Another AC anomaly?
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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/
From: Dik T. Winter on 21 Dec 2009 08:17 In article <2fac8bb1-4c90-4421-b559-1ea7f0301d4f(a)e27g2000yqd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > 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. Oh, is it often required? > T talk about all in case there > is no last is silly. And I think it is silly to require there being a last to be able to talk about all. > > > > > > Right, but there is no finite initial segment that contains them > > > > > > all. .... > > > > Sorry, I have no knowledge of the bible. But live without that axiom > > > > when you can't stomach it. And do not attack mathematicians who > > > > live with that axiom. > > > > > > To live with that axiom does not create uncountability. See the proof > > > here: > > >http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb= > ... > > > > Where is the proof there? I see only you writing a bit of nonsense and > > two rebuttals. > > 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. So what? That is not contested and it does not show in *any* way that the axiom of infinity does not create uncountability. So no proof at all. > Just what I said. And just wat I said: see the quote above: > > > > > > Right, but there is no finite initial segment that contains them > > > > > > all. which you contested. > > > > 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? Does not matter. The axiom of infinity does *not* construct infinite paths in your tree, beacuse you stated that your tree did not contain infinite paths a priori. So those infinite things are not paths by your statement. Neither does the axiom of infinity establish a finite set N of all finite numbers. > > > 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. Only minds completely disformed by set theory could try to > defend the obviously false position that there were uncountably many > paths. But: if you consider only finite sequences of nodes as paths, there *are* countably many paths. You continuously confuse what you consider being a path and what others consider a path. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 Dec 2009 08:23
In article <9ad63bf1-549b-4822-bf86-839dd7c64d58(a)j4g2000yqe.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 18 Dez., 15:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > No, it is a matter of convention. In mathematics > > > > a, b, c, ..., z > > > > means a, b, c, continue this way until you reach z. But starting > > > > {1}, {1, 2}, {1, 2, 3} > > > > and going on you never reach > > > > {1, 2, 3, ...} > > > > > > That is true. Therefore it does not exist. > > > > That you can not get there step by step does not mean that it does not > > exist. > > That you cannot get step by step to 1/0 does not mean that it does not > exist? Indeed. On the projective line (that precedes Cantor by quite some time as far as I know) it does exist. > > > However, see Cantor, > > > collected works, p 445: > > > 0, 1, 2, 3, ... w_0, w_0 + 1, ..., gamma, ..., > > > He seems to reach far more. > > > > Right, he uses a convention that is no longer used. > > Wrong, it is used presently, for instance by myself. But you are not a mathematician. > > > No. But the union contains two paths. > > > > Wrong. If we look at the paths as sets, they are sets of nodes. Their > > union is a set of nodes, not a set of paths. And as a set of nodes we > > can form from them seven different paths. > > Wrong. The nodes of two paths give exactly two paths. Darn, the paths 0.000 and 0.100 contain the following nodes: 0.000 = {0., 0.0, 0.00, 0.000} and 0.100 = {0., 0.1, 0.10, 0.100} where a node is named by the path leading to it. Their union contains the following paths: 0., 0.0, 0.00, 0.000, 0.1, 0.10, 0.100 and I count seven. > > > Let every finite path of every infinite path be mapped on the elements > > > of omega. That was simple. > > > > By your statements infinite paths do not exist. But pray give such a > > mapping. Until now you have only asserted that such a mapping exists > > without showing that. > > Do you accept the mapping from omega on SUM{k = 1 to n} 3*10^-k yields > the infinite decimal expansion of 1/3? *What* mapping? Do you mean from n in omega -> SUM...? In that case the infinite decimal expansion of 1/3 is unmapped. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |