constructible = unconstructible
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From: Virgil on 10 Jun 2010 16:58 In article <4a76ac0d-009f-4ca8-8e1b-74cd3234c5b3(a)z10g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Jun 10, 11:32�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 10 Jun., 16:00, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On Jun 10, 10:33�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 10 Jun., 14:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > On Jun 9, 10:37�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > > It is well known that the set of finite descriptions > > > > > > > > is countable (though not computable). > > > > > > > > > The set of all names which unavoidably include all numbers is > > > > > > > computable. > > > > > > > > The set of all names, S is computable. > > > > It is well known that the set of computable numbers > > > > is countable (though not computable). > > > > <important context restored> > > > > The set of all computable numbers, R is a subset of S. > > A subset of a computable set may or may not be computable. > > R is not computable. > > > > > And what is the real part or numerical value of an uncomputable real > > > number? > > > > Something that cannot be computed. �If you take the > > position that something that cannot be computed > > does not exist > > There is no position to take. If something that does only exist by > computing it, cannot be computed, then it does not exist. That is not > a position but is simply the truth. > > > then neither an uncomputable number > > nor a list of all computable numbers exist. > > I showed you a computed list that contains all computable numbers (and > some other names). So yout assertion is false. Wm now claims, in effect, that because {1,3,5,7} is a list of odd naturals then any list of which it is a sublist must also be a list of odd naturals. Including {1,2,3,4,5,6,7,8} must now, in WM's world, be a list of odd numbers. > In addition: Cantor's diagonal numbers are computed numbers. Only after a list of numbers has been provided, and if that list were complete, it would contain those uncomputable numbers. > Therefore > it is meaningless to defend his false conclusion by uncomputable > numbers. Cantor's "proof" proves the uncountability of a countable > set. Whether or not a list of those numbers can be computed is > irrelevant because we know it is countable. That particular mish-mash of illogicisms is typical of WM's doublethink.
From: Virgil on 10 Jun 2010 17:04 In article <98a0bbea-0379-497a-b65d-744d167bccb9(a)j8g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Jun 10, 11:32�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 10 Jun., 16:00, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On Jun 10, 10:33�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 10 Jun., 14:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > On Jun 9, 10:37�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > > It is well known that the set of finite descriptions > > > > > > > > is countable (though not computable). > > > > > > > > > The set of all names which unavoidably include all numbers is > > > > > > > computable. > > > > > > > > The set of all names, S is computable. > > > > It is well known that the set of computable numbers > > > > is countable (though not computable). > > > > <important context restored> > > > > The set of all computable numbers, R is a subset of S. > > A subset of a computable set may or may not be computable. > > R is not computable. > > > > > And what is the real part or numerical value of an uncomputable real > > > number? > > > > Something that cannot be computed. �If you take the > > position that something that cannot be computed > > does not exist > > Something that can only exist as a computed entity does not exist, of > course, when it cannot be computed. That is not a position, but that > is simply the truth. There is nothing, or at least nothing that I am aware of, in any definition of "number" that either states explicitly or even implies that it must be 'computable'. And in the real number system, which is quite well defined, there is good reason to suppose otherwise. > > > > then neither an uncomputable number > > nor a list of all computable numbers exist. > > I showed you the list that contains all computable numbers (and some > more names). So your claim is wrong. Then, by WM's own argument, WM must insist that {1,2,3,4,5,6,7,8} is a LIST OF ODD NUMBERS because some of them are odd.
From: Owen Jacobson on 10 Jun 2010 23:02 On 2010-06-10 09:30:27 -0400, WM said: > On 10 Jun., 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: >> On Jun 10, 7:34�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: >> >> <snip> >> >>> if we divide 1 by 9, then 0.111... does exist according >>> to the above definition. >>> If we write it in a single line, nobody will disagree. >>> If we write it such that every next 1 is placed in a new line, like >> >>> 0.1 >>> 0.11 >>> 0.111 >>> ... >> >>> then everybody will disagree that all 1 are in one line but all 1 >>> "must be there somehow". >> >> Yes and outside Wolkenmuekenheim, everyone is correct. >> However inside Wolenmuekenheim there is a 1 in 0.111... >> that is not in the list. > > What is flaming good for? My proof uses complete induction. One can > believe in it or not. Your proof reaches nonsensical conclusions in the context in which you present it. Therefore there must be a flaw in it. Finding this flaw is decidedly *your* problem. Let's frame your argument slightly differently. Presuppose N exists, and has the properties normally associated with it, for the sake of the following. For all elements n of N, let F_n be a function from N to {0, 1} as follows: F_n(x) = 1 if x <= n F_n(x) = 0 if x > n Let G be a function from N to {0, 1} as follows: G(x) = 1 Let L be a function from N to {F_0, F_1, ...} as follows: L(x) = F_x Which of the following statements accurately characterizes your argument? 1. If N exists, then there exists some natural number z in N such that L(z) = F_z = G. 2. G cannot exist even given the presupposition that N exists as above. -o
From: WM on 11 Jun 2010 07:18 On 11 Jun., 05:02, Owen Jacobson <angrybald...(a)gmail.com> wrote: > On 2010-06-10 09:30:27 -0400, WM said: > > > > > > > On 10 Jun., 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > >> On Jun 10, 7:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > >> <snip> > > >>> if we divide 1 by 9, then 0.111... does exist according > >>> to the above definition. > >>> If we write it in a single line, nobody will disagree. > >>> If we write it such that every next 1 is placed in a new line, like > > >>> 0.1 > >>> 0.11 > >>> 0.111 > >>> ... > > >>> then everybody will disagree that all 1 are in one line but all 1 > >>> "must be there somehow". > > >> Yes and outside Wolkenmuekenheim, everyone is correct. > >> However inside Wolenmuekenheim there is a 1 in 0.111... > >> that is not in the list. > > > What is flaming good for? My proof uses complete induction. One can > > believe in it or not. > > Your proof reaches nonsensical conclusions in the context in which you > present it. Therefore there must be a flaw in it. Finding this flaw is > decidedly *your* problem. The flaw has been found long time ago. It consists in the assumption that infinity can be finished, entertained by set theorists. > > Let's frame your argument slightly differently. Presuppose N exists, That's the flaw. > and has the properties normally associated with it, for the sake of the > following. > > For all elements n of N, let F_n be a function from N to {0, 1} as follows: > > F_n(x) = 1 if x <= n > F_n(x) = 0 if x > n > > Let G be a function from N to {0, 1} as follows: > > G(x) = 1 > > Let L be a function from N to {F_0, F_1, ...} as follows: > > L(x) = F_x > > Which of the following statements accurately characterizes your argument? > > 1. If N exists, then there exists some natural number z in N such that > L(z) = F_z = G. Yes. But z does not and cannot exist. Therefore N cannot exists as a completed entity. > 2. G cannot exist even given the presupposition that N exists as above. Given N then G must exist. G consists merely of indices n attached to the ever repeating 1. The result of my proof is that G cannot exist as an infinite sequence. G exists as a finite definition "1/9" or "0.111..." (this is a finite definition using 8 symbols) or "G". But there are only countably many finite definitons. Regards, WM
From: WM on 11 Jun 2010 07:23
On 10 Jun., 17:20, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 10, 12:07 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > <snip> > > > > then neither an uncomputable number > > > nor a list of all computable numbers exist. > > > I showed you a computed list that contains all computable numbers (and > > some other names). So yout assertion is false. > > You did not show me a list that contains all computable > numbers and only computable numbers. Showing a computable > list that contains "all computable numbers > (and some other names)" is not enough to demonstrate > the existence of a computable list that contains all > computable numbers and only computable numbers. Why should I try to find such a list when I know that this list would be countable? That's the same with the paths of the binary tree. I cannot enumerate them. But I can estimate that all together form a set that is not larger than a countable set. Regards, WM |