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

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From: |-|ercules on 28 Jun 2010 22:54

"Sylvia Else" <sylvia(a)not.here.invalid> wrote...
> On 29/06/2010 12:13 PM, |-|ercules wrote:
>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>> On 29/06/2010 10:42 AM, |-|ercules wrote:
>>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>>> On 28/06/2010 11:15 PM, |-|ercules wrote:
>>>>>
>>>>>> You prove case n, by logical inspection, that the property holds
>>>>>> for all
>>>>>> elements in the list of size n. (a whole object)
>>>>>>
>>>>>> I prove case n, by logical inspection, that the property holds for
>>>>>> prefixes of width n. (a subset of digits)
>>>>>>
>>>>>> You prove, by induction, that all (finite) sizes of lists the property
>>>>>> holds.
>>>>>>
>>>>>> I prove, by induction, that the property holds for all digit widths.
>>>>>> (all digits).
>>>>>
>>>>> You only prove that it holds for finite digit widths. Somehow you want
>>>>> to be allowed to extrapolate from finite to infinite when it suits
>>>>> you, but not for me to allowed to do the same thing when it doesn't
>>>>> suit you.
>>>>>
>>>>
>>>> Nope! Your prove it for increasing different objects.
>>>>
>>>> I sample larger and larger sizes of the one object. Different style of
>>>> proof!
>>>>
>>>
>>> <sigh> So we're sampling now, not constructing. Fine.
>>
>> I used the term sample a week ago, and that the computable reals
>> are deterministic and I'm not constructing per se, etc. etc.
>>
>> You can say I construct subcomponents too, as opposed to your construction
>> of individual objects.
>>
>>
>>>
>>> By sampling, I suppose you mean that you look through the list of
>>> computables to find sequences with the required prefixes.
>>>
>>> I question whether your inductive reasoning even works with that
>>> approach. It is certainly possible to find in the list of computables
>>> a sequence that starts with any finite prefix, but that doesn't mean
>>> that the inductive step is valid, merely that the result is true.
>>>
>>> An inductive step would need to take the form:
>>>
>>> ---
>>> If a sequence with a specific prefix p is in the list of computables,
>>> then prefixes constituting p followed by 0, p followed by 1, p
>>> followed by 2, etc. are also in the list.
>>
>>
>> Yes, you could have saved yourself a lot of time by reading the proof.
>>
>> -> there are 10 computable copies of the
>> -> complete permutations of width w
>> -> each ending in each of digits 0..9 (at position w+1)
>> -> which generates a set larger than width w
>>
>>
>>
>>>
>>> Then prove (by inspection presumably) that the prefixes 0 thru 9 exist.
>>
>>
>> Yes, in your wording you start with a set of complete prefixes and add
>> each of 0..9 to the end of each real,
>> my wording is equivalent, but I use a "block append" operation 10 times..
>
> That doesn't make the proof inductive, which was my point. You haven't
> proved any relation between the existence of prefixes of length n, and
> the existence of prefixes of length n+1. As it happens, we know from
> other considerations that they all exist in the list, but you haven't
> proved it by an inductive process.

There's a very clear P(n) -> P(n+1)

In your wording and mine.

>
>>
>>
>>
>>> ---
>>>
>>> If the proof merely consists of observing that since prefixes one
>>> digit longer than p are also of finite length, and therefore exist in
>>> the list of computables, then that is not an inductive step (except in
>>> the very limited sense that if n is finite, then so is n+1).
>>
>> your n is the digits that the property holds for,
>> all you're doing is reminding everyone that digit positions (natural
>> numbers) are finite.
>>
>>
>>>
>>> That aside, since the list of computables contains all finite
>>> prefixes, then when you've sampled the list to find sequences starting
>>> with all prefixes of length n digits, with the sample constituting a
>>> sublist[*] of length 10^n, I can sample the list of computables to
>>> find a sequence with a prefix of length 10^n which is not in your
>>> sublist. This remains true for all finite n.
>>>
>>> [*] I use the term loosely. It's really a listing of the subset.
>>>
>>> Sylvia.
>>
>>
>> So, it doesn't work for length n antidiagonals as all prefixes of length
>> n digits are counted for,
>> we've done enough analogies of bogus proofs.
>
> It doesn't have to work for length n antidiagonals, it merely has to
> work for some finite number, which it does. As long as your prefixes are
> of finite length, there is a sequence that is not in your sublist.

The sequences' width grows slower than the list length, your example proves nothing,
you could just as well get a 10 length list and a 20 digit antidiagonal and use that
to prove transfiniteness.

>
> You want that awkward sequence to vanish when you start dealing with
> infinite prefixes, but you've offered no reason to think it does.

This is sidetracking that the proof has been fully justified, please make your objections
to steps 1, 2, 3, or 4, or better still read how I justified the induction for all digit positions.

You haven't commented on the difference between induction over a single data structure
and induction over numerous data structures.

Herc

From: Sylvia Else on 28 Jun 2010 23:27

On 29/06/2010 12:54 PM, |-|ercules wrote:
> "Sylvia Else" <sylvia(a)not.here.invalid> wrote...
>> On 29/06/2010 12:13 PM, |-|ercules wrote:
>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>> On 29/06/2010 10:42 AM, |-|ercules wrote:
>>>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>>>>> On 28/06/2010 11:15 PM, |-|ercules wrote:
>>>>>>
>>>>>>> You prove case n, by logical inspection, that the property holds
>>>>>>> for all
>>>>>>> elements in the list of size n. (a whole object)
>>>>>>>
>>>>>>> I prove case n, by logical inspection, that the property holds for
>>>>>>> prefixes of width n. (a subset of digits)
>>>>>>>
>>>>>>> You prove, by induction, that all (finite) sizes of lists the
>>>>>>> property
>>>>>>> holds.
>>>>>>>
>>>>>>> I prove, by induction, that the property holds for all digit widths.
>>>>>>> (all digits).
>>>>>>
>>>>>> You only prove that it holds for finite digit widths. Somehow you
>>>>>> want
>>>>>> to be allowed to extrapolate from finite to infinite when it suits
>>>>>> you, but not for me to allowed to do the same thing when it doesn't
>>>>>> suit you.
>>>>>>
>>>>>
>>>>> Nope! Your prove it for increasing different objects.
>>>>>
>>>>> I sample larger and larger sizes of the one object. Different style of
>>>>> proof!
>>>>>
>>>>
>>>> <sigh> So we're sampling now, not constructing. Fine.
>>>
>>> I used the term sample a week ago, and that the computable reals
>>> are deterministic and I'm not constructing per se, etc. etc.
>>>
>>> You can say I construct subcomponents too, as opposed to your
>>> construction
>>> of individual objects.
>>>
>>>
>>>>
>>>> By sampling, I suppose you mean that you look through the list of
>>>> computables to find sequences with the required prefixes.
>>>>
>>>> I question whether your inductive reasoning even works with that
>>>> approach. It is certainly possible to find in the list of computables
>>>> a sequence that starts with any finite prefix, but that doesn't mean
>>>> that the inductive step is valid, merely that the result is true.
>>>>
>>>> An inductive step would need to take the form:
>>>>
>>>> ---
>>>> If a sequence with a specific prefix p is in the list of computables,
>>>> then prefixes constituting p followed by 0, p followed by 1, p
>>>> followed by 2, etc. are also in the list.
>>>
>>>
>>> Yes, you could have saved yourself a lot of time by reading the proof.
>>>
>>> -> there are 10 computable copies of the
>>> -> complete permutations of width w
>>> -> each ending in each of digits 0..9 (at position w+1)
>>> -> which generates a set larger than width w
>>>
>>>
>>>
>>>>
>>>> Then prove (by inspection presumably) that the prefixes 0 thru 9 exist.
>>>
>>>
>>> Yes, in your wording you start with a set of complete prefixes and add
>>> each of 0..9 to the end of each real,
>>> my wording is equivalent, but I use a "block append" operation 10
>>> times..
>>
>> That doesn't make the proof inductive, which was my point. You haven't
>> proved any relation between the existence of prefixes of length n, and
>> the existence of prefixes of length n+1. As it happens, we know from
>> other considerations that they all exist in the list, but you haven't
>> proved it by an inductive process.
>
>
> There's a very clear P(n) -> P(n+1)
>
> In your wording and mine.
>
>
>
>>
>>>
>>>
>>>
>>>> ---
>>>>
>>>> If the proof merely consists of observing that since prefixes one
>>>> digit longer than p are also of finite length, and therefore exist in
>>>> the list of computables, then that is not an inductive step (except in
>>>> the very limited sense that if n is finite, then so is n+1).
>>>
>>> your n is the digits that the property holds for,
>>> all you're doing is reminding everyone that digit positions (natural
>>> numbers) are finite.
>>>
>>>
>>>>
>>>> That aside, since the list of computables contains all finite
>>>> prefixes, then when you've sampled the list to find sequences starting
>>>> with all prefixes of length n digits, with the sample constituting a
>>>> sublist[*] of length 10^n, I can sample the list of computables to
>>>> find a sequence with a prefix of length 10^n which is not in your
>>>> sublist. This remains true for all finite n.
>>>>
>>>> [*] I use the term loosely. It's really a listing of the subset.
>>>>
>>>> Sylvia.
>>>
>>>
>>> So, it doesn't work for length n antidiagonals as all prefixes of length
>>> n digits are counted for,
>>> we've done enough analogies of bogus proofs.
>>
>> It doesn't have to work for length n antidiagonals, it merely has to
>> work for some finite number, which it does. As long as your prefixes
>> are of finite length, there is a sequence that is not in your sublist.
>
>
> The sequences' width grows slower than the list length, your example
> proves nothing,
> you could just as well get a 10 length list and a 20 digit antidiagonal
> and use that
> to prove transfiniteness.

I'm not proving, nor seeking to prove, anything about infinities. I'm
just pointing out that your proof is flawed, because to get anywhere you
have to explain how the anti-diagonal that's present at every stage,
disappears when the prefix length goes to infinity.

But your reference to the different finite lengths makes me wonder - are
you suggesting that infinite list's length is smaller than the
anti-diagonal sequence's infinite length?

>
>
>>
>> You want that awkward sequence to vanish when you start dealing with
>> infinite prefixes, but you've offered no reason to think it does.
>
>
> This is sidetracking that the proof has been fully justified, please
> make your objections
> to steps 1, 2, 3, or 4, or better still read how I justified the
> induction for all digit positions.
>
> You haven't commented on the difference between induction over a single
> data structure
> and induction over numerous data structures.
>

I have questioned whether what you're doing is induction. P(n) -> P(n+1)
is the result of an inductive proof, not the proof itself.

Sylvia.

From: |-|ercules on 28 Jun 2010 23:57

"Sylvia Else" <sylvia(a)not.here.invalid> wrote
>> You haven't commented on the difference between induction over a single
>> data structure
>> and induction over numerous data structures.
>>
>
> I have questioned whether what you're doing is induction. P(n) -> P(n+1)
> is the result of an inductive proof, not the proof itself.
>
> Sylvia.

Here's a story that tells of the difference that's more your level.

You and I start a landscaping business Herc And Syl's Landscaping Ad Infinitium

I handle all the mowing, and you do the fencing.

We get a call from Mr Fenceme and Mrs Mowme Blockheads.

We drive to the property which appears to be divided into 2 blocks, both infinite
rectangular lawns.

On one block, you start doing the fencing for Mr Fenceme, completing the perimeters of larger and larger concentric
rectangular paddocks.

I get to the mowing for Mrs Mowme, completing larger and larger rectangular mown lawn areas, each building
upon the earlier smaller rectangular lawn area.

Being a man, I'm well on my way to mowing the whole lawn.

Because you're a woman, you never ever come close to fencing the entire lawn! ;-)

Herc

From: Sylvia Else on 29 Jun 2010 00:39

On 29/06/2010 12:18 PM, herbzet wrote:
>
>
> Sylvia Else wrote:
>> The Raven wrote:
>>
>>> Sylvia and all the other followers of this thread, could you please keep it
>>> out of aus.tv and take it back to sci.math? It's off topic to aus.tv.
>>>
>>> Please (I'm asking politely) :-)
>>
>> I can't stop Herc posting them to aus.tv in the first place, and if I
>> remove groups from my replies, he just puts them back again.
>
> Oh, you poor dear! Oh, you poor, poor dear thing! Being *forced*
> by that wicked Herc to crosspost to ngs that have no interest in
> this topic at all!
>
> You poor, poor thing, being *forced* into this discourteous behaviour
> by that wicked Herc, who's not really wicked because he's so kuh-RAZY,
> the poor dear thing. What can a reasonable person do, other than
> suggest that dozens or perhaps hundreds of other people go to the
> trouble of filtering out *your* off-topic noise?
>
> Let me help you out, darling.
>
> 1) Set your own filter to kill any post with aus.tv in the "Newsgroups"
> field, and inform Herc that you won't see any such posts.
>
> 2) If your newsreader does not allow filtering on that field (I know
> mine certainly doesn't) just be your own filter -- announce firmly
> that you will not respond to posts with aus.tv in the Newsgroups
> field.
>
> Like this:
>
> I will not wittingly respond to posts about foundations of math
> that are posted to the aus.tv newsgroup, as that is just off-topic
> spam to them. I will not indulge Herc's narcissistic desire to
> keep his name in front of his countrymen's eyes.
>
> Herc may find me in sci.math if he wishes to converse with me.
>
> See how easy it is? All the people at aus.tv will not find it necessary
> to forward your posts to abuse(a)individual.net , prepending the sentence
> "Off-topic for aus.tv" to the body of the forwarded post, and otherwise
> following the guidelines at
>
> http://individual.net/faq.php#5.5 .

Let's see - does it mention replies to off-topic posts? No.

OK, technically, a reply is itself a post, but it's the initial
off-topic posting that the nuisance, not the replies.

I'm reading this thread in sci.math. I'm not going to engage in a
process of manual filtering based on the newsgroups that are being
posted to just to appease those who can't be bothered to kill the thread
of Herc's initial posting.

But what I can do, if you like, is to killfile you. Then you'll know
that there's really no point in your posting the kind of comment above,
and in consequence won't feel any need so to do, thus saving you much time.

Sylvia.

From: Sylvia Else on 29 Jun 2010 00:45

On 29/06/2010 1:57 PM, |-|ercules wrote:
> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>>> You haven't commented on the difference between induction over a single
>>> data structure
>>> and induction over numerous data structures.
>>>
>>
>> I have questioned whether what you're doing is induction. P(n) ->
>> P(n+1) is the result of an inductive proof, not the proof itself.
>>
>> Sylvia.
>
>
>
> Here's a story that tells of the difference that's more your level.
>
> You and I start a landscaping business Herc And Syl's Landscaping Ad
> Infinitium
>
> I handle all the mowing, and you do the fencing.
>
> We get a call from Mr Fenceme and Mrs Mowme Blockheads.
>
> We drive to the property which appears to be divided into 2 blocks, both
> infinite rectangular lawns.
>
> On one block, you start doing the fencing for Mr Fenceme, completing the
> perimeters of larger and larger concentric
> rectangular paddocks.
>
> I get to the mowing for Mrs Mowme, completing larger and larger
> rectangular mown lawn areas, each building
> upon the earlier smaller rectangular lawn area.
>
> Being a man, I'm well on my way to mowing the whole lawn.
>
> Because you're a woman, you never ever come close to fencing the entire
> lawn! ;-)
> Herc
>

Since neither of us makes the smallest dent in our respective infinite
tasks, you cannot meaningfully say which of us is closer to completing it.

What has that to do with induction anyway?

Sylvia.