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

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From: Graham Cooper on 23 Jun 2010 09:04

On Jun 23, 10:02 pm, Sylvia Else <syl...(a)not.here.invalid> wrote:
> On 23/06/2010 8:28 PM, Graham Cooper wrote:
>
>
>
>
>
> > On Jun 23, 8:12 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >> On 23/06/2010 7:50 PM, Graham Cooper wrote:
>
> >>> On Jun 23, 7:41 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>> On 23/06/2010 7:32 PM, Graham Cooper wrote:
>
> >>>>>>> start with an assumption the computable
> >>>>>>> reals has a finite maximum to the digit
> >>>>>>> width of COMPLETE permutation set.
>
> >>>>>> That's garbled. Try again.
>
> >>>>>> Sylvia.
>
> >>>>> Dingo can comprehend it. You try again.
>
> >>>> I can find no evidence that Dingo can comprehend it.
>
> >>>> Anyway, you're trying to prove something to me, and I cannot parse that
> >>>> sentence.
>
> >>>> Sylvia.
>
> >>> Ok let's define complete permutation set.
>
> >>> With an example!!
>
> >>> 00
> >>> 01
> >>> 10
> >>> 11
>
> >>> this is a complete permutation set of digit width 2.
>
> >>> Does that help?
>
> >> It's all the different ways in which the digits 0 and 1 can be placed
> >> into a sequence of length 2. If you're confining yourself to just those
> >> two digits (which you can do without loss of generality), then I can
> >> accept that as the definition of "complete permutation set of digit
> >> width 2". That is, the expression "complete permutation set of digit
> >> width n" is all the combinations of 0 and 1 in a sequence of length n.
> >> Indeed there are 2^n of them.
>
> >> Conversely, if the complete permutation set contains 2^n sequences, then
> >> the digit width is defined to be n.
>
> >> So far so good.
>
> >> Next...
>
> >> Sylvia.
>
> > Is there a complete permutation set with digit width 1,000,000
> > in the list of computable reals? Use base 10.
>
> I take that to mean: Is the complete permutation set (using digits 0
> thru 9) of digit width 1,000,000 a subset of the set of computable reals?
>
> The answer is yes.
>
> I'll add that it's also yes if any other finite positive integer is
> substituted for 1,000,000.
>
> Next....
>
> Sylvia.

Is the maximum digit width finite?

Herc

From: Sylvia Else on 23 Jun 2010 09:45

On 23/06/2010 11:04 PM, Graham Cooper wrote:
> On Jun 23, 10:02 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>> On 23/06/2010 8:28 PM, Graham Cooper wrote:
>>
>>
>>
>>
>>
>>> On Jun 23, 8:12 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>> On 23/06/2010 7:50 PM, Graham Cooper wrote:
>>
>>>>> On Jun 23, 7:41 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
>>>>>> On 23/06/2010 7:32 PM, Graham Cooper wrote:
>>
>>>>>>>>> start with an assumption the computable
>>>>>>>>> reals has a finite maximum to the digit
>>>>>>>>> width of COMPLETE permutation set.
>>
>>>>>>>> That's garbled. Try again.
>>
>>>>>>>> Sylvia.
>>
>>>>>>> Dingo can comprehend it. You try again.
>>
>>>>>> I can find no evidence that Dingo can comprehend it.
>>
>>>>>> Anyway, you're trying to prove something to me, and I cannot parse that
>>>>>> sentence.
>>
>>>>>> Sylvia.
>>
>>>>> Ok let's define complete permutation set.
>>
>>>>> With an example!!
>>
>>>>> 00
>>>>> 01
>>>>> 10
>>>>> 11
>>
>>>>> this is a complete permutation set of digit width 2.
>>
>>>>> Does that help?
>>
>>>> It's all the different ways in which the digits 0 and 1 can be placed
>>>> into a sequence of length 2. If you're confining yourself to just those
>>>> two digits (which you can do without loss of generality), then I can
>>>> accept that as the definition of "complete permutation set of digit
>>>> width 2". That is, the expression "complete permutation set of digit
>>>> width n" is all the combinations of 0 and 1 in a sequence of length n.
>>>> Indeed there are 2^n of them.
>>
>>>> Conversely, if the complete permutation set contains 2^n sequences, then
>>>> the digit width is defined to be n.
>>
>>>> So far so good.
>>
>>>> Next...
>>
>>>> Sylvia.
>>
>>> Is there a complete permutation set with digit width 1,000,000
>>> in the list of computable reals? Use base 10.
>>
>> I take that to mean: Is the complete permutation set (using digits 0
>> thru 9) of digit width 1,000,000 a subset of the set of computable reals?
>>
>> The answer is yes.
>>
>> I'll add that it's also yes if any other finite positive integer is
>> substituted for 1,000,000.
>>
>> Next....
>>
>> Sylvia.
>
>
> Is the maximum digit width finite?

No.

I'm beginning to get bad feelings about this. This is another proof
(well, pretty much the same one, actually) of the undisputed fact that
the width is infinite isn't it?

Anyway, next....

Sylvia.

From: Graham Cooper on 23 Jun 2010 10:01

On Jun 23, 11:45 pm, Sylvia Else <syl...(a)not.here.invalid> wrote:
> On 23/06/2010 11:04 PM, Graham Cooper wrote:
>
>
>
>
>
> > On Jun 23, 10:02 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >> On 23/06/2010 8:28 PM, Graham Cooper wrote:
>
> >>> On Jun 23, 8:12 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>> On 23/06/2010 7:50 PM, Graham Cooper wrote:
>
> >>>>> On Jun 23, 7:41 pm, Sylvia Else<syl...(a)not.here.invalid> wrote:
> >>>>>> On 23/06/2010 7:32 PM, Graham Cooper wrote:
>
> >>>>>>>>> start with an assumption the computable
> >>>>>>>>> reals has a finite maximum to the digit
> >>>>>>>>> width of COMPLETE permutation set.
>
> >>>>>>>> That's garbled. Try again.
>
> >>>>>>>> Sylvia.
>
> >>>>>>> Dingo can comprehend it. You try again.
>
> >>>>>> I can find no evidence that Dingo can comprehend it.
>
> >>>>>> Anyway, you're trying to prove something to me, and I cannot parse that
> >>>>>> sentence.
>
> >>>>>> Sylvia.
>
> >>>>> Ok let's define complete permutation set.
>
> >>>>> With an example!!
>
> >>>>> 00
> >>>>> 01
> >>>>> 10
> >>>>> 11
>
> >>>>> this is a complete permutation set of digit width 2.
>
> >>>>> Does that help?
>
> >>>> It's all the different ways in which the digits 0 and 1 can be placed
> >>>> into a sequence of length 2. If you're confining yourself to just those
> >>>> two digits (which you can do without loss of generality), then I can
> >>>> accept that as the definition of "complete permutation set of digit
> >>>> width 2". That is, the expression "complete permutation set of digit
> >>>> width n" is all the combinations of 0 and 1 in a sequence of length n.
> >>>> Indeed there are 2^n of them.
>
> >>>> Conversely, if the complete permutation set contains 2^n sequences, then
> >>>> the digit width is defined to be n.
>
> >>>> So far so good.
>
> >>>> Next...
>
> >>>> Sylvia.
>
> >>> Is there a complete permutation set with digit width 1,000,000
> >>> in the list of computable reals? Use base 10.
>
> >> I take that to mean: Is the complete permutation set (using digits 0
> >> thru 9) of digit width 1,000,000 a subset of the set of computable reals?
>
> >> The answer is yes.
>
> >> I'll add that it's also yes if any other finite positive integer is
> >> substituted for 1,000,000.
>
> >> Next....
>
> >> Sylvia.
>
> > Is the maximum digit width finite?
>
> No.
>
> I'm beginning to get bad feelings about this. This is another proof
> (well, pretty much the same one, actually) of the undisputed fact that
> the width is infinite isn't it?
>
> Anyway, next....
>
> Sylvia.

Can you parse 'start with the assumption' paragraph yet?

If you can compute all permutations infinitely wide then
isn't that all reals?

That's all from me I'm homeless in a few hours so I'll need
my iPhone battery to check my bank account.
Herc

From: Transfer Principle on 23 Jun 2010 17:47

On Jun 21, 10:15 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote:
> Transfer Principle wrote:
> > He should be allowed to
> > oppose Cantor's Theorem without five-letter insults.
> Five-letter insult or not, he's not just opposing Cantor's
> Theorem and he's not just opposing the whole FOL proof machinery:
> he's inconsistent in reasoning! He either should accept FOL
> proof machinery or disregard it entirely, but once he accepts
> it he has to accept the proof of Cantor's theorem and that
> has nothing to do with his seeing the theorem as useful or useless.
> We should NOT support any kind of inconsistency in reasoning.

This is one of several posts by several different posters who
point out that if one accepts several axioms and accepts the
underlying logic, then one automatically accepts any statement
that is derived from such, whether one likes it or not. Another
poster pointed out that Cantor's Theorem is a finitistic
statement that has a finite proof in ZFC. I don't dispute that
"ZFC proves Cantor's Theorem" is uncontroversial and not open
to debate, but there's a big difference between "ZFC proves
Cantor's Theorem" and "Cantor's theorem is true." One can
accept the former yet reject the latter.

I will consider a poster who makes a claim of the form "I
accept A, and I accept A->B, (and I accept Modus Ponens), yet
I don't accept B," to have crossed the line beyond which they
deserve five-letter insults.

But so far, I have yet to see Herc cross that line. If one
accepts all the axioms of ZFC and all of FOL, then one must
accept Cantor's Theorem whether they like it or not, but Herc
apparently _doesn't_ accept all of the axioms of ZFC (at least
not enough axioms to prove Cantor). On the contrary, Herc
considers ZFC to be a "religion."

So I agree with Nguyen that we shouldn't support inconsistency
in reasoning, but until Herc claims that he accepts enough
axioms and logic from which to derive Cantor, he hasn't posted
any inconsistency in reasoning yet.

From: Mike Terry on 23 Jun 2010 18:53

"herbzet" <herbzet(a)gmail.com> wrote in message
news:4C21A102.E118F4CB(a)gmail.com...
>
>
> Mike Terry wrote:
> > "Sylvia Else" wrote:
> > > herbzet wrote:
> > > > Sylvia Else wrote:
> > > >> herbzet wrote:
> > > >>
> > > >>> Herc is a troll who is HAVING A BALL jerking all the "smart guys"
> > > >>> around.
> > > >>
> > > >> Or not. Herc is a paranoid schizophrenic, and subject to a variety
of
> > > >> delusions.
> > > >
> > > > None of which implies that he is not also a troll.
> > > >
> > > >> What isn't clear is whether this Cantor stuff is a
> > > >> conventional misunderstanding, or yet another delusion.
> > > >
> > > > It's the same old tired Cantor troll b.s.
> > >
> > > I didn't realise before how long this has been going on for.
> > >
> > > But I don't think he's a troll - he appears to have a genuine belief
> > > that the world's mathematicians have got this wrong. If it's a
> > > conventional misunderstanding, he might yet be persuaded that he is
> > > mistaken.
> >
> > Personally I can't see this ever happening. When I started off with
Herc
> > (years ago), it seemed like he was just making a simple mistake, and so
it
> > should be easy enough to show where this mistake was. (And indeed it is
> > easy in a mathematical sense...)
> >
> > As I went further, I realised Herc knows nothing of normal mathematical
> > definitions (like um.. like the ones used in Cantor's proofs which he is
> > discussing), and nothing of mathematical reasoning (proofs starting from
> > definitions etc.). Also he has his own unclear (contradictory maybe?)
> > definitions for words he uses. So obviously a bit more work than I
first
> > thought! :)
> >
> > Still, I thought if I break everything down into smaller and smaller
steps,
> > explain exactly all the definitions involved, get Herc to clarify his
own
> > definitions to make them precise etc., then I could still get him to
realise
> > he's mistaken.
> >
> > But there is a much more basic problem - Herc actually refuses to engage
in
> > "normal mathematical dialog". What I mean is that if you and I
discussed
> > something, and I didn't understand a step in your proof, I'd point out
what
> > I didn't understand, and you'd go away and expand the proof until I was
> > happy. Similarly, if I used a vague term, you could ask me to clarify
it,
> > and I would break it down into well understood basic notions,
quantifiers,
> > etc., and we'd move on... Neither of us would be offended by the
process or
> > think we were being insulted, it's just business as usual for
communicating
> > mathematics.
> >
> > Actually, I've never really thought of this as a "mathematical" skill,
as
> > I've always thought of mathematics as being the interesting stuff we do
on
> > top of all that. It's a basic skill which I'm sure I had around the age
of
> > 10 (once I'd read simple proofs like the infinitude of the primes etc.),
> > although clearly at that age I didn't understand many definitions.
> >
> > Anyway, it's to be expected that posters won't all have the same level
of
> > knowledge of working definitions, which is why we have "normal
mathematical
> > dialog" to get along! I believe it's impossible to "talk maths" with
> > someone who simply refuses to engage in this behaviour.
> >
> > This includes Herc - I don't believe he will ever respond to a request
to
> > clarify something into simpler terms. (Maybe some people's brains just
> > don't work in that analytic way?, and so they don't understand the need
for
> > it?) And if you suggest a precise definition for something vague Herc
is
> > saying, he will neither confirm nor deny that that is what he meant.
(He
> > may even scold you for introducing irrelevent factors into the argument,
and
> > suggest you should just ask him to explain, but if you do that of course
you
> > won't get much of a clarification!)
> >
> > So what will Herc actually do if you follow my earlier idea of
explaining in
> > greater and greater detail, asking for clarifications, refusing to go
along
> > with vague confusing terminology until it is clarified and so on? [I
> > thought that surely if I did this thoroughly enough, Herc would have NO
> > CHOICE but to agree where he was wrong, or at least he would have to
reply
> > in such a way that it was obvious to himself and others that he was not
able
> > to answer the questions and support his claims.]
> >
> > The answer is that Herc will just ignore all your efforts and respond
with
> > something vague, unrelated to the detail of your postings. E.g. he will
> > ignore your questions and ask you to "go away and work out all the
possible
> > antidiagonals", or something. Perhaps he will write a piece at the end
of
> > your post telling you where YOU are going wrong, and repeat his demand
that
> > you answer some ambiguous or irrelevent question. (And yes, with enough
> > persistence he will become abusive.) What he WILL NOT do is respond
> > meaningfully to any requests for mathematical clarification! Later on
he'll
> > start another thread using the same unclear terminology, and nothing
will
> > have moved on.
> >
> > I think Herc's problem with Cantor's are only sustainable while he is
> > allowed to confuse himself with his
> > ambiguous/contradictory/plain-old-incorrect terminology, but while he
will
> > not engage in "meaningful mathematical dialogue" I don't see how
anything
> > will change...
>
> Good stuff, Mike.
>
> But in your YEARS of dealing with Herc, has the idea not occurred
> to you that Herc is arguing in bad faith? That he is a TROLL who
> is HAVING A BALL jerking around a dumb sucker for YEARS AT A STRETCH?
>
> Has that idea not occurred to you, Mike?

Yes, but I don't think that's what Herc's doing.
(Could be wrong I suppose :)

>
> --
> hz