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

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From: Sylvia Else on 26 Jun 2010 08:09

On 26/06/2010 8:45 PM, |-|ercules wrote:
> "Sylvia Else" <sylvia(a)not.here.invalid> wrote
>> On 26/06/2010 3:40 AM, Graham Cooper wrote:
>>> On Jun 25, 8:23 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>> On 25/06/2010 7:07 PM, Graham Cooper wrote:
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>> On Jun 25, 6:54 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>> OK.
>>>>
>>>>>> "1 start with an infinite list of all computable reals".
>>>>
>>>>>> That is any list of all the computable reals, howsoever constructed.
>>>>
>>>>>> "2 let w = the maximum width of complete permutation sets"
>>>>
>>>>>> Where a complete permutation set is all the possible combinations of
>>>>>> some finite number of digits. So this step doesn't involve doing
>>>>>> anything with the list described in step 1? It's a completely
>>>>>> independent step?
>>>>
>>>>>> Sylvia.
>>>>
>>>>> Hmmm. Did you consider that the CPS found in the list of
>>>>> step 1 was what I meant. Step 1 - consider this list...
>>>>
>>>> I'm reluctant to assume you mean anything unless it's stated. It seems
>>>> to cause difficulties. However, apparently CPS is an abbreviation for
>>>> "complete permutation set".
>>>>
>>>> So the list of all computable reals contains as a subset complete
>>>> permutation sets whose width is unbounded. Slightly rewording 2,
>>>> gives us:
>>>>
>>>> "2 let w = the maximum width of those complete permutation sets"
>>>>
>>>> and the next step is
>>>>
>>>> "3 contradict 2"
>>>>
>>>> How is it to be contradicted?
>>>>
>>>> Sylvia.
>>>
>>> There is no (finite) maximum.
>>
>> So there is no finite maximum. How does that advance your proof?
>
>
> There seems to be 2 possibilities.
>
> 1 All finite digit permutations occur to infinite length.

It's certainly true that any digit permutation of finite length will be
on the list, but that's not a proof that all infinite sequences are on
the list.

Certainly some infinite sequences are on the list, because they are
computable, but you haven't proved that all infinite sequences are on
the list.

>
> 2 It would be very difficult to come up with a new sequence of digits
> that wasn't on the computable reals list

The anti-diagonal wouldn't be on it. Now, you might feel that the
anti-diagonal is obviously computable, in that each digit can be
obtained algorithmically from the list. But that presupposed that a list
of computable reals is itself computable, which you haven't proved. We
know that it is countable, but that's not the same thing.

Sylvia.

From: George Greene on 26 Jun 2010 12:55

On Jun 26, 6:45 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> There seems to be 2 possibilities.
>
> 1 All finite digit permutations occur to infinite length.

This IS NOT a possibility because for a FINITE permutation to occur
"to infinite length" IS JUST MEANINGLESS!
"To EVERY FINITE length" DOES NOT MEAN "to infinite length",
yet you CONTINUE TO TALK as though it did!
This has just gone on LONG ENOUGH!

Here is another point: if you are only talking about FINITE lists of
digits,
then "permutation" is IRRELEVANT because ANY permutation of any
finite list of digits IS JUST ANOTHER FINITE LIST OF DIGITS, that
WAS ALREADY IN the class (of finite lists of digits) that you were
talking about!
Finite SEQUENCES of digits or finite LISTS of digits are what are
actually being
dealt with here. A permutation IS A CHANGE OF ORDER and would require
some
notion of WHAT AN "original" order was for a given finite basket of
digits!
There IS NO such notion for you here! STOP SAYING "permutation"!
YOU DON'T KNOW what a permutation is!

From: George Greene on 26 Jun 2010 12:58

On Jun 26, 8:09 am, Sylvia Else <syl...(a)not.here.invalid> wrote:
> >> So there is no finite maximum. How does that advance your proof?

There IS an INFINITE maximum, obviously.
THAT IS his whole proof! That matching occurs "to infinite length"!
He keeps saying that!

Seriously, do you really think you are going to be able to teach Herc
the concept of a "least upper bound"??
I personally remain quite confident that "max" will prove to be "the
max of" his
capacity for understanding.

From: George Greene on 26 Jun 2010 13:01

> On 26/06/2010 8:45 PM, |-|ercules wrote:
> > 2 It would be very difficult to come up with a new sequence of digits
> > that wasn't on the computable reals list

On Jun 26, 8:09 am, Sylvia Else <syl...(a)not.here.invalid> wrote:
> The anti-diagonal wouldn't be on it. Now, you might feel that the
> anti-diagonal is obviously computable, in that each digit can be
> obtained algorithmically from the list.

Of course we feel that! It IS IMPORTANT that anti-diagonalization (of
any
square list) IS computable! THERE IS a TM that does that!

> But that presupposed that a list
> of computable reals is itself computable, which you haven't proved. We
> know that it is countable, but that's not the same thing.

PLEASE! That is NOT the point! The point IS that EVEN IF the list of
computable reals is computable, THE FACT THAT the anti-diagonal
IS THEN ALSO computable *produces*a*CONTRADICTION* because
the anti-diagonal both 1) Must Be On the list, since it is computable,
and
2) CAN'T BE ON the list, BECAUSE it IS the ANTI-diagonal!

So what is Herc to conclude FROM THAT??

>
> Sylvia.

From: George Greene on 26 Jun 2010 13:15

On Jun 25, 10:37 pm, Transfer Principle <lwal...(a)lausd.net> wrote:
> Originally, Herc's arguments involved taking a list of ternary reals
> and _shuffling_ them to produce almost any real on the diagonal
> of the shuffled list.

NO, THEY DIDN'T.
That may have been the point AT WHICH *YOU* started paying attention,
but if it was, THEN YOU WERE LATE. This was NOT original!
ORIGINALLY, Herc DID NOT HAVE SENSE enough to use a small base.
THIS WHOLE thing OUGHT to be being done IN BINARY, but Herc will not
go down!
He was talking about digit-lists from 0..9, ORIGINALLY.

But what is "almost any real" supposed to mean??
Obviously, THERE IS a permutation that will produce ANY real,
IF THE ORIGINAL list was rich enough, so why "almost"??
The whole point of refuting THAT is that THE ORIGINAL list could be
TRIVIAL,
being nowhere NEAR as complex as "all computable", AND STILL ALLOW
this!
The original list could have NO infinitely wide elements on it at all,
or no elements
with multiple digits (just enough copies of ALL 1s, or ALL 0s, or
whatever his third
digit was) AND STILL ALLOW this! The permutations just DON'T MEAN
anything.
You should have tried harder to crush them.

> In this case, we really are talking about a permutation

Yes, he was, IN THAT case, but THAT, AS IS USUAL with 80%
of the stuff Herc brings up, WAS JUST IRRELEVANT TO the question!
The other 20% is on the question but IS WRONG.