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

Prev: Anders, Ebihara Re: additive versus multiplicative creation: Dirac's new radioactivities Chapt 5 #180; ATOM TOTALITY
Next: combinations of additive and multiplicative creation: Dirac's new radioactivities Chapt 5 #181; ATOM TOTALITY

From: |-|ercules on 28 Jun 2010 20:42

"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!

>>>
>>> BTW, it occurs to me that if you have a list containing all the
>>> prefixes obtained by permuting 1, 2, 3, ... n digits, followed in each
>>> case by infinite suffixes, then we can use the anti-diagonal technique
>>> to construct an infinite sequence A whose first k digits differ from
>>> the corresponding diagonal digits of your list. That sequence A isn't
>>> in your list.
>>>
>>> When you move from n digit prefixes to n+1 digit prefixes, you add
>>> 10^n+1 new sequences, and we can change the next 10^n+1 digits of A so
>>> that it is still not in the list.
>>>
>>> Thus at each induction step, you increase the number of prefixes in
>>> the list, but you retain a sequence A that is not in the list. This
>>> remains true for any finite number of steps.
>>>
>>> That is, no induction step extinguishes sequence A which is not in the
>>> list. It is hardly reasonable to argue that extending this to an
>>> infinite number of steps creates a list with every inifinite sequence,
>>> but in the process mysteriously extinguishes sequence A.
>>>
>>> Sylvia.
>>
>> That's just a proof that finite lists are incomplete, and that sequences
>> are shorter than
>> the number of sequences to cover every digit sequence.
>> Similar to the favorite argument of ANTI-Cantorians!
>>
>> A list contains all finite prefixes.
>>
>> Step 1: a 1 digit anti-diagonal is not a new sequence
>> Step n: a n digit anti-diagonal is not a new sequence
>> Step n+1: a n+1 digit anti-diagonal is not a new sequence (ok IF n is
>> then n+1 is, but no base step!)
>
> You're talking about finite length anti-diagonals. I clearly said
> "construct an *infinite* sequence A whose *first* k digits differ from
> the corresponding diagonal digits of your list."
>
> As I said, A is not in your list, and it never will be no matter how far
> you take the induction.
>
> Sylvia.

As I said, your proof of finite incompleteness and short sequence lengths both
have more trivial proofs, and induction on the diagonal has a similar (but different)
proof that anti-diagonals contain no unique sequence.

Herc

From: Sylvia Else on 28 Jun 2010 21:19

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.

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.

Then prove (by inspection presumably) that the prefixes 0 thru 9 exist.
---

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).

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.

From: |-|ercules on 28 Jun 2010 22:13

"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..

> ---
>
> 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.

Herc

From: herbzet on 28 Jun 2010 22:18

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 .

You don't have to be an intercontinental pest if you don't want to be, darling.

You really don't.

You're welcome, dear.

--
hz

P.S. -- I won't wittingly be responding to posts with aus.tv in
in the Newsgroups field any further.

From: herbzet on 28 Jun 2010 22:24

Jim Burns wrote:
> 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

Heck, consider a list L of sequences of the digits
'5' and '7'.

Your program is just

If L(n,n) = 5 then output(n) = 7
else output(n) = 5.

The number being computed simply is not on the list.

You can't even get a list of all sequences of *two* digits,
much less a list of all sequences of *ten* digits, of which
the list of sequences of just *two* digits is a proper part.

--
hz