Another AC anomaly?
in [Logic]
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From: Jesse F. Hughes on 21 Dec 2009 14:17 WM <mueckenh(a)rz.fh-augsburg.de> writes: >> But you are not a mathematician. > > Do you think so? I studied mathematics and a university council > appointed me to teach mathematical lessons. Have you better insights > than they? Or have you only a different definition of mathematics? Do > you mean that I am not a matheologian? Then you are right! That you teach mathematics is a damned shame. I'm not sure why you draw attention to this sad mistake and I don't know why the administration hasn't the sense to fix their error. -- Jesse F. Hughes "[I]f gravel cannot make itself into an animal in a year, how could it do it in a million years? The animal would be dead before it got alive." --The Creation Evolution Encyclopedia
From: Virgil on 21 Dec 2009 16:34 In article <65fa0fa6-3723-4b81-ad46-1c3ab274fccc(a)t42g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 21 Dez., 14:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article > > <2fac8bb1-4c90-4421-b559-1ea7f0301...(a)e27g2000yqd.googlegroups.com> WM > > <mueck...(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? > > Except in matheology it is always required. What WM calls "Matheology" is merely a few of those parts of standard mathematics that WM is too inept to deal with > > > > �> � � � � � � � � � � � � � � � � � � � � 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. (those parts of standard mathematics that WM is too inept to deal with) > > > > �> > �> > �> > 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. > > 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. > > Even a matheologian should understand that. We understand it, but find it quite irrelevant to whether 1/3 can be identified with an infinite sequence of digits, which it can, in many ways. > > > > �> 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. > > I did not contest it. I said, if there is a sequence that identifies > 1/3, then the identifying digits must be 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. > > > > �> > �> > 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. � > > The union of finite �nitial segments cannot ield an infinite initial > segment? Does the sequence of 1/3 not consist of a union of all finite > initial segments? > > Regards, WM
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/ |