Another AC anomaly?
in [Logic]
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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/
From: WM on 21 Dec 2009 12:02 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. > > > 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. > > > > > > > > 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: 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. > > > 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 ínitial 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: WM on 21 Dec 2009 12:15 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? > > > > > 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? 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! > > > > > 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. > > > > > 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. Regards, WM
From: Marshall on 21 Dec 2009 13:20
On Dec 21, 9:15 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 21 Dez., 14:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > 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? I studied mathematics and a university council > appointed me to teach mathematical lessons. Have you better insights > than they? Obviously. Marshall |