How Wide Do All The Possible Computed Digit Sequences Go?
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From: Tim Little on 24 Jun 2010 23:41 On 2010-06-24, Sylvia Else <sylvia(a)not.here.invalid> wrote: > On 24/06/2010 1:12 PM, Tim Little wrote: >> The list of all computable reals is exactly such a list. Most >> importantly, no property specific to pi is used here. Every real >> would be "contained in" that list if we were to use Herc's broken idea >> of "contained in". > > Ok, but his next step - all finite prefixes implies all infinite > sequences - is false. So all it means is that his 'proof' contains two > invalid steps rather than just one. I think they're both examples of the same invalid use of "L contains x". If you replaced Herc's incorrect statements of "L contains x" with "L contains a sequence with limit x" then both steps would be true. If L "contains" all finite prefixes in the above broken sense, then it provably does "contain" all infinite sequences in the same sense. True, but irrelevant to Cantor's proof (which uses the ordinary mathematical meaning) and everything else he's ranting about though. - Tim
From: Graham Cooper on 25 Jun 2010 00:17 On Jun 25, 1:41 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-24, Sylvia Else <syl...(a)not.here.invalid> wrote: > > > On 24/06/2010 1:12 PM, Tim Little wrote: > >> The list of all computable reals is exactly such a list. Most > >> importantly, no property specific to pi is used here. Every real > >> would be "contained in" that list if we were to use Herc's broken idea > >> of "contained in". > > > Ok, but his next step - all finite prefixes implies all infinite > > sequences - is false. So all it means is that his 'proof' contains two > > invalid steps rather than just one. > > I think they're both examples of the same invalid use of "L contains x". > > If you replaced Herc's incorrect statements of "L contains x" with > "L contains a sequence with limit x" then both steps would be true. > If L "contains" all finite prefixes in the above broken sense, then it > provably does "contain" all infinite sequences in the same sense. > > True, but irrelevant to Cantor's proof (which uses the ordinary > mathematical meaning) and everything else he's ranting about though. > > - Tim I won bother with repeated explanations on the "contains" dilema as you ignore my counter questions and continue that it's meaningless. Herc
From: Graham Cooper on 25 Jun 2010 00:20 On Jun 25, 12:55 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 25/06/2010 6:10 AM, Graham Cooper wrote: > > > > > > > On Jun 24, 11:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> On 24/06/2010 5:09 PM, Graham Cooper wrote: > > >>> On Jun 24, 5:00 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>> On 24/06/2010 1:12 PM, Tim Little wrote: > > >>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>>>> On 23/06/2010 5:00 PM, Tim Little wrote: > >>>>>>> It can't be conceded, as it is simply false. The predicate "list L > >>>>>>> contains x" means exactly that there exists n in N such that L_n = x. > >>>>>>> The list does not contain pi since there is no such n. It really is > >>>>>>> that simple. > > >>>>>> Well, OK. Though I can't see how it makes any difference to Herc's > >>>>>> argument, since I could never see what role it played anyway. > > >>>>> It is the very core of his argument: if pi were "contained in" the > >>>>> infinite list > > >>>>> 3 > >>>>> 3.1 > >>>>> 3.14 > >>>>> 3.141 > >>>>> ... > > >>>>> then it would also be "contained in" any other list sharing the "all > >>>>> prefixes" property, such as > > >>>>> 3.0000000 > >>>>> 3.1000000 > >>>>> 3.1428571. . . > >>>>> 3.1414141 > >>>>> ... > > >>>>> It would even be "contained in" any list having that as a sublist, e.g. > > >>>>> 0.0000000 > >>>>> 3.0000000 > >>>>> 3.1000000 > >>>>> 1.0000000 > >>>>> 3.1428571. . . > >>>>> 2.0934953 > >>>>> 0.5829345 > >>>>> 3.1414141 > >>>>> ... > > >>>>> The list of all computable reals is exactly such a list. Most > >>>>> importantly, no property specific to pi is used here. Every real > >>>>> would be "contained in" that list if we were to use Herc's broken idea > >>>>> of "contained in". > > >>>> Ok, but his next step - all finite prefixes implies all infinite > >>>> sequences - is false. So all it means is that his 'proof' contains two > >>>> invalid steps rather than just one. > > >>>> Sylvia. > > >>> I've told you atleast 3 times specifically that is NOT an implication > >>> of hc3. I put it in a implication formula -> are you amnesiac? > > >> I never said it was an implication *of* hc3. I say it's an implication > >> *in* hc3. > > >>> As long as all permutations oo wide are in the set, (segmented, > >>> appended to hitlers number, inverted, imputed, with any other > >>> numbers you can think of) I don't care! > > >> You need to prove that they're all in the set, or you have nothing. > > >>> But one of these days > >>> one of is going to confirm oo digits of every sequence are ALL THERE > > >> Sylvia. > > > That is no excuse you are still wrong. Something that is implied > > is implied. You corrected the correction in place of admitting > > you have been told > > > all finite prefixes -> all oo long sequences > > > is N O T hc3 > > > you've been told 5 times now plus I explained it several > > more times in this thread and you still carry on. > > > Learn to read! > > > Herc > > I think it's best if we kill this subthread, and instead try to deal > with the entire issue in the subthread where I've questioned your step 1. > > Sylvia. Mike Terry countered your objections to step 1 so maybe he can clarify it for you. I certainly can't! Herc
From: Sylvia Else on 25 Jun 2010 00:25 On 25/06/2010 2:20 PM, Graham Cooper wrote: > On Jun 25, 12:55 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >> On 25/06/2010 6:10 AM, Graham Cooper wrote: >> >> >> >> >> >>> On Jun 24, 11:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>> On 24/06/2010 5:09 PM, Graham Cooper wrote: >> >>>>> On Jun 24, 5:00 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>>>> On 24/06/2010 1:12 PM, Tim Little wrote: >> >>>>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>>>>>> On 23/06/2010 5:00 PM, Tim Little wrote: >>>>>>>>> It can't be conceded, as it is simply false. The predicate "list L >>>>>>>>> contains x" means exactly that there exists n in N such that L_n = x. >>>>>>>>> The list does not contain pi since there is no such n. It really is >>>>>>>>> that simple. >> >>>>>>>> Well, OK. Though I can't see how it makes any difference to Herc's >>>>>>>> argument, since I could never see what role it played anyway. >> >>>>>>> It is the very core of his argument: if pi were "contained in" the >>>>>>> infinite list >> >>>>>>> 3 >>>>>>> 3.1 >>>>>>> 3.14 >>>>>>> 3.141 >>>>>>> ... >> >>>>>>> then it would also be "contained in" any other list sharing the "all >>>>>>> prefixes" property, such as >> >>>>>>> 3.0000000 >>>>>>> 3.1000000 >>>>>>> 3.1428571. . . >>>>>>> 3.1414141 >>>>>>> ... >> >>>>>>> It would even be "contained in" any list having that as a sublist, e.g. >> >>>>>>> 0.0000000 >>>>>>> 3.0000000 >>>>>>> 3.1000000 >>>>>>> 1.0000000 >>>>>>> 3.1428571. . . >>>>>>> 2.0934953 >>>>>>> 0.5829345 >>>>>>> 3.1414141 >>>>>>> ... >> >>>>>>> The list of all computable reals is exactly such a list. Most >>>>>>> importantly, no property specific to pi is used here. Every real >>>>>>> would be "contained in" that list if we were to use Herc's broken idea >>>>>>> of "contained in". >> >>>>>> Ok, but his next step - all finite prefixes implies all infinite >>>>>> sequences - is false. So all it means is that his 'proof' contains two >>>>>> invalid steps rather than just one. >> >>>>>> Sylvia. >> >>>>> I've told you atleast 3 times specifically that is NOT an implication >>>>> of hc3. I put it in a implication formula -> are you amnesiac? >> >>>> I never said it was an implication *of* hc3. I say it's an implication >>>> *in* hc3. >> >>>>> As long as all permutations oo wide are in the set, (segmented, >>>>> appended to hitlers number, inverted, imputed, with any other >>>>> numbers you can think of) I don't care! >> >>>> You need to prove that they're all in the set, or you have nothing. >> >>>>> But one of these days >>>>> one of is going to confirm oo digits of every sequence are ALL THERE >> >>>> Sylvia. >> >>> That is no excuse you are still wrong. Something that is implied >>> is implied. You corrected the correction in place of admitting >>> you have been told >> >>> all finite prefixes -> all oo long sequences >> >>> is N O T hc3 >> >>> you've been told 5 times now plus I explained it several >>> more times in this thread and you still carry on. >> >>> Learn to read! >> >>> Herc >> >> I think it's best if we kill this subthread, and instead try to deal >> with the entire issue in the subthread where I've questioned your step 1. >> >> Sylvia. > > > Mike Terry countered your objections to step 1 He most certainly did not. Sylvia.
From: Graham Cooper on 25 Jun 2010 01:24
On Jun 25, 2:25 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 25/06/2010 2:20 PM, Graham Cooper wrote: > > > > > > > On Jun 25, 12:55 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> On 25/06/2010 6:10 AM, Graham Cooper wrote: > > >>> On Jun 24, 11:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>> On 24/06/2010 5:09 PM, Graham Cooper wrote: > > >>>>> On Jun 24, 5:00 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>>>> On 24/06/2010 1:12 PM, Tim Little wrote: > > >>>>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>>>>>> On 23/06/2010 5:00 PM, Tim Little wrote: > >>>>>>>>> It can't be conceded, as it is simply false. The predicate "list L > >>>>>>>>> contains x" means exactly that there exists n in N such that L_n = x. > >>>>>>>>> The list does not contain pi since there is no such n. It really is > >>>>>>>>> that simple. > > >>>>>>>> Well, OK. Though I can't see how it makes any difference to Herc's > >>>>>>>> argument, since I could never see what role it played anyway. > > >>>>>>> It is the very core of his argument: if pi were "contained in" the > >>>>>>> infinite list > > >>>>>>> 3 > >>>>>>> 3.1 > >>>>>>> 3.14 > >>>>>>> 3.141 > >>>>>>> ... > > >>>>>>> then it would also be "contained in" any other list sharing the "all > >>>>>>> prefixes" property, such as > > >>>>>>> 3.0000000 > >>>>>>> 3.1000000 > >>>>>>> 3.1428571. . . > >>>>>>> 3.1414141 > >>>>>>> ... > > >>>>>>> It would even be "contained in" any list having that as a sublist, e.g. > > >>>>>>> 0.0000000 > >>>>>>> 3.0000000 > >>>>>>> 3.1000000 > >>>>>>> 1.0000000 > >>>>>>> 3.1428571. . . > >>>>>>> 2.0934953 > >>>>>>> 0.5829345 > >>>>>>> 3.1414141 > >>>>>>> ... > > >>>>>>> The list of all computable reals is exactly such a list. Most > >>>>>>> importantly, no property specific to pi is used here. Every real > >>>>>>> would be "contained in" that list if we were to use Herc's broken idea > >>>>>>> of "contained in". > > >>>>>> Ok, but his next step - all finite prefixes implies all infinite > >>>>>> sequences - is false. So all it means is that his 'proof' contains two > >>>>>> invalid steps rather than just one. > > >>>>>> Sylvia. > > >>>>> I've told you atleast 3 times specifically that is NOT an implication > >>>>> of hc3. I put it in a implication formula -> are you amnesiac? > > >>>> I never said it was an implication *of* hc3. I say it's an implication > >>>> *in* hc3. > > >>>>> As long as all permutations oo wide are in the set, (segmented, > >>>>> appended to hitlers number, inverted, imputed, with any other > >>>>> numbers you can think of) I don't care! > > >>>> You need to prove that they're all in the set, or you have nothing. > > >>>>> But one of these days > >>>>> one of is going to confirm oo digits of every sequence are ALL THERE > > >>>> Sylvia. > > >>> That is no excuse you are still wrong. Something that is implied > >>> is implied. You corrected the correction in place of admitting > >>> you have been told > > >>> all finite prefixes -> all oo long sequences > > >>> is N O T hc3 > > >>> you've been told 5 times now plus I explained it several > >>> more times in this thread and you still carry on. > > >>> Learn to read! > > >>> Herc > > >> I think it's best if we kill this subthread, and instead try to deal > >> with the entire issue in the subthread where I've questioned your step 1. > > >> Sylvia. > > > Mike Terry countered your objections to step 1 > > He most certainly did not. > > Sylvia. I can't quote a long post on my iPhone but search "definite index position" and "bona-fide list" kudos to Mike the first person to Take action to end my torture. Could be a sign!!!!! Herc |