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

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From: Sylvia Else on 23 Jun 2010 04:21

On 23/06/2010 5:45 PM, Graham Cooper wrote:
> On Jun 23, 5:29 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>> On 23/06/2010 4:57 PM, Graham Cooper wrote:
>>
>>
>>
>>
>>
>>> On Jun 23, 4:46 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>> On 23/06/2010 4:37 PM, Graham Cooper wrote:
>>
>>>>> On Jun 23, 4:04 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>> On 23/06/2010 3:03 PM, Graham Cooper wrote:
>>
>>>>>>> On Jun 23, 3:00 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
>>>>>>>> On Jun 23, 2:57 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>
>>>>>>>>> On 23/06/2010 2:30 PM, Graham Cooper wrote:
>>
>>>>>>>>>> On Jun 23, 1:02 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
>>>>>>>>>>> On Jun 23, 12:56 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>
>>>>>>>>>>>> On 23/06/2010 12:45 PM, Graham Cooper wrote:
>>
>>>>>>>>>>>>> On Jun 23, 12:25 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>>>>>>>>>> On 23/06/2010 10:09 AM, Sylvia Else wrote:
>>
>>>>>>>>>>>>>>> On 22/06/2010 4:49 PM, Graham Cooper wrote:
>>
>>>>>>>>>>>>>>>> IN FACT
>>
>>>>>>>>>>>>>>>> 3 It takes 10^x reals to list every permutation of digits x digits
>>>>>>>>>>>>>>>> wide
>>>>>>>>>>>>>>>> So with infinite reals you can list Every permutation of digits
>>>>>>>>>>>>>>>> infinite digits wide.
>>
>>>>>>>>>>>>>>> That's just an assertion. Let's see your proof. You might think it's
>>>>>>>>>>>>>>> obvious, but in Maths, obvious doesn't count.
>>
>>>>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>>>>> Are you going to ignore this Herc? Let's see the colour of your money.
>>>>>>>>>>>>>> If you can prove it, do so.
>>
>>>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>>>> You agreed the width of all permutations approached oo
>>>>>>>>>>>>> since the list of reals is considered infinitely long
>>>>>>>>>>>>> your claim is that the limit does not equal the infinite case
>>
>>>>>>>>>>>> The width is not in question. What you have failed to prove is that
>>>>>>>>>>>> every permutation can be *listed*. Since that's the core issue in your
>>>>>>>>>>>> entire attack on Cantor, you cannot be allowed to get away with merely
>>>>>>>>>>>> asserting it. Prove it!
>>
>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>> Consider the list of computable reals.
>>
>>>>>>>>>>> Let w = the digit width of the largest set
>>>>>>>>>>> of complete permutations
>>
>>>>>>>>>>> assume w is finite
>>>>>>>>>>> there are 10 computable copies of the
>>>>>>>>>>> complete permutations of width w
>>>>>>>>>>> each ending in each of digits 0..9
>>>>>>>>>>> which generates a set larger than width w
>>>>>>>>>>> so finite w cannot be the maximum size
>>
>>>>>>>>>>> therefore w is infinite
>>
>>>>>>>>>>> Herc
>>
>>>>>>>>>> Where I say a sequence ends in a new
>>>>>>>>>> digit I meant that new digit is at position w+1
>>>>>>>>>> appended to the sequence
>>
>>>>>>>>>> Herc
>>
>>>>>>>>> And yet another proof that w is infinite when I'm clearly asking for a
>>>>>>>>> proof that every permutation can be *listed*.
>>
>>>>>>>>> Let me ask this as a direct question - are you of the opinion that
>>>>>>>>> infinite length implies listability?
>>
>>>>>>>>> Sylvia.
>>
>>>>>>>> Exactly what younare asking.
>>
>>>>>>>> Are you shifting the goals to whether an infinite list exists?
>>
>>>>>>>> Herc
>>
>>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
>>>>>>>> infinitely wide permutations on a countable list.
>>
>>>>>>> iPhones are difficult to type
>>
>>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
>>>>>>>> infinitely wide permutations on a countable list.
>>
>>>>>>> Exactly what you are asking .......
>>
>>>>>> It seemed a reasonable question. I ask for a proof of listability, and
>>>>>> you provide a proof that the width is infinite.
>>
>>>>>> Leaving that aside, perhaps you're under the impression that this
>>>>>> process, copied from another posting of yours
>>
>>>>>> ---
>>
>>>>>> Given a set of complete permutations w digits wide
>>
>>>>>> eg
>>
>>>>>> 00
>>>>>> 01
>>>>>> 10
>>>>>> 11
>>
>>>>>> make 2 copies and append each of 0,1
>>
>>>>>> 00+0
>>>>>> 01+0
>>>>>> 10+0
>>>>>> 11+0
>>
>>>>>> 00+1
>>>>>> 01+1
>>>>>> 10+1
>>>>>> 11+1
>>>>>> ----
>>
>>>>>> and extended indefinitely, ultimately lists all permutations.
>>
>>>>>> It's certainly an infinite list of permutations, but you haven't proved
>>>>>> that it contains all of them. Since it's infinite in length, you can't
>>>>>> go through them to check. Instead you need to identify an algorithm that
>>>>>> will allow you to take any permutation and determine, in finite time,
>>>>>> the finite number that defines its position in the list. With such an
>>>>>> algorithm you could then say that since you can identify the position in
>>>>>> the list of any permutation, the list must contain them all.
>>
>>>>>> With a list of rationals constructed using a diagonal method this is
>>>>>> straight forward.
>>
>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>> With a given rational expressed in decimal, you try multiplying it by
>>>>>> sucessively higher prime numbers until the result is an integer. Since
>>>>>> the original divisor must be finite, this will be achieved in finite
>>>>>> time. This gives you the two numbers that form the ratio. The number of
>>>>>> the position in the list is then just the number of steps through a
>>>>>> diagonal chart required to reach that pair of numbers (the red ones are
>>>>>> not counted, because the two numbers are not co-prime).
>>
>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>> Since it's obvious that any pair of numbers can be reached after a
>>>>>> finite number of steps, this proves that all the rationals are in the list.
>>
>>>>>> To prove that all permutations are in the list, you need to do something
>>>>>> similar. So far you haven't.
>>
>>>>>> Sylvia.
>>
>>>>> The proof I just gave to count all permutations of oo width
>>>>> is a different nature to counting all rationals.
>>
>>>> Except that you didn't actually give a proof.
>>
>>>> In a *very* informal sense, the set of numbers to be permuted at the
>>>> next digit grows faster than you're processing them - the end is forever
>>>> getting further and further away. So the task cannot be completed even
>>>> in infinite time. Now, this is hardly a mathematical proof that the
>>>> permutations cannot be listed, but it must at least give some pause.
>>
>>>>> Proving all permutations oo digits wide and proving it can be
>>>>> done on a countable list is the one same proof.
>>
>>>> So you say, but it's far from obvious. A proof of that would be nice,
>>>> and even in the absence of Cantro's work, no one, apart from you, would
>>>> be convinced without one.
>>
>>>> Sylvia.
>>
>>> No it's clearly obvious that's exactly what the proof does.
>>
>> It's not obvious to me, and it doesn't seem to have been obvious to any
>> other readers. Anyway, you should be able to break it down into smaller
>> steps so as to make the proof clear.
>>
>>
>>
>>> You seem to follow it proves the permutations are oo digits
>>> wide. Do you know that all permutations implies every
>>> digit sequence?
>>
>> Of course it does. But that's not the issue, the issue, for the
>> umpteenth time, is whether they can be listed.
>>
>> Are these not computable?
>>
>> There's certainly no reason to think they're computable. For a number to
>> be computable there has to be an algorithm that will, given n, in finite
>> time, provide you with the nth digit of the number.
>>
>> But the only way you have in this case of specifying which number you
>> want the digit of is to provide the number. An algorithm that provides
>> the nth digit when provided with an infinite sequence of digits
>> including the nth digit is hardly an algorithm for the purpose of the
>> definition of computable. If it were, then all numbers would be
>> computable by definition, and the definition would be useless.
>>
>> Sylvia.
>
> The proof gives an algorithm for generating a bigger permutation
> set from a smaller one.
>
> If a set of digit permutations are listed on the comp. Set of reals
> then an algorithm exists to duplicate that set 10 times and append
> digits 0..9 to each duplicate set.
>
> This constructs algorithmically a countable list that contains
> 'full permutation' oo digits wide.

You say it's countable, but you haven't explained how it assigns a
natural number to every sequence.

>
> If you don't follow that addresses your concern then you
> simply don't understand the proof.

Just saying it's countable hardly constitutes a proof. If you're ever to
prevail in your view, you'll need to provide a concrete irrefutable
proof. Not just your say-so.

>
> What do you expect mike and herzbert to fess up that the
> permutations are all listed. They'll stick to ad hom mathematics
> no matter what proof is presented.

Why? If you were right, and they believed you were right, what
motivation would they have for staying on what would eventually have to
be the losing side?

Sylvia.

From: Graham Cooper on 23 Jun 2010 04:31

On Jun 23, 6:21 pm, Sylvia Else <syl...(a)not.here.invalid> wrote:
> On 23/06/2010 5:45 PM, Graham Cooper wrote:
>
>
>
>
>
> > On Jun 23, 5:29 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >> On 23/06/2010 4:57 PM, Graham Cooper wrote:
>
> >>> On Jun 23, 4:46 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>> On 23/06/2010 4:37 PM, Graham Cooper wrote:
>
> >>>>> On Jun 23, 4:04 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>>>> On 23/06/2010 3:03 PM, Graham Cooper wrote:
>
> >>>>>>> On Jun 23, 3:00 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
> >>>>>>>> On Jun 23, 2:57 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>
> >>>>>>>>> On 23/06/2010 2:30 PM, Graham Cooper wrote:
>
> >>>>>>>>>> On Jun 23, 1:02 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
> >>>>>>>>>>> On Jun 23, 12:56 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>
> >>>>>>>>>>>> On 23/06/2010 12:45 PM, Graham Cooper wrote:
>
> >>>>>>>>>>>>> On Jun 23, 12:25 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>>>>>>>>>>>> On 23/06/2010 10:09 AM, Sylvia Else wrote:
>
> >>>>>>>>>>>>>>> On 22/06/2010 4:49 PM, Graham Cooper wrote:
>
> >>>>>>>>>>>>>>>> IN FACT
>
> >>>>>>>>>>>>>>>> 3 It takes 10^x reals to list every permutation of digits x digits
> >>>>>>>>>>>>>>>> wide
> >>>>>>>>>>>>>>>> So with infinite reals you can list Every permutation of digits
> >>>>>>>>>>>>>>>> infinite digits wide.
>
> >>>>>>>>>>>>>>> That's just an assertion. Let's see your proof. You might think it's
> >>>>>>>>>>>>>>> obvious, but in Maths, obvious doesn't count.
>
> >>>>>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>>>>> Are you going to ignore this Herc? Let's see the colour of your money.
> >>>>>>>>>>>>>> If you can prove it, do so.
>
> >>>>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>>>> You agreed the width of all permutations approached oo
> >>>>>>>>>>>>> since the list of reals is considered infinitely long
> >>>>>>>>>>>>> your claim is that the limit does not equal the infinite case
>
> >>>>>>>>>>>> The width is not in question. What you have failed to prove is that
> >>>>>>>>>>>> every permutation can be *listed*. Since that's the core issue in your
> >>>>>>>>>>>> entire attack on Cantor, you cannot be allowed to get away with merely
> >>>>>>>>>>>> asserting it. Prove it!
>
> >>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>> Consider the list of computable reals.
>
> >>>>>>>>>>> Let w = the digit width of the largest set
> >>>>>>>>>>> of complete permutations
>
> >>>>>>>>>>> assume w is finite
> >>>>>>>>>>> there are 10 computable copies of the
> >>>>>>>>>>> complete permutations of width w
> >>>>>>>>>>> each ending in each of digits 0..9
> >>>>>>>>>>> which generates a set larger than width w
> >>>>>>>>>>> so finite w cannot be the maximum size
>
> >>>>>>>>>>> therefore w is infinite
>
> >>>>>>>>>>> Herc
>
> >>>>>>>>>> Where I say a sequence ends in a new
> >>>>>>>>>> digit I meant that new digit is at position w+1
> >>>>>>>>>> appended to the sequence
>
> >>>>>>>>>> Herc
>
> >>>>>>>>> And yet another proof that w is infinite when I'm clearly asking for a
> >>>>>>>>> proof that every permutation can be *listed*.
>
> >>>>>>>>> Let me ask this as a direct question - are you of the opinion that
> >>>>>>>>> infinite length implies listability?
>
> >>>>>>>>> Sylvia.
>
> >>>>>>>> Exactly what younare asking.
>
> >>>>>>>> Are you shifting the goals to whether an infinite list exists?
>
> >>>>>>>> Herc
>
> >>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
> >>>>>>>> infinitely wide permutations on a countable list.
>
> >>>>>>> iPhones are difficult to type
>
> >>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
> >>>>>>>> infinitely wide permutations on a countable list.
>
> >>>>>>> Exactly what you are asking .......
>
> >>>>>> It seemed a reasonable question. I ask for a proof of listability, and
> >>>>>> you provide a proof that the width is infinite.
>
> >>>>>> Leaving that aside, perhaps you're under the impression that this
> >>>>>> process, copied from another posting of yours
>
> >>>>>> ---
>
> >>>>>> Given a set of complete permutations w digits wide
>
> >>>>>> eg
>
> >>>>>> 00
> >>>>>> 01
> >>>>>> 10
> >>>>>> 11
>
> >>>>>> make 2 copies and append each of 0,1
>
> >>>>>> 00+0
> >>>>>> 01+0
> >>>>>> 10+0
> >>>>>> 11+0
>
> >>>>>> 00+1
> >>>>>> 01+1
> >>>>>> 10+1
> >>>>>> 11+1
> >>>>>> ----
>
> >>>>>> and extended indefinitely, ultimately lists all permutations.
>
> >>>>>> It's certainly an infinite list of permutations, but you haven't proved
> >>>>>> that it contains all of them. Since it's infinite in length, you can't
> >>>>>> go through them to check. Instead you need to identify an algorithm that
> >>>>>> will allow you to take any permutation and determine, in finite time,
> >>>>>> the finite number that defines its position in the list. With such an
> >>>>>> algorithm you could then say that since you can identify the position in
> >>>>>> the list of any permutation, the list must contain them all.
>
> >>>>>> With a list of rationals constructed using a diagonal method this is
> >>>>>> straight forward.
>
> >>>>>>http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>
> >>>>>> With a given rational expressed in decimal, you try multiplying it by
> >>>>>> sucessively higher prime numbers until the result is an integer. Since
> >>>>>> the original divisor must be finite, this will be achieved in finite
> >>>>>> time. This gives you the two numbers that form the ratio. The number of
> >>>>>> the position in the list is then just the number of steps through a
> >>>>>> diagonal chart required to reach that pair of numbers (the red ones are
> >>>>>> not counted, because the two numbers are not co-prime).
>
> >>>>>>http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>
> >>>>>> Since it's obvious that any pair of numbers can be reached after a
> >>>>>> finite number of steps, this proves that all the rationals are in the list.
>
> >>>>>> To prove that all permutations are in the list, you need to do something
> >>>>>> similar. So far you haven't.
>
> >>>>>> Sylvia.
>
> >>>>> The proof I just gave to count all permutations of oo width
> >>>>> is a different nature to counting all rationals.
>
> >>>> Except that you didn't actually give a proof.
>
> >>>> In a *very* informal sense, the set of numbers to be permuted at the
> >>>> next digit grows faster than you're processing them - the end is forever
> >>>> getting further and further away. So the task cannot be completed even
> >>>> in infinite time. Now, this is hardly a mathematical proof that the
> >>>> permutations cannot be listed, but it must at least give some pause.
>
> >>>>> Proving all permutations oo digits wide and proving it can be
> >>>>> done on a countable list is the one same proof.
>
> >>>> So you say, but it's far from obvious. A proof of that would be nice,
> >>>> and even in the absence of Cantro's work, no one, apart from you, would
> >>>> be convinced without one.
>
> >>>> Sylvia.
>
> >>> No it's clearly obvious that's exactly what the proof does.
>
> >> It's not obvious to me, and it doesn't seem to have been obvious to any
> >> other readers. Anyway, you should be able to break it down into smaller
> >> steps so as to make the proof clear.
>
> >>> You seem to follow it proves the permutations are oo digits
> >>> wide. Do you know that all permutations implies every
> >>> digit sequence?
>
> >> Of course it does. But that's not the issue, the issue, for the
> >> umpteenth time, is whether they can be listed.
>
> >> Are these not computable?
>
> >> There's certainly no reason to think they're computable. For a number to
> >> be computable there has to be an algorithm that will, given n, in finite
> >> time, provide you with the nth digit of the number.
>
> >> But the only way you have in this case of specifying which number you
> >> want the digit of is to provide the number. An algorithm that provides
> >> the nth digit when provided with an infinite sequence of digits
> >> including the nth digit is hardly an algorithm for the purpose of the
> >> definition of computable. If it were, then all numbers would be
> >> computable by definition, and the definition would be useless.
>
> >> Sylvia.
>
> > The proof gives an algorithm for generating a bigger permutation
> > set from a smaller one.
>
> > If a set of digit permutations are listed on the comp. Set of reals
> > then an algorithm exists to duplicate that set 10 times and append
> > digits 0..9 to each duplicate set.
>
> > This constructs algorithmically a countable list that contains
> > 'full permutation' oo digits wide.
>
> You say it's countable, but you haven't explained how it assigns a
> natural number to every sequence.
>
>
>
> > If you don't follow that addresses your concern then you
> > simply don't understand the proof.
>
> Just saying it's countable hardly constitutes a proof. If you're ever to
> prevail in your view, you'll need to provide a concrete irrefutable
> proof. Not just your say-so.
>
>
>
> > What do you expect mike and herzbert to fess up that the
> > permutations are all listed. They'll stick to ad hom mathematics
> > no matter what proof is presented.
>
> Why? If you were right, and they believed you were right, what
> motivation would they have for staying on what would eventually have to
> be the losing side?
>
> Sylvia.

In your words what does the proof establish?

Nobody else is disputing the proof.

Sci.math usually take 24 hours to comment on new ideas.

Herc

From: Graham Cooper on 23 Jun 2010 04:34

On Jun 23, 6:21 pm, Sylvia Else <syl...(a)not.here.invalid> wrote:
> On 23/06/2010 5:45 PM, Graham Cooper wrote:
>
>
>
>
>
> > On Jun 23, 5:29 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >> On 23/06/2010 4:57 PM, Graham Cooper wrote:
>
> >>> On Jun 23, 4:46 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>> On 23/06/2010 4:37 PM, Graham Cooper wrote:
>
> >>>>> On Jun 23, 4:04 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>>>> On 23/06/2010 3:03 PM, Graham Cooper wrote:
>
> >>>>>>> On Jun 23, 3:00 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
> >>>>>>>> On Jun 23, 2:57 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>
> >>>>>>>>> On 23/06/2010 2:30 PM, Graham Cooper wrote:
>
> >>>>>>>>>> On Jun 23, 1:02 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
> >>>>>>>>>>> On Jun 23, 12:56 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>
> >>>>>>>>>>>> On 23/06/2010 12:45 PM, Graham Cooper wrote:
>
> >>>>>>>>>>>>> On Jun 23, 12:25 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>>>>>>>>>>>> On 23/06/2010 10:09 AM, Sylvia Else wrote:
>
> >>>>>>>>>>>>>>> On 22/06/2010 4:49 PM, Graham Cooper wrote:
>
> >>>>>>>>>>>>>>>> IN FACT
>
> >>>>>>>>>>>>>>>> 3 It takes 10^x reals to list every permutation of digits x digits
> >>>>>>>>>>>>>>>> wide
> >>>>>>>>>>>>>>>> So with infinite reals you can list Every permutation of digits
> >>>>>>>>>>>>>>>> infinite digits wide.
>
> >>>>>>>>>>>>>>> That's just an assertion. Let's see your proof. You might think it's
> >>>>>>>>>>>>>>> obvious, but in Maths, obvious doesn't count.
>
> >>>>>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>>>>> Are you going to ignore this Herc? Let's see the colour of your money.
> >>>>>>>>>>>>>> If you can prove it, do so.
>
> >>>>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>>>> You agreed the width of all permutations approached oo
> >>>>>>>>>>>>> since the list of reals is considered infinitely long
> >>>>>>>>>>>>> your claim is that the limit does not equal the infinite case
>
> >>>>>>>>>>>> The width is not in question. What you have failed to prove is that
> >>>>>>>>>>>> every permutation can be *listed*. Since that's the core issue in your
> >>>>>>>>>>>> entire attack on Cantor, you cannot be allowed to get away with merely
> >>>>>>>>>>>> asserting it. Prove it!
>
> >>>>>>>>>>>> Sylvia.
>
> >>>>>>>>>>> Consider the list of computable reals.
>
> >>>>>>>>>>> Let w = the digit width of the largest set
> >>>>>>>>>>> of complete permutations
>
> >>>>>>>>>>> assume w is finite
> >>>>>>>>>>> there are 10 computable copies of the
> >>>>>>>>>>> complete permutations of width w
> >>>>>>>>>>> each ending in each of digits 0..9
> >>>>>>>>>>> which generates a set larger than width w
> >>>>>>>>>>> so finite w cannot be the maximum size
>
> >>>>>>>>>>> therefore w is infinite
>
> >>>>>>>>>>> Herc
>
> >>>>>>>>>> Where I say a sequence ends in a new
> >>>>>>>>>> digit I meant that new digit is at position w+1
> >>>>>>>>>> appended to the sequence
>
> >>>>>>>>>> Herc
>
> >>>>>>>>> And yet another proof that w is infinite when I'm clearly asking for a
> >>>>>>>>> proof that every permutation can be *listed*.
>
> >>>>>>>>> Let me ask this as a direct question - are you of the opinion that
> >>>>>>>>> infinite length implies listability?
>
> >>>>>>>>> Sylvia.
>
> >>>>>>>> Exactly what younare asking.
>
> >>>>>>>> Are you shifting the goals to whether an infinite list exists?
>
> >>>>>>>> Herc
>
> >>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
> >>>>>>>> infinitely wide permutations on a countable list.
>
> >>>>>>> iPhones are difficult to type
>
> >>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
> >>>>>>>> infinitely wide permutations on a countable list.
>
> >>>>>>> Exactly what you are asking .......
>
> >>>>>> It seemed a reasonable question. I ask for a proof of listability, and
> >>>>>> you provide a proof that the width is infinite.
>
> >>>>>> Leaving that aside, perhaps you're under the impression that this
> >>>>>> process, copied from another posting of yours
>
> >>>>>> ---
>
> >>>>>> Given a set of complete permutations w digits wide
>
> >>>>>> eg
>
> >>>>>> 00
> >>>>>> 01
> >>>>>> 10
> >>>>>> 11
>
> >>>>>> make 2 copies and append each of 0,1
>
> >>>>>> 00+0
> >>>>>> 01+0
> >>>>>> 10+0
> >>>>>> 11+0
>
> >>>>>> 00+1
> >>>>>> 01+1
> >>>>>> 10+1
> >>>>>> 11+1
> >>>>>> ----
>
> >>>>>> and extended indefinitely, ultimately lists all permutations.
>
> >>>>>> It's certainly an infinite list of permutations, but you haven't proved
> >>>>>> that it contains all of them. Since it's infinite in length, you can't
> >>>>>> go through them to check. Instead you need to identify an algorithm that
> >>>>>> will allow you to take any permutation and determine, in finite time,
> >>>>>> the finite number that defines its position in the list. With such an
> >>>>>> algorithm you could then say that since you can identify the position in
> >>>>>> the list of any permutation, the list must contain them all.
>
> >>>>>> With a list of rationals constructed using a diagonal method this is
> >>>>>> straight forward.
>
> >>>>>>http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>
> >>>>>> With a given rational expressed in decimal, you try multiplying it by
> >>>>>> sucessively higher prime numbers until the result is an integer. Since
> >>>>>> the original divisor must be finite, this will be achieved in finite
> >>>>>> time. This gives you the two numbers that form the ratio. The number of
> >>>>>> the position in the list is then just the number of steps through a
> >>>>>> diagonal chart required to reach that pair of numbers (the red ones are
> >>>>>> not counted, because the two numbers are not co-prime).
>
> >>>>>>http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>
> >>>>>> Since it's obvious that any pair of numbers can be reached after a
> >>>>>> finite number of steps, this proves that all the rationals are in the list.
>
> >>>>>> To prove that all permutations are in the list, you need to do something
> >>>>>> similar. So far you haven't.
>
> >>>>>> Sylvia.
>
> >>>>> The proof I just gave to count all permutations of oo width
> >>>>> is a different nature to counting all rationals.
>
> >>>> Except that you didn't actually give a proof.
>
> >>>> In a *very* informal sense, the set of numbers to be permuted at the
> >>>> next digit grows faster than you're processing them - the end is forever
> >>>> getting further and further away. So the task cannot be completed even
> >>>> in infinite time. Now, this is hardly a mathematical proof that the
> >>>> permutations cannot be listed, but it must at least give some pause.
>
> >>>>> Proving all permutations oo digits wide and proving it can be
> >>>>> done on a countable list is the one same proof.
>
> >>>> So you say, but it's far from obvious. A proof of that would be nice,
> >>>> and even in the absence of Cantro's work, no one, apart from you, would
> >>>> be convinced without one.
>
> >>>> Sylvia.
>
> >>> No it's clearly obvious that's exactly what the proof does.
>
> >> It's not obvious to me, and it doesn't seem to have been obvious to any
> >> other readers. Anyway, you should be able to break it down into smaller
> >> steps so as to make the proof clear.
>
> >>> You seem to follow it proves the permutations are oo digits
> >>> wide. Do you know that all permutations implies every
> >>> digit sequence?
>
> >> Of course it does. But that's not the issue, the issue, for the
> >> umpteenth time, is whether they can be listed.
>
> >> Are these not computable?
>
> >> There's certainly no reason to think they're computable. For a number to
> >> be computable there has to be an algorithm that will, given n, in finite
> >> time, provide you with the nth digit of the number.
>
> >> But the only way you have in this case of specifying which number you
> >> want the digit of is to provide the number. An algorithm that provides
> >> the nth digit when provided with an infinite sequence of digits
> >> including the nth digit is hardly an algorithm for the purpose of the
> >> definition of computable. If it were, then all numbers would be
> >> computable by definition, and the definition would be useless.
>
> >> Sylvia.
>
> > The proof gives an algorithm for generating a bigger permutation
> > set from a smaller one.
>
> > If a set of digit permutations are listed on the comp. Set of reals
> > then an algorithm exists to duplicate that set 10 times and append
> > digits 0..9 to each duplicate set.
>
> > This constructs algorithmically a countable list that contains
> > 'full permutation' oo digits wide.
>
> You say it's countable, but you haven't explained how it assigns a
> natural number to every sequence.

Are you going on the tangent that computable reals can't
be listed tangent?

Herc

From: Sylvia Else on 23 Jun 2010 04:41

On 23/06/2010 6:31 PM, Graham Cooper wrote:
> On Jun 23, 6:21 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:

>>>>>>>> Leaving that aside, perhaps you're under the impression that this
>>>>>>>> process, copied from another posting of yours
>>
>>>>>>>> ---
>>
>>>>>>>> Given a set of complete permutations w digits wide
>>
>>>>>>>> eg
>>
>>>>>>>> 00
>>>>>>>> 01
>>>>>>>> 10
>>>>>>>> 11
>>
>>>>>>>> make 2 copies and append each of 0,1
>>
>>>>>>>> 00+0
>>>>>>>> 01+0
>>>>>>>> 10+0
>>>>>>>> 11+0
>>
>>>>>>>> 00+1
>>>>>>>> 01+1
>>>>>>>> 10+1
>>>>>>>> 11+1
>>>>>>>> ----
>>
>>>>>>>> and extended indefinitely, ultimately lists all permutations.
>>
>>>>>>>> It's certainly an infinite list of permutations, but you haven't proved
>>>>>>>> that it contains all of them. Since it's infinite in length, you can't
>>>>>>>> go through them to check. Instead you need to identify an algorithm that
>>>>>>>> will allow you to take any permutation and determine, in finite time,
>>>>>>>> the finite number that defines its position in the list. With such an
>>>>>>>> algorithm you could then say that since you can identify the position in
>>>>>>>> the list of any permutation, the list must contain them all.
>>
>>>>>>>> With a list of rationals constructed using a diagonal method this is
>>>>>>>> straight forward.
>>
>>>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>>>> With a given rational expressed in decimal, you try multiplying it by
>>>>>>>> sucessively higher prime numbers until the result is an integer. Since
>>>>>>>> the original divisor must be finite, this will be achieved in finite
>>>>>>>> time. This gives you the two numbers that form the ratio. The number of
>>>>>>>> the position in the list is then just the number of steps through a
>>>>>>>> diagonal chart required to reach that pair of numbers (the red ones are
>>>>>>>> not counted, because the two numbers are not co-prime).
>>
>>>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>>>> Since it's obvious that any pair of numbers can be reached after a
>>>>>>>> finite number of steps, this proves that all the rationals are in the list.
>>
>>>>>>>> To prove that all permutations are in the list, you need to do something
>>>>>>>> similar. So far you haven't.
>>
>>>>>>>> Sylvia.
>>
>>>>>>> The proof I just gave to count all permutations of oo width
>>>>>>> is a different nature to counting all rationals.
>>
>>>>>> Except that you didn't actually give a proof.
>>
>>>>>> In a *very* informal sense, the set of numbers to be permuted at the
>>>>>> next digit grows faster than you're processing them - the end is forever
>>>>>> getting further and further away. So the task cannot be completed even
>>>>>> in infinite time. Now, this is hardly a mathematical proof that the
>>>>>> permutations cannot be listed, but it must at least give some pause.
>>
>>>>>>> Proving all permutations oo digits wide and proving it can be
>>>>>>> done on a countable list is the one same proof.
>>
>>>>>> So you say, but it's far from obvious. A proof of that would be nice,
>>>>>> and even in the absence of Cantro's work, no one, apart from you, would
>>>>>> be convinced without one.
>>
>>>>>> Sylvia.
>>
>>>>> No it's clearly obvious that's exactly what the proof does.
>>
>>>> It's not obvious to me, and it doesn't seem to have been obvious to any
>>>> other readers. Anyway, you should be able to break it down into smaller
>>>> steps so as to make the proof clear.
>>
>>>>> You seem to follow it proves the permutations are oo digits
>>>>> wide. Do you know that all permutations implies every
>>>>> digit sequence?
>>
>>>> Of course it does. But that's not the issue, the issue, for the
>>>> umpteenth time, is whether they can be listed.
>>
>>>> Are these not computable?
>>
>>>> There's certainly no reason to think they're computable. For a number to
>>>> be computable there has to be an algorithm that will, given n, in finite
>>>> time, provide you with the nth digit of the number.
>>
>>>> But the only way you have in this case of specifying which number you
>>>> want the digit of is to provide the number. An algorithm that provides
>>>> the nth digit when provided with an infinite sequence of digits
>>>> including the nth digit is hardly an algorithm for the purpose of the
>>>> definition of computable. If it were, then all numbers would be
>>>> computable by definition, and the definition would be useless.
>>
>>>> Sylvia.
>>
>>> The proof gives an algorithm for generating a bigger permutation
>>> set from a smaller one.
>>
>>> If a set of digit permutations are listed on the comp. Set of reals
>>> then an algorithm exists to duplicate that set 10 times and append
>>> digits 0..9 to each duplicate set.
>>
>>> This constructs algorithmically a countable list that contains
>>> 'full permutation' oo digits wide.
>>
>> You say it's countable, but you haven't explained how it assigns a
>> natural number to every sequence.
>>
>>
>>
>>> If you don't follow that addresses your concern then you
>>> simply don't understand the proof.
>>
>> Just saying it's countable hardly constitutes a proof. If you're ever to
>> prevail in your view, you'll need to provide a concrete irrefutable
>> proof. Not just your say-so.
>>
>>
>>
>>> What do you expect mike and herzbert to fess up that the
>>> permutations are all listed. They'll stick to ad hom mathematics
>>> no matter what proof is presented.
>>
>> Why? If you were right, and they believed you were right, what
>> motivation would they have for staying on what would eventually have to
>> be the losing side?
>>
>> Sylvia.
>
>
> In your words what does the proof establish?

That the permutations of an infinite number of digits has infinite width.

And that is all. It does *not* prove that the set of all permutations
can be listed, or, equivalently, is countable.

>
> Nobody else is disputing the proof.

Maybe no one else is reading it either.

>
> Sci.math usually take 24 hours to comment on new ideas.

New? I gather you've been on about this for years.

Sylvia.

From: Sylvia Else on 23 Jun 2010 04:47

On 23/06/2010 6:34 PM, Graham Cooper wrote:
> On Jun 23, 6:21 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>> On 23/06/2010 5:45 PM, Graham Cooper wrote:
>>
>>
>>
>>
>>
>>> On Jun 23, 5:29 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>> On 23/06/2010 4:57 PM, Graham Cooper wrote:
>>
>>>>> On Jun 23, 4:46 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>> On 23/06/2010 4:37 PM, Graham Cooper wrote:
>>
>>>>>>> On Jun 23, 4:04 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>>>> On 23/06/2010 3:03 PM, Graham Cooper wrote:
>>
>>>>>>>>> On Jun 23, 3:00 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
>>>>>>>>>> On Jun 23, 2:57 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>
>>>>>>>>>>> On 23/06/2010 2:30 PM, Graham Cooper wrote:
>>
>>>>>>>>>>>> On Jun 23, 1:02 pm, Graham Cooper<grahamcoop...(a)gmail.com> wrote:
>>>>>>>>>>>>> On Jun 23, 12:56 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>
>>>>>>>>>>>>>> On 23/06/2010 12:45 PM, Graham Cooper wrote:
>>
>>>>>>>>>>>>>>> On Jun 23, 12:25 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>>>>>>>>>>>> On 23/06/2010 10:09 AM, Sylvia Else wrote:
>>
>>>>>>>>>>>>>>>>> On 22/06/2010 4:49 PM, Graham Cooper wrote:
>>
>>>>>>>>>>>>>>>>>> IN FACT
>>
>>>>>>>>>>>>>>>>>> 3 It takes 10^x reals to list every permutation of digits x digits
>>>>>>>>>>>>>>>>>> wide
>>>>>>>>>>>>>>>>>> So with infinite reals you can list Every permutation of digits
>>>>>>>>>>>>>>>>>> infinite digits wide.
>>
>>>>>>>>>>>>>>>>> That's just an assertion. Let's see your proof. You might think it's
>>>>>>>>>>>>>>>>> obvious, but in Maths, obvious doesn't count.
>>
>>>>>>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>>>>>>> Are you going to ignore this Herc? Let's see the colour of your money.
>>>>>>>>>>>>>>>> If you can prove it, do so.
>>
>>>>>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>>>>>> You agreed the width of all permutations approached oo
>>>>>>>>>>>>>>> since the list of reals is considered infinitely long
>>>>>>>>>>>>>>> your claim is that the limit does not equal the infinite case
>>
>>>>>>>>>>>>>> The width is not in question. What you have failed to prove is that
>>>>>>>>>>>>>> every permutation can be *listed*. Since that's the core issue in your
>>>>>>>>>>>>>> entire attack on Cantor, you cannot be allowed to get away with merely
>>>>>>>>>>>>>> asserting it. Prove it!
>>
>>>>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>>>>> Consider the list of computable reals.
>>
>>>>>>>>>>>>> Let w = the digit width of the largest set
>>>>>>>>>>>>> of complete permutations
>>
>>>>>>>>>>>>> assume w is finite
>>>>>>>>>>>>> there are 10 computable copies of the
>>>>>>>>>>>>> complete permutations of width w
>>>>>>>>>>>>> each ending in each of digits 0..9
>>>>>>>>>>>>> which generates a set larger than width w
>>>>>>>>>>>>> so finite w cannot be the maximum size
>>
>>>>>>>>>>>>> therefore w is infinite
>>
>>>>>>>>>>>>> Herc
>>
>>>>>>>>>>>> Where I say a sequence ends in a new
>>>>>>>>>>>> digit I meant that new digit is at position w+1
>>>>>>>>>>>> appended to the sequence
>>
>>>>>>>>>>>> Herc
>>
>>>>>>>>>>> And yet another proof that w is infinite when I'm clearly asking for a
>>>>>>>>>>> proof that every permutation can be *listed*.
>>
>>>>>>>>>>> Let me ask this as a direct question - are you of the opinion that
>>>>>>>>>>> infinite length implies listability?
>>
>>>>>>>>>>> Sylvia.
>>
>>>>>>>>>> Exactly what younare asking.
>>
>>>>>>>>>> Are you shifting the goals to whether an infinite list exists?
>>
>>>>>>>>>> Herc
>>
>>>>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
>>>>>>>>>> infinitely wide permutations on a countable list.
>>
>>>>>>>>> iPhones are difficult to type
>>
>>>>>>>>>> You're a nutter Sylvia. I gave a procedure for iterating
>>>>>>>>>> infinitely wide permutations on a countable list.
>>
>>>>>>>>> Exactly what you are asking .......
>>
>>>>>>>> It seemed a reasonable question. I ask for a proof of listability, and
>>>>>>>> you provide a proof that the width is infinite.
>>
>>>>>>>> Leaving that aside, perhaps you're under the impression that this
>>>>>>>> process, copied from another posting of yours
>>
>>>>>>>> ---
>>
>>>>>>>> Given a set of complete permutations w digits wide
>>
>>>>>>>> eg
>>
>>>>>>>> 00
>>>>>>>> 01
>>>>>>>> 10
>>>>>>>> 11
>>
>>>>>>>> make 2 copies and append each of 0,1
>>
>>>>>>>> 00+0
>>>>>>>> 01+0
>>>>>>>> 10+0
>>>>>>>> 11+0
>>
>>>>>>>> 00+1
>>>>>>>> 01+1
>>>>>>>> 10+1
>>>>>>>> 11+1
>>>>>>>> ----
>>
>>>>>>>> and extended indefinitely, ultimately lists all permutations.
>>
>>>>>>>> It's certainly an infinite list of permutations, but you haven't proved
>>>>>>>> that it contains all of them. Since it's infinite in length, you can't
>>>>>>>> go through them to check. Instead you need to identify an algorithm that
>>>>>>>> will allow you to take any permutation and determine, in finite time,
>>>>>>>> the finite number that defines its position in the list. With such an
>>>>>>>> algorithm you could then say that since you can identify the position in
>>>>>>>> the list of any permutation, the list must contain them all.
>>
>>>>>>>> With a list of rationals constructed using a diagonal method this is
>>>>>>>> straight forward.
>>
>>>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>>>> With a given rational expressed in decimal, you try multiplying it by
>>>>>>>> sucessively higher prime numbers until the result is an integer. Since
>>>>>>>> the original divisor must be finite, this will be achieved in finite
>>>>>>>> time. This gives you the two numbers that form the ratio. The number of
>>>>>>>> the position in the list is then just the number of steps through a
>>>>>>>> diagonal chart required to reach that pair of numbers (the red ones are
>>>>>>>> not counted, because the two numbers are not co-prime).
>>
>>>>>>>> http://en.wikipedia.org/wiki/File:Diagonal_argument.svg
>>
>>>>>>>> Since it's obvious that any pair of numbers can be reached after a
>>>>>>>> finite number of steps, this proves that all the rationals are in the list.
>>
>>>>>>>> To prove that all permutations are in the list, you need to do something
>>>>>>>> similar. So far you haven't.
>>
>>>>>>>> Sylvia.
>>
>>>>>>> The proof I just gave to count all permutations of oo width
>>>>>>> is a different nature to counting all rationals.
>>
>>>>>> Except that you didn't actually give a proof.
>>
>>>>>> In a *very* informal sense, the set of numbers to be permuted at the
>>>>>> next digit grows faster than you're processing them - the end is forever
>>>>>> getting further and further away. So the task cannot be completed even
>>>>>> in infinite time. Now, this is hardly a mathematical proof that the
>>>>>> permutations cannot be listed, but it must at least give some pause.
>>
>>>>>>> Proving all permutations oo digits wide and proving it can be
>>>>>>> done on a countable list is the one same proof.
>>
>>>>>> So you say, but it's far from obvious. A proof of that would be nice,
>>>>>> and even in the absence of Cantro's work, no one, apart from you, would
>>>>>> be convinced without one.
>>
>>>>>> Sylvia.
>>
>>>>> No it's clearly obvious that's exactly what the proof does.
>>
>>>> It's not obvious to me, and it doesn't seem to have been obvious to any
>>>> other readers. Anyway, you should be able to break it down into smaller
>>>> steps so as to make the proof clear.
>>
>>>>> You seem to follow it proves the permutations are oo digits
>>>>> wide. Do you know that all permutations implies every
>>>>> digit sequence?
>>
>>>> Of course it does. But that's not the issue, the issue, for the
>>>> umpteenth time, is whether they can be listed.
>>
>>>> Are these not computable?
>>
>>>> There's certainly no reason to think they're computable. For a number to
>>>> be computable there has to be an algorithm that will, given n, in finite
>>>> time, provide you with the nth digit of the number.
>>
>>>> But the only way you have in this case of specifying which number you
>>>> want the digit of is to provide the number. An algorithm that provides
>>>> the nth digit when provided with an infinite sequence of digits
>>>> including the nth digit is hardly an algorithm for the purpose of the
>>>> definition of computable. If it were, then all numbers would be
>>>> computable by definition, and the definition would be useless.
>>
>>>> Sylvia.
>>
>>> The proof gives an algorithm for generating a bigger permutation
>>> set from a smaller one.
>>
>>> If a set of digit permutations are listed on the comp. Set of reals
>>> then an algorithm exists to duplicate that set 10 times and append
>>> digits 0..9 to each duplicate set.
>>
>>> This constructs algorithmically a countable list that contains
>>> 'full permutation' oo digits wide.
>>
>> You say it's countable, but you haven't explained how it assigns a
>> natural number to every sequence.
>
>
> Are you going on the tangent that computable reals can't
> be listed tangent?
>

No. The computable reals are countable. I'm saying that you haven't
proved that all the members of the set of all permutations are computable.

For this purpose, "computable" has the meaning it usually has in the
statement "The computable reals are countable". You may, if you wish,
define computable in some other way, but if you do, you cannot then also
claim that the set of computables so defined are countable unless you
can prove it.

Sylvia.