Another AC anomaly?
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From: Virgil on 21 Dec 2009 16:49 In article <b484e377-9dc2-424b-80c3-2912165f636c(a)a32g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Dez., 14:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article > > <9ad63bf1-549b-4822-bf86-839dd7c64...(a)j4g2000yqe.googlegroups.com> WM > > <mueck...(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. > > But does the projective line exist? To mathematicians it does. It is hardly relevant whether it exists for putzers like WM. > > > > �> > �> � � � � � � � � � � � � � � � � � 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. > > Do you think so? Present evidence supports Dik on this issue. >I studied mathematics and a university council > appointed me to teach mathematical lessons. Attempting is not the same as succeeding. And unless that university council included at least one bona fide mathematicain, they were in no position to tell whether WM was one or not. > Have you better insights > than they? We have the insight into your inadequacies provided by your postings here, which that council unfortunately lacked. > Or have you only a different definition of mathematics? Do > you mean that I am not a matheologian? Then you are right! Since what you call matheology is what actual mathematicians consider a part of standard mathematics, your rejection of it marks you as not a mathematician at all. > > > > �> > �> 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. > > The nodes of two maximal paths give two maximal paths. You count > initial segments. The nodes of two distinct maximal paths do not form a path at all, since there must be some node among those nodes with two children among those nodes. > > > > �> > �> 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. > > The sum of all finite segments is not the infinite path. Interesting. The UNION of the set of all finite initial segments may be a path, but the alleged sum of them diverges.
From: Dik T. Winter on 22 Dec 2009 09:17 In article <8baa45a8-2ba4-464b-9598-5656c43a7456(a)j19g2000yqk.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 18 Dez., 15:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <dd852490-e622-4bc4-b858-dfcc3b142...(a)l13g2000yqb.googlegroups= > .com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > > ... > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > No. Suppose we have the paths 0.000 and 0.100, what is their > > > > > > union? > > > > > > And is it a path? > > > > > > > > > No. But the union contains two paths. And an infinite union of that > > > > > kind may contain the path 0.111... > > > > > > > > So according to WM a union of sets may contain an object not > > > > contained in any of the sets being unioned. > > > > > > For paths and initial segments to contain and to be is the same. > > > > Eh? paths contain nodes and initial segments contain numbers. > > But be aware that you use the word 'contains' with two different meanings > > at different times. > > For paths we have: Paths contain initial segments and paths are > initial segments. Yes, you are using "contains" with two different meanings: "be an element of" and "be a subset of". In many cases you do not distinguish them and that leads to misunderstandings. > > > For paths and initial segments to contain and to be is the same. In > > > fact: > > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > > and also > > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. > > > > In the mathematical sense the union contains numbers, not sets. > > The union contains subsets. > > > This is so according to set theory. Of course it is rubbish. > > > > Well, if you want to use terminology in a different meaning than > > standard, of course. > > To contain as a subset is not correct in English? It is correct English but misleading in a set theoretic context. -- 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 22 Dec 2009 09:28 In article <65fa0fa6-3723-4b81-ad46-1c3ab274fccc(a)t42g2000yqd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Dez., 14:17, "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? > > Except in matheology it is always required. I did not know of that requirement. Can you provide for a reference where that requirement is mentioned? > > > 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. > > That's why you love matheology. I would have thought that you would be able to provide for a textbook where that requirement is mentioned. So give me one. > > > > > 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. > > It may create what you like. Either 1/3 can be identified at a finite > digit or 1/3 cannot be identified at a finite digit. Not. > Even a matheologian should understand that: If there is no digit at a > finite place up to that the sequence 0.333... identifies the number > 1/3, then there is no digit at a finite place up to that the number > 1/3 can be identified. Right, there is no digit at a finite place up to that the number 1/3 can be identified. And as there are no digits at infinite places that appears to you to be a paradox. It is not. There is *no* finite sequence of digits that identifies 1/3. But there is an *infinite* sequence of digits that does so. > > And just wat I said: see the quote above: > > > > > > > > Right, but there is no finite initial segment that contains > > > > > > > > them all. > > > > which you contested. > > I did not contest it. Why then did you reply with: > That is pure opinion, believd by the holy bible (Dominus regnabit in > aeternum et ultra. [2. Buch Moses: Exodus 15 Vers 18]) or forced upon > us by the men-made axiom of infinity. it that is not contesting it? > I said, if there is a sequence that identifies > 1/3, then the identifying digits must be at finite places. Right, all identifying digits (there are infinitely many) are at finite places. > But we know > that for every finite place d_n, there is a sequence d_1, ..., d_n > that is not 1/3 but is identical to the sequence of 1/3. Therefore we > can conclude that there is no sequence identifying the number 1/3 by > means of digits at finite places only. You can only conclude that there is no *finite* sequence that identifies the number 1/3. You here exclude the possibility of an infinite sequence of digits at finite places only, i.e. assuming that what you want to prove. > > > > 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. > > The union of finite initial segments cannot ield an infinite initial > segment? Yes. But as you have stated that your tree contained finite paths only, such an infinite initial segment is not (according to *your* definition) a path. > Does the sequence of 1/3 not consist of a union of all finite > initial segments? It is, but also (according to *your* definition) it is not 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 22 Dec 2009 09:47 In article <b484e377-9dc2-424b-80c3-2912165f636c(a)a32g2000yqm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Dez., 14:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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. > > But does the projective line exist? Exist in what sense? But it is a concept from projective geometry, but perhaps you think that is also nonsense? I may note that the point at infinity was developed by Kepler and Desargues in the 17th century. > > > > > 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. > > Do you think so? Yes, I think so and your inability to provide definitions and proper proofs shows it. > I studied mathematics and a university council > appointed me to teach mathematical lessons. Well, the same did hold for my father, but he never considered himself a mathematician, but a physicist. > > > > 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. > > The nodes of two maximal paths give two maximal paths. You count > initial segments. I did not count segments, I counted paths. You stated there are only two paths in the union, I count seven paths. Or are those seven things suddenly not paths? > > > > 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. > > The sum of all finite segments is not the infinite path. Interesting. You have stated that your paths are finite, so I consider the mentioning of an infinite path strange. Moreover, you are now talking about sums, not about paths. But *what* n in omega maps to the infinite decimal expansion? You have not stated that. And if you think the infinite expansion is mapped you should be able to state an element of omega that maps to that infinite 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: Virgil on 22 Dec 2009 16:10
In article <Kv251u.sI(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > > The union contains subsets. > > > > > > This is so according to set theory. Of course it is rubbish. > > > > > > Well, if you want to use terminology in a different meaning than > > > standard, of course. > > > > To contain as a subset is not correct in English? > > It is correct English but misleading in a set theoretic context. A set can contain something as a subset OR contain something as a member, but these are different containment relationships, neither of which necessitates the other. WM has the bad habit of using "contain" ambiguously. It often seems that he is being purposely ambiguous about it. Proper mathematical usage requires that "contain" be used unambiguously, whenever there can be any doubt, by explicitly stating in each such instance whether it means the subset or the membership relation. |