Cantor Confusion
in [Math]
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From: Dik T. Winter on 15 Feb 2007 08:05 In article <1171469941.993167.166620(a)h3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 14 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > > Numbers can express properties? You have lost me here. > > > > > > > > > > To have three elements is a property of a set. > > > > > > > > Oh. Your terminology is unfathomable. Indeed the number 3, when seen > > > > as a set, can have three elements. In the von Neumann model. However, > > > > I remember to also having seen another model, where the number three was > > > > {{{}}} > > > > > > It was page 93 of my book. > > > > I have seen it earlier than that. > > By the way, above is only number 2 given. Indeed, my error. So your comment: > When seen as a set of curly brackets it has 3 at the left sinde and 3 > at the right. was actually completely irrelevant. Let's get on with the actual representation of 3: {{{{}}}}. In this model, 3, when seen as a set, has only one element. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Feb 2007 08:03 In article <1171470343.811036.193110(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 14 Feb., 02:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > This union contains (as a subset) all sets p(n). Their union is, as yo > > > say, p(oo). They all are in the tree. Their union is in the tree. > > > Hence p(oo) s in the tree. > > > > But you want it in the union of the P(i). Again you fail to answer the > > basic question. How should it be possible that p(oo) (an infinite path) > > is in the union of the P(i) (sets of finite paths), when each P(i) > > contains only finite paths as elements. > > It need not be in the union. You say it is the union of {p(0)} U > {p(1)} U {p(2)}. If so, then it is the union of a subset of P(0) U > P(1) U P(2) U ... I never stated that p(oo) = {p(0)} U {p(1)} U ...; that can not be the case, because on the left hand side we have a set of nodes, and on the right hand side we have a set of paths. And a path is not a node. Moreover, {p(0)} U {p(1)} U ... is not the union of a subset of P(0) U P(1) U ...; it *is* a subset. You are again confusing two things. If you replace in your paragraph {p(0)} U {p(1)} U ... by p(0) U p(1) U ... it becomes correct. But you stated that P(oo) was a subset of P(0)U P(1) U ..., and that is trivially false, because p(0) U p(1) U ... is an element of P(oo) but not of P(0) U P(1) U ... . > > > > Bit with > > > > this definition above, p(oo) is *not* established by: > > > > {p(0)} U {p(1)} U {p(2)} U ... > > > > because the paths are not united, it is the sets containing paths as a > > > > single element that are united. > > > > > > The tree T(oo) contains all pats p(n), hence the tree conains the > > > union. > > > > We were not talking about the tree T(oo), we were talking about: > > {p(0)} U {p(1)} U {p(2)} U ... > > which according to you did "establish" p(oo), while I show above that > > that is not the case (according to the definition for "establish" you > > agreed with). > > That is a formal result. Indeed, so your statement that is was established by that union was false. > Take he union of this union, then you have > p(oo). This union is done in the tree. That does have no bearing on the correctness of your statement. > > > The tree T(oo) contains all finite trees T(n). Do you think if we > > > don't form the union, then T(oo) does not exist? It is established by > > > the presence of all finite trees. Same is valid for the paths. > > > > > > But if you think so, then, please, form the union. Then you have T(oo) > > > and all paths p(oo), q(oo), ... > > > > Yes. Why do you think I disagree with all this? I have stated again > > and again that I *did* agree with all this. What I disagree with is > > that P(oo) (the set of paths in T(oo)) is a subset of the union of the > > P(i) (the sets of paths in T(i)). > > Every element of P(oo) is the union of a subset of the union of the > sets P(i). Yes. But why do you think that P(oo) is a subset of that union? For that each element of P(oo) must be an *element* of that union. > > And I also stated, again and again, > > that that is trivial to prove because none of the P(i) contains an > > infinite path, and so their union also does not contain an infinite > > path, but P(oo) contains infinite paths. (Contains meaning: having > > as an element.) > > > > Pray keep your focus on what is discussed. > > Tha is: "to estalish" means from now on "being the union" of a set or > a subset. > p(oo) is the union of a subset of the union of the sets P(i). So it has no bearing on whether P(oo) is a subset of the P(i) or not. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: G. Frege on 15 Feb 2007 09:27 On Thu, 15 Feb 2007 13:05:34 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >> >> When seen as a set of curly brackets it has 3 at the left sinde and 3 >> at the right. >> Seems that WM is mixing up the name with the named object again. :-) F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 15 Feb 2007 13:02 On 14 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171366247.193486.239...(a)v45g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 13 Feb., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > The abstract > > > number three is not a representation. > > > > I asked what has the abstract number 3, that III has not? (I did not > > ask what the abstract number 3 has not.) > > It is abstract and does not depend on convention. > > > > > > Numbers can express properties? You have lost me here. > > > > > > > > To have three elements is a property of a set. > > > > > > Oh. Your terminology is unfathomable. Indeed the number 3, when seen > > > as a set, can have three elements. In the von Neumann model. However, > > > I remember to also having seen another model, where the number three was > > > {{{}}} > > > > It was page 93 of my book. > > I have seen it earlier than that. > > > >(but I think now you can only model the natural numbers). Anyhow, > > > within this model 3 when seen as a set has only one element. > > > > When seen as a set of curly brackets it has 3 at the left sinde and 3 > > at the right. > > Yeah. So 3 has the property that it has two sets of three elements. > So 3 has properties that depend on the representation. One set of curly brackets would be enough to be number 3. But it is correct that there are more obvious and less obvious numbers: .... or ||| is number 3 in a more obvious way than {{{ while this is number 3 in a more obvious way than abc .. Regards, WM
From: Virgil on 15 Feb 2007 14:05
In article <1171562525.973762.49300(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 14 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171366247.193486.239...(a)v45g2000cwv.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 13 Feb., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > The abstract > > > > number three is not a representation. > > > > > > I asked what has the abstract number 3, that III has not? (I did not > > > ask what the abstract number 3 has not.) > > > > It is abstract and does not depend on convention. > > > > > > > > Numbers can express properties? You have lost me here. > > > > > > > > > > To have three elements is a property of a set. > > > > > > > > Oh. Your terminology is unfathomable. Indeed the number 3, when seen > > > > as a set, can have three elements. In the von Neumann model. > > > > However, > > > > I remember to also having seen another model, where the number three > > > > was > > > > {{{}}} > > > > > > It was page 93 of my book. > > > > I have seen it earlier than that. > > > > > >(but I think now you can only model the natural numbers). Anyhow, > > > > within this model 3 when seen as a set has only one element. > > > > > > When seen as a set of curly brackets it has 3 at the left sinde and 3 > > > at the right. > > > > Yeah. So 3 has the property that it has two sets of three elements. > > So 3 has properties that depend on the representation. > > > One set of curly brackets would be enough to be number 3. But it is > correct that there are more obvious and less obvious numbers: > ... > or > ||| > is number 3 in a more obvious way than > {{{ > while this is number 3 in a more obvious way than > abc WM is still conflating the name with the named and maps with territories. |