Another AC anomaly?
in [Logic]
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From: Ross A. Finlayson on 19 Dec 2009 18:21 On Dec 19, 12:14 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <2fac8bb1-4c90-4421-b559-1ea7f0301...(a)e27g2000yqd.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > Aha, you are clearly a mindreader. Well, as far as I know mindreading > > > > > is not part of mathematics. Anyhow, I can think of numbers larger > > > than > > > > > that path. > > > > > But that is completely irrelevant. I am able to think about a set > > > that > > > > > contains all natural numbers, you apparently are not. > > > > > > > > How do you know that without confirming it by thinking the last too? > > > > Why need I to think about a last one (which there isn't) to be able to > > > think > > > about a set that contains all natural numbers? Apparently you have some > > > knowledge about how my mind works that I do not have. > > > Yes. A very convincing and often required proof of completenes of a > > linear set is to know the last element. > > Knowing the allegedly last element in a set does not work unless one has > imposed a linear ordering on the set, which is in no wise necessary for > sethood. Which is the "last" point of the set of points on a circle? > > > > > One of the rebuttals has meanwhile been changed. Peter Webb > > recognized: It is true that you cannot show pi as a finite decimal, > > but you can't show 1/3 as a finite decimal either. > > > Just what I said. > > > > > > > The tree contains all paths that can be constructed by nodes, using > > > > > > the axiom of infinity. Which one would be missing? > > > > > > > > > > The infinite paths because you stated a priori that your tree did not > > > > > contain infinite paths. So it is impossible to construct in your tree > > > > > infinite paths by the axiom of infinity. > > > > > > > > The axiom of infinity establishes the set N from finite numbers. > > > > It establishes the *existence* of a set N of finite numbers. > > > What else should be established? > > > > > It establishes the infinite paths as well in my tree from finite > > > > paths. > > > > No. That is impossible because you stated that the paths were finite. > > > What it *does* establish is the extistence of a set P of finite paths.. > > > It is rather silly to argue about the uncountability of the set of > > paths. > > Then stop doing it. There is a perfectly adequate proof that one cannot > have a list of paths that contains all paths, or , equivalently, one > cannot have a list of all subset of N. > Since being capable of being listed is a necessary requirement for > countability, that proof eliminates countability for the set of all > paths s well as for the power set of any infinite set. > > > Only minds completely disformed by set theory could try to > > defend the obviously false position that there were uncountably many > > paths. > > Minds thus "deformed" by set theory are often capable of creating valid > proofs that WM not only cannot create but also cannot even follow. > > > > > But "10 Questions" will give you the answer why there are not > > uncountably many paths. There are no infinite decimal expansions of > > real numbers. There are not due paths in the tree. > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > ZF there are infinitely many more of them. > > I know not what a "due path" is, but in my infinite binary tree, whose > nodes are the members of N, every path is an infinite subset of N, and > there are more of them than even WM can count. > > > > > It is as impossible to express any real number by an infinite decimal > > expansion as it is impossible to express 0 by a unit fraction. The > > rest will be explained in "10 Questions". Therefore I will stop with > > this topic here. > > Then in WM's world, 1.000... is not a real number. > > Odd of him, but he has always been odd. There you assume the consistency of the statement "Dedekind/Cauchy is complete", where it's a constructive argument. (Union of countables.) Otherwise there might be systems where that is not so yet constructively irrelevant. Ross Finlayson
From: Virgil on 19 Dec 2009 20:14 In article <0b77d31b-1160-4c36-8935-58859e094d1c(a)j19g2000yqk.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Dez., 21:14, Virgil <Vir...(a)home.esc> wrote: > > > > Yes. A very convincing and often required proof of completenes of a > > > linear set is to know the last element. > > > > Knowing the allegedly last element in a set does not work unless one has > > imposed a linear ordering on the set, which is in no wise necessary for > > sethood. Which is the "last" point of the set of points on a circle? > > I am interested in linear sets. Every set with more than one element has more than one "linearity" or ordering possible, and any set non-empty set can be ordered so as to have a first and a last member according to that ordering, though for one-element sets they are not distinct. > > > > It is rather silly to argue about the uncountability of the set of > > > paths. > > > > Then stop doing it. There is a perfectly adequate proof that one cannot > > have a list of paths that contains all paths, or , equivalently, one > > cannot have a list of all subset of N. > > You are wrong. The "perfectly adequate proof" is wrong. You keep claiming that but have never come up with convincing reasons for your claim. > > > > Only minds completely disformed by set theory could try to > > > defend the obviously false position that there were uncountably many > > > paths. > > > > Minds thus "deformed" by set theory are often capable of creating valid > > proofs that WM not only cannot create but also cannot even follow. > > There is one simple proof: Every node of the binary tree splits one > path into two paths. the number of paths cannot surpass the number of > nodes. First of all, every node of a COMPLETE INFINITE BINARY TREE splits one infinite set of infinite paths into two disjoint but equally infinite subsets of equally infinite paths. Anything less and the original tree, as is usual with WM's trees, cannot have been a complete infinite binary tree. > Nobody would ever have doubted that without the unfortunate > development of mathematics induced by Cantor. You have just conceded that it was a MATHEMATICAL developement, so it is now, and ever will remain, a part of mathematics. > > > > > > > > > But "10 Questions" will give you the answer why there are not > > > uncountably many paths. There are no infinite decimal expansions of > > > real numbers. There are not due paths in the tree. > > > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > > ZF there are infinitely many more of them. > > From the sequence of digits you cannot obtain 1. First you must be > sure that there never will appear a digit other than 0. But unless you > know the last digit, you cannot be sure. And you cannot know the last > digit. I can know that there ISN'T a last digit, at which point I can and do know that all of the zero digits are redundant. > > Of course you can know a formula saying digit d_n = 0 for all n in N. > But that is a finite formula. It may be a finite formula, but it defines infinitely many digit positions quite satisfactorily, at least for mathematicians. > It is not an infinite sequence of digits > that informes you about the due number. Numbers can exist quite satisfactorily in mathematics without any particular form of numeral being able to name them all. WM's peculiar theory of numerals is irrelevant to mathematics.
From: Virgil on 19 Dec 2009 20:16 In article <1396bed2-02c8-4a39-b6aa-b1f23fa0b4c6(a)u25g2000prh.googlegroups.com>, Marshall <marshall.spight(a)gmail.com> wrote: > On Dec 19, 1:36�pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > From the sequence of digits you cannot obtain 1. First you must be > > sure that there never will appear a digit other than 0. But unless you > > know the last digit, you cannot be sure. And you cannot know the last > > digit. > > > > Of course you can know a formula saying digit d_n = 0 for all n in N. > > But that is a finite formula. It is not an infinite sequence of digits > > that informes you about the due number. > > I think that may be your lamest argument yet. > > > Marshall Just wait. WM will ever exceed himself in that respect.
From: Virgil on 19 Dec 2009 20:20 In article <500d09c4-6322-417e-a4d9-f9716b19ee8a(a)c34g2000yqn.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson(a)gmail.com> wrote: > On Dec 19, 12:14�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <2fac8bb1-4c90-4421-b559-1ea7f0301...(a)e27g2000yqd.googlegroups.com>, > > > > > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > �> > Aha, you are clearly a mindreader. �Well, as far as I know > > > > mindreading > > > > �> > is not part of mathematics. �Anyhow, I can think of numbers larger > > > > than > > > > �> > that path. > > > > �> > But that is completely irrelevant. �I am able to think about a set > > > > that > > > > �> > contains all natural numbers, you apparently are not. > > > > �> > > > > �> How do you know that without confirming it by thinking the last too? > > > > > > Why need I to think about a last one (which there isn't) to be able to > > > > think > > > > about a set that contains all natural numbers? �Apparently you have > > > > some > > > > knowledge about how my mind works that I do not have. > > > > > Yes. A very convincing and often required proof of completenes of a > > > linear set is to know the last element. > > > > Knowing the allegedly last element in a set does not work unless one has > > imposed a linear ordering on the set, which is in no wise necessary for > > sethood. Which is the "last" point of the set of points on a circle? > > > > > > > > > One of the rebuttals has meanwhile been changed. Peter Webb > > > recognized: It is true that you cannot show pi as a finite decimal, > > > but you can't show 1/3 as a finite decimal either. > > > > > Just what I said. > > > > > > �> > �> The tree contains all paths that can be constructed by nodes, > > > > using > > > > �> > �> the axiom of infinity. Which one would be missing? > > > > �> > > > > > �> > The infinite paths because you stated a priori that your tree did > > > > not > > > > �> > contain infinite paths. �So it is impossible to construct in your > > > > tree > > > > �> > infinite paths by the axiom of infinity. > > > > �> > > > > �> The axiom of infinity establishes the set N from finite numbers. > > > > > > It establishes the *existence* of a set N of finite numbers. > > > > > What else should be established? > > > > > > �> It establishes the infinite paths as well in my tree from finite > > > > �> paths. > > > > > > No. �That is impossible because you stated that the paths were finite. > > > > What it *does* establish is the extistence of a set P of finite paths. > > > > > It is rather silly to argue about the uncountability of the set of > > > paths. > > > > Then stop doing it. There is a perfectly adequate proof that one cannot > > have a list of paths that contains all paths, or , equivalently, one > > cannot have a list of all subset of N. > > �Since being capable of being listed is a necessary requirement for > > countability, that proof eliminates countability for the set of all > > paths s well as for the power set of any infinite set. > > > > > Only minds completely disformed by set theory could try to > > > defend the obviously false position that there were uncountably many > > > paths. > > > > Minds thus "deformed" by set theory are often capable of creating valid > > proofs that WM not only cannot create but also cannot even follow. > > > > > > > > > But "10 Questions" will give you the answer why there are not > > > uncountably many paths. There are no infinite decimal expansions of > > > real numbers. There are not due paths in the tree. > > > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > > ZF there are infinitely many more of them. > > > > I know not what a "due path" is, but in my infinite binary tree, whose > > nodes are the members of N, every path is an infinite subset of N, and > > there are more of them than even WM can count. > > > > > > > > > It is as impossible to express any real number by an infinite decimal > > > expansion as it is impossible to express 0 by a unit fraction. The > > > rest will be explained in "10 Questions". Therefore I will stop with > > > this topic here. > > > > Then in WM's world, 1.000... is not a real number. > > > > Odd of him, but he has always been odd. > > There you assume the consistency of the statement "Dedekind/Cauchy is > complete" Not at all! I merely note that every rational integer is included among the reals, in both the Dedekind construction and the Cauchy construction, and 1 is nicely both rational and integral.
From: Dik T. Winter on 21 Dec 2009 08:08
In article <QeCdnWPXUMnJqLHWnZ2dnUVZ_o-dnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:Kuuqrs.FJ7(a)cwi.nl... .... > Let A_n and B_n be two sequences of sets of the form {X_n}. > Let A_s = lim sup A_n and A_i = lim inf A_n, similar for B_s > and B_i. Let C_n be the sequence defined as C_2n = A_n and > C_(2n+1) = B_n. > > Theorem: > Since A_s = {a_s} and B_s = {b_s} > > lim sup C_n = {a_s \/ b_s} > > lim inf C_n = {a_i /\ b_i} Let A_n = {{a}} for some object a, let B_n = {{n}} with n natural. With your definition: A_s = {{a}}, B_s = {{N}} A_i = {{a}}, B_i = {{N}} but C_s = {{a}} and C_i = {{}}. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |