How Wide Do All The Possible Computed Digit Sequences Go?
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From: Graham Cooper on 23 Jun 2010 04:43 On Jun 23, 5:35 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 5:03 PM, Graham Cooper wrote: > > > > > > > On Jun 23, 5:00 pm, Tim Little<t...(a)little-possums.net> wrote: > >> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >>> On the face of it, line n contains the n digits of pie, > >>> sequentially, and in order. I suppose it can be conceded that the > >>> infinite list contains Pi. > > >> 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. > > >> It does satisfy a much looser property: there exists a sublist S such > >> that lim S = pi. In general you can form a set of real numbers > >> closure(L) = { x in R | exists sublist S of L such that lim S = x } > >> which you could call the "closure" of a list L. > > >> Then you could say that pi is in the closure of the list, but pi is > >> certainly not in the list itself. > > >> Herc does not know the difference. > > >> - Tim > > > How many digits in order of pi are below this line > > if interpreted mathematically? > > What does that question mean? In particular, what does "digits in order > of pi" mean? > > Perhaps you could give some example lists, with the answer in each case. > > Sylvia. > > > > > > > ____________ > > > 3 > > 31 > > 314 > > ... Huh? I'm not explaining trivial items to you. Your "mutilation" of herc_can't_3 was proved erronous and it stands. There are other terms than contains where it holds so what is your argument? Your disproof was wrong and you won't admit you were wrong. Now you are shifting your erronous claim to a termi ology issue. Either give a proper disproof or ceasevyour eternal complaints. Hc3 stands. The ball is in your court to prove otherwise. Herc
From: Sylvia Else on 23 Jun 2010 04:55 On 23/06/2010 6:43 PM, Graham Cooper wrote: > On Jun 23, 5:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >> On 23/06/2010 5:03 PM, Graham Cooper wrote: >> >> >> >> >> >>> On Jun 23, 5:00 pm, Tim Little<t...(a)little-possums.net> wrote: >>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: >> >>>>> On the face of it, line n contains the n digits of pie, >>>>> sequentially, and in order. I suppose it can be conceded that the >>>>> infinite list contains Pi. >> >>>> 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. >> >>>> It does satisfy a much looser property: there exists a sublist S such >>>> that lim S = pi. In general you can form a set of real numbers >>>> closure(L) = { x in R | exists sublist S of L such that lim S = x } >>>> which you could call the "closure" of a list L. >> >>>> Then you could say that pi is in the closure of the list, but pi is >>>> certainly not in the list itself. >> >>>> Herc does not know the difference. >> >>>> - Tim >> >>> How many digits in order of pi are below this line >>> if interpreted mathematically? >> >> What does that question mean? In particular, what does "digits in order >> of pi" mean? >> >> Perhaps you could give some example lists, with the answer in each case. >> >> Sylvia. >> >> >> >> >> >>> ____________ >> >>> 3 >>> 31 >>> 314 >>> ... > > > > Huh? I'm not explaining trivial items to you. The purpose was to clarify the intent of "How many digits in order of pi are below this line if interpreted mathematically?" Don't you want it clarified? > > Your "mutilation" of herc_can't_3 was proved erronous > and it stands. Where was it proved erroneous? > There are other terms than contains where it holds > so what is your argument? herc_cant_3 contained a proposition of the form If X has property A, then it also has property B. Such a proposition is demonstrable false if a possible X is demonstrated to have property A, and demonstrated not to have property B. It doesn't matter than there are some Xs with both A and B. The proposition is falsified by a single counter-example. Since the proposition in herc_cant_3 has a counter-example, it cannot be correct, and without a substitute proposition, herc_cant_3 cannot stand as a theorem. > Your disproof was wrong and you > won't admit you were wrong. Now you are shifting your erronous > claim to a termi ology issue. No - I'm just trying to understand what your question was intended to mean. > Either give a proper disproof > or ceasevyour eternal complaints. Hc3 stands. The ball > is in your court to prove otherwise. Sylvia.
From: Graham Cooper on 23 Jun 2010 05:01 On Jun 23, 6:55 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 6:43 PM, Graham Cooper wrote: > > > > > > > On Jun 23, 5:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> On 23/06/2010 5:03 PM, Graham Cooper wrote: > > >>> On Jun 23, 5:00 pm, Tim Little<t...(a)little-possums.net> wrote: > >>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >>>>> On the face of it, line n contains the n digits of pie, > >>>>> sequentially, and in order. I suppose it can be conceded that the > >>>>> infinite list contains Pi. > > >>>> 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. > > >>>> It does satisfy a much looser property: there exists a sublist S such > >>>> that lim S = pi. In general you can form a set of real numbers > >>>> closure(L) = { x in R | exists sublist S of L such that lim S = x } > >>>> which you could call the "closure" of a list L. > > >>>> Then you could say that pi is in the closure of the list, but pi is > >>>> certainly not in the list itself. > > >>>> Herc does not know the difference. > > >>>> - Tim > > >>> How many digits in order of pi are below this line > >>> if interpreted mathematically? > > >> What does that question mean? In particular, what does "digits in order > >> of pi" mean? > > >> Perhaps you could give some example lists, with the answer in each case. > > >> Sylvia. > > >>> ____________ > > >>> 3 > >>> 31 > >>> 314 > >>> ... > > > Huh? I'm not explaining trivial items to you. > > The purpose was to clarify the intent of "How many digits in order of pi > are below this line if interpreted mathematically?" Don't you want it > clarified? > > > > > Your "mutilation" of herc_can't_3 was proved erronous > > and it stands. > > Where was it proved erroneous? > > > There are other terms than contains where it holds > > so what is your argument? > > herc_cant_3 contained a proposition of the form > > If X has property A, then it also has property B. > > Such a proposition is demonstrable false if a possible X is demonstrated > to have property A, and demonstrated not to have property B. It doesn't > matter than there are some Xs with both A and B. The proposition is > falsified by a single counter-example. > > Since the proposition in herc_cant_3 has a counter-example, it cannot be > correct, and without a substitute proposition, herc_cant_3 cannot stand > as a theorem. > > > Your disproof was wrong and you > > won't admit you were wrong. Now you are shifting your erronous > > claim to a termi ology issue. > > No - I'm just trying to understand what your question was intended to mean. > > > Either give a proper disproof > > or ceasevyour eternal complaints. Hc3 stands. The ball > > is in your court to prove otherwise. > > Sylvia. Hc3 does not state all finite prefixes -> all inf sequences so your counter example is moot herc
From: Sylvia Else on 23 Jun 2010 05:06 On 23/06/2010 7:01 PM, Graham Cooper wrote: > On Jun 23, 6:55 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >> On 23/06/2010 6:43 PM, Graham Cooper wrote: >> >> >> >> >> >>> On Jun 23, 5:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: >>>> On 23/06/2010 5:03 PM, Graham Cooper wrote: >> >>>>> On Jun 23, 5:00 pm, Tim Little<t...(a)little-possums.net> wrote: >>>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: >> >>>>>>> On the face of it, line n contains the n digits of pie, >>>>>>> sequentially, and in order. I suppose it can be conceded that the >>>>>>> infinite list contains Pi. >> >>>>>> 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. >> >>>>>> It does satisfy a much looser property: there exists a sublist S such >>>>>> that lim S = pi. In general you can form a set of real numbers >>>>>> closure(L) = { x in R | exists sublist S of L such that lim S = x } >>>>>> which you could call the "closure" of a list L. >> >>>>>> Then you could say that pi is in the closure of the list, but pi is >>>>>> certainly not in the list itself. >> >>>>>> Herc does not know the difference. >> >>>>>> - Tim >> >>>>> How many digits in order of pi are below this line >>>>> if interpreted mathematically? >> >>>> What does that question mean? In particular, what does "digits in order >>>> of pi" mean? >> >>>> Perhaps you could give some example lists, with the answer in each case. >> >>>> Sylvia. >> >>>>> ____________ >> >>>>> 3 >>>>> 31 >>>>> 314 >>>>> ... >> >>> Huh? I'm not explaining trivial items to you. >> >> The purpose was to clarify the intent of "How many digits in order of pi >> are below this line if interpreted mathematically?" Don't you want it >> clarified? >> >> >> >>> Your "mutilation" of herc_can't_3 was proved erronous >>> and it stands. >> >> Where was it proved erroneous? >> >>> There are other terms than contains where it holds >>> so what is your argument? >> >> herc_cant_3 contained a proposition of the form >> >> If X has property A, then it also has property B. >> >> Such a proposition is demonstrable false if a possible X is demonstrated >> to have property A, and demonstrated not to have property B. It doesn't >> matter than there are some Xs with both A and B. The proposition is >> falsified by a single counter-example. >> >> Since the proposition in herc_cant_3 has a counter-example, it cannot be >> correct, and without a substitute proposition, herc_cant_3 cannot stand >> as a theorem. >> >>> Your disproof was wrong and you >>> won't admit you were wrong. Now you are shifting your erronous >>> claim to a termi ology issue. >> >> No - I'm just trying to understand what your question was intended to mean. >> >>> Either give a proper disproof >>> or ceasevyour eternal complaints. Hc3 stands. The ball >>> is in your court to prove otherwise. >> >> Sylvia. > > > Hc3 does not state > > all finite prefixes -> all inf sequences > > so your counter example is moot herc_cant_3 states "Given the increasing finite prefixes of ALL infinite expansions, that list contains every digit (in order) of every infinite expansion." If that's not what that statement is intended to mean, then what, *exactly* is it intended to mean? Sylvia.
From: Graham Cooper on 23 Jun 2010 06:00
On Jun 23, 7:06 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 7:01 PM, Graham Cooper wrote: > > > > > > > On Jun 23, 6:55 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> On 23/06/2010 6:43 PM, Graham Cooper wrote: > > >>> On Jun 23, 5:35 pm, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>> On 23/06/2010 5:03 PM, Graham Cooper wrote: > > >>>>> On Jun 23, 5:00 pm, Tim Little<t...(a)little-possums.net> wrote: > >>>>>> On 2010-06-23, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >>>>>>> On the face of it, line n contains the n digits of pie, > >>>>>>> sequentially, and in order. I suppose it can be conceded that the > >>>>>>> infinite list contains Pi. > > >>>>>> 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. > > >>>>>> It does satisfy a much looser property: there exists a sublist S such > >>>>>> that lim S = pi. In general you can form a set of real numbers > >>>>>> closure(L) = { x in R | exists sublist S of L such that lim S = x } > >>>>>> which you could call the "closure" of a list L. > > >>>>>> Then you could say that pi is in the closure of the list, but pi is > >>>>>> certainly not in the list itself. > > >>>>>> Herc does not know the difference. > > >>>>>> - Tim > > >>>>> How many digits in order of pi are below this line > >>>>> if interpreted mathematically? > > >>>> What does that question mean? In particular, what does "digits in order > >>>> of pi" mean? > > >>>> Perhaps you could give some example lists, with the answer in each case. > > >>>> Sylvia. > > >>>>> ____________ > > >>>>> 3 > >>>>> 31 > >>>>> 314 > >>>>> ... > > >>> Huh? I'm not explaining trivial items to you. > > >> The purpose was to clarify the intent of "How many digits in order of pi > >> are below this line if interpreted mathematically?" Don't you want it > >> clarified? > > >>> Your "mutilation" of herc_can't_3 was proved erronous > >>> and it stands. > > >> Where was it proved erroneous? > > >>> There are other terms than contains where it holds > >>> so what is your argument? > > >> herc_cant_3 contained a proposition of the form > > >> If X has property A, then it also has property B. > > >> Such a proposition is demonstrable false if a possible X is demonstrated > >> to have property A, and demonstrated not to have property B. It doesn't > >> matter than there are some Xs with both A and B. The proposition is > >> falsified by a single counter-example. > > >> Since the proposition in herc_cant_3 has a counter-example, it cannot be > >> correct, and without a substitute proposition, herc_cant_3 cannot stand > >> as a theorem. > > >>> Your disproof was wrong and you > >>> won't admit you were wrong. Now you are shifting your erronous > >>> claim to a termi ology issue. > > >> No - I'm just trying to understand what your question was intended to mean. > > >>> Either give a proper disproof > >>> or ceasevyour eternal complaints. Hc3 stands. The ball > >>> is in your court to prove otherwise. > > >> Sylvia. > > > Hc3 does not state > > > all finite prefixes -> all inf sequences > > > so your counter example is moot > > herc_cant_3 states > > "Given the increasing finite prefixes of ALL infinite expansions, > that list contains every digit (in order) of every infinite expansion." > > If that's not what that statement is intended to mean, then what, > *exactly* is it intended to mean? > > Sylvia. Don't you listen? I told you every digit in order does not mean a single infinite sequence. 3 31 314 .... How many digits of pi are in that list? Herc |