the following proposition:
>>>>>
>>>>> For any *finite* list of infinite digit sequences,
>>>>> one can use the anti-diagonal method to produce
>>>>> a sequence that is not in the list.
>>>>>
>>>>> Do you have any difficulty with that?
>>>>
>>>> No. This is precisely my point.
>>>>
>>>> 123
>>>> 456
>>>> 789
>>>>
>>>> Diag = 159
>>>> Anti-Diag = 260
>>>>
>>>> 260 is a NEW DIGIT SEQUENCE.
>>>
>>> OK, so you accept that it is true for any finite list.
>>> That is, that for any finite list there is a sequence
>>> that is not in the list, or to put it another way,
>>> all finite lists omit at least one sequence.
>>> Note that the requirement that the length of the list
>>> be finite doesn't impose any maximum on the length.
>>
>> Please pardon my interruption.
>>
>> I would like to point out that you don't need to extrapolate
>> to the infinite case here, because every real on the list is at
>> a finite position.
>>
>> If we assume that the anti-diagonal is on the infinite list,
>> then it must be on the list for some finite position n*.
>> If we then truncate the list after n*, we have a
>> finite list which (by the same argument) /we have agreed/
>> cannot have the anti-diagonal anywhere, including position n*.
>> Except that it does. Contradiction.
>>
>> And so the anti-diagonal cannot be on the infinite list, either.
>>
>> Jim Burns
>>
>
>
> Isn't Sylvia making a NON correct proof to illustrate a point about MY
> proof?
> Works on all finite cases --/--> Works on infinite case

Yes.

>
> There is another "mid-way" example where you form a finite diagonal on
> the infinite list,
> more akin to my induction of prefixes on an infinite string, which would
> not work in your case.

I cannot for the life of me see how you regard that as more akin to your
induction. It looks to me as if you recognised the issue I raised, and
looked for a non-analagous failing case in an attempt to save your position.

Sylvia.