CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: Sylvia Else on 23 Jun 2010 20:20 On 24/06/2010 12:01 AM, Graham Cooper wrote: > 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? <sigh> I was right. All permutations infinitely wide is all reals. But that was not the issue. The question was whether they could be listed, which you still haven't proved. I'm at a loss to understand why you think that proving they're infinitely wide proves that they can be listed. > That's all from me I'm homeless in a few hours so I'll need > my iPhone battery to check my bank account. With all that income from camgirls.com, your bank account shouldn't be a problem. Sylvia.
From: Sylvia Else on 23 Jun 2010 20:22 On 24/06/2010 7:47 AM, Transfer Principle wrote: > 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. That's largely because his posts are devoid of reasoning. He just asserts things, and them blames the reader for not recognising the truth of the things asserted. Sylvia.
From: Arsène Lupin on 23 Jun 2010 21:10 Why people bother replying?
From: Graham Cooper on 23 Jun 2010 21:12 On Jun 24, 10:20 am, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 24/06/2010 12:01 AM, Graham Cooper wrote: > > > > > > > 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? > > <sigh> I was right. > > All permutations infinitely wide is all reals. But that was not the > issue. The question was whether they could be listed, which you still > haven't proved. I'm at a loss to understand why you think that proving > they're infinitely wide proves that they can be listed. > > > That's all from me I'm homeless in a few hours so I'll need > > my iPhone battery to check my bank account. > > With all that income from camgirls.com, your bank account shouldn't be a > problem. > > Sylvia. For the 10th time the proof shows how to list all permutations of digits oo wide. What do you think the list of computable reals is? A list! Herc
From: Graham Cooper on 23 Jun 2010 21:14
On Jun 24, 11:10 am, Arsène Lupin <deten...(a)gmail.com> wrote: > Why people bother replying? The more important question is why the proof that computable reals contain all digit permutations oo long is ignored. Herc |