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

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From: Sylvia Else on 23 Jun 2010 02:46

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.

From: Sylvia Else on 23 Jun 2010 02:50

On 23/06/2010 4:38 PM, herbzet wrote:
>
>
> Sylvia Else wrote:
>> herbzet wrote:
>>> Sylvia Else wrote:
>>>> herbzet wrote:
>>>>> Sylvia Else wrote:
>>>>>> Graham Cooper wrote:
>>>>>>> Sylvia Else wrote:
>>>>>>>> Sylvia Else wrote:
>>>>>>>>> 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.
>>>>>>>
>>>>>>> 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!
>>>>>
>>>>> Idiot -- is it not clear by now that he's not going to do that?
>>>>>
>>>>> Do you imagine that you're going to "corner" Herc somehow into
>>>>> displaying some rigor? Do you enjoying chasing after that dollar
>>>>> bill on a string?
>>>>
>>>> I can do it until I get bored, can't I? Sometimes it's not the
>>>> destination that's important, but the journey.
>>>
>>> Well, there's plenty of abusive people around for you to waste
>>> your time on, if that's your thing.
>>
>> His abusiveness is unfortunate, and I wish he'd control himself. On the
>> other hand, he provides useful practice at arguing with abusive people.
>> The particular lesson is to remember what it is one is trying to
>> achieve, and not to allow oneself to be provoked.
>
> Rationalize away.
>
> I've no more time to waste on an apologist for Herc.
>

I've described him as a paranoid schizophrenic, cited a situation where
he apparently threatened to contaminate food, and referred to a number
of bizarre delusions that he holds.

Anyone in their right mind would want to be a troll, rather than be in
Herc's situation.

If I'm an apologist for him, he certainly doesn't need any critics.

Sylvia.

From: Graham Cooper on 23 Jun 2010 02:57

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.

You seem to follow it proves the permutations are oo digits
wide. Do you know that all permutations implies every
digit sequence? Are these not computable?

Herc

From: Sylvia Else on 23 Jun 2010 03:29

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.

From: Graham Cooper on 23 Jun 2010 03:45

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.

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

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.
Herc