Cantor Confusion
in [Math]
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From: Virgil on 13 Feb 2007 15:47 In article <1171366968.518530.122710(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 13 Feb., 02:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171284657.431713.275...(a)l53g2000cwa.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 12 Feb., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > You repeatadly misunderstand: > > > > (1) The union of sets of finite elements does contain an infinite > > > > element > > > > with > > > > (2) The union of finite sets is infinite. > > > > The first is false, the second is true. > > > > > > > > But you want to conclude (1) from (2). > > > > > > No. I do not conclude that p(oo) is in the union of P(0) U P(1) U P(2) > > > U ... and I do not conclude that p(oo) is in the union of {p(0)} U > > > {p(1)} U {p(2)} U .... > > > > > > What I conclude and what reasonably must be concluded is: > > > {p(0)} U {p(1)} U {p(2)} U .... c P(0) U P(1) U P(2) U ... > > > > But, indeed, that is true. > > > > > If p(oo) is established by {p(0)} U {p(1)} U {p(2)} U .... > > > Then p(oo) is also established by P(0) U P(1) U P(2) U ... > > > > But again, I ask you what you mean with "established" here. It is neither > > an element of those unions, nor is it one of those unions. So what is > > your meaning? As I have stated again and again: > > p(oo) = p(0) U p(1) U p(2) U ... > > but that one is quite different from > > {p(0)} U {p(1)} U {p(2)} U ... > > 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. WM again conflates A U B with {A} U {B}. They are not the same. > > The tree T(oo) contains all pats p(n), hence the tree conains the > union. How does T(oo) "contain" all p(n)? As members? As members of members? WM carefully glosses over all the essential structural questions which are necessary to determine what paths can be part of of T(oo) and in what way. > 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. Then what is its tree structure? If it is a maximal infinite binary tree then every maximal sequence of nodes from the root along successive edges must represent a path, and the number of such sequences is well known and easily proven to be uncountable. If this is NOT the tree that WM envisions, he should show us what he does envision more clearly.
From: William Hughes on 13 Feb 2007 20:21 On Feb 13, 11:58 am, mueck...(a)rz.fh-augsburg.de wrote: > On 10 Feb., 15:31, "William Hughes" <wpihug...(a)hotmail.com> wrote: And you have snipped again. I conclude: You now agree the statement Every set of finite even numbers contains numbers which are larger than the cardinal number of the set. is false. > > > > > If we decide to call the sparrow of E a number, then it > > > > is not a natural number > > > > But it does not mean that this sparrow is alive. > > > If we decide to call a number between 1 and 2 a natural number, then > > > this is a wrong definition. > > > And if we decide to call the sparrow of E a natural number then > > this is a wrong definition. So " If we decide to call the sparrow of > > E a number, then it is not a natural number" > > Correct. And it can be shown that the sparrow is not lager than any > natural number. No. The sparrow of E is an equivalence class. This equivalence class is different than any equivalence class containing a set with cardinality a (fixed) finite number. If we extend injection and surjection in the same way we extended bijection (in-transforms, and sur- transforms) we can put a ordering on the sparrows such that the sparrow of E is larger than the sparrow of any set whose cardinality is a natural number. > Why should we decide to call it a number? Well, i is > also called a number, but would we name it a number, if we had the > choice today? > > > > > No statement you make about things that are true > > > > of every set with finite cardinality, or things that > > > > are true for every natural number, can be used > > > > to show something about the sparrow of E. > > > > The set E is not a set with finite cardinality > > > > and the sparrow of E is not a natural number. > > > > The fact that E is composed of sets with finite > > > > cardinality does not mean that E is a set > > > > with finite cardinality. > > > > The finite cardinality has been proved by complete induction. > > > No the finite cardinality of the components of E has been > > proved by induction. E is not one of the components > > of E, > > E is all components of E. What holds for all components (not only for > each component!) holds for E. No. E.g. All components of E have a fixed maximum. E does not have a fixed maximum. What holds for all components of E does not necessarily hold for E. > > > and the fact that the compenents of E have a given > > property does not mean that E has that property > > More precisely: If all initial segments of E have that property and if > no element of E is outside of every initial segment, then E has that > property. > No. E.g. All inital segments of E have a fixed maximum. No element of E is outside of every initial segment. However, E does not have a fixed maximum. > > > A limit of a sequence (a_n) has to have the Cauchy property. > > > Piffle. We are not talking about the convergence of real numbers > > to a real number. [Note also: it is the sequence that has the Cauchy > > propery, not the limit] > > Here you are right. >I meant the property |a_n - a| < eps but, in fact, > Cauchy property is |a_n - a_m| < eps. > > > > omega - n > > > = omega does not satisfy it. > > namely |n - omega| < eps Piffle! omega is not a real number. We are not talking about the convergence of a sequence of real numbers to a real number! > > > > > No. One possible definition for the cardinality of E is the sparrow > > of E. > > The sparrow of E is a fixed equivalence class. > > Correct. Note that you have just agreed that the sparrow of E is fixed. > It is the class of stes which have a natural number as > cardinality which can grow. > Your claim is that E has a maximum, but this maximum is not fixed, and that E has a cardinality that is a natural number but this cardinality is not fixed. This changes what we usually think of as a maximum or a finite cardinality, but what the hey, it's your looking glass world. However, using maximum or finite cardinality is a non-standard way does not change the properties of E. In particular the sparrow of E is fixed. We can call the sparrow of E a number greater than any natural number (and everywhere but Wolkenmueckenheim, the sparrow of E is known as the cardinal of E). There is no problem is calling an equivalence class of sets a number. - William Hughes
From: Dik T. Winter on 13 Feb 2007 20:22 In article <1171366247.193486.239080(a)v45g2000cwv.googlegroups.com> mueckenh(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. > > > > > of them is number 3. > > > > > > > > Well, except that in a number of languages using the Latin script, "= > abc" > > > > are *not* the first three letters. And I have *no* idea what you are > > > > meaning with a statement that numbers can express something. > > > > > > If you think that numbers cannot expres anything, then you can use the > > > empty set for all of them. > > > > I have *no* idea what you are meaning with a statement that numbers can > > express something. I do not deny it, because I do not understand the > > meaning. > > "sie [die Zahlen] dienen als ein Mittel, um die Verschiedenheit der > Dinge leichter und sch�rfer aufzufassen." Dedekind. But that is not mathematics but philosophy. Moreover, it does not state that they "express" something, but only that they can be used for some purpose. > > > A number can express how many wheels your car has, for instance. > > > > Perhaps, but I would not use this strange terminology. > > Why not? Because I think it is strange terminology. I would use "indicate" rather than "express". -- 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 13 Feb 2007 20:34 In article <1171366968.518530.122710(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 13 Feb., 02:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > If p(oo) is established by {p(0)} U {p(1)} U {p(2)} U .... > > > Then p(oo) is also established by P(0) U P(1) U P(2) U ... > > > > But again, I ask you what you mean with "established" here. It is neither > > an element of those unions, nor is it one of those unions. So what is > > your meaning? As I have stated again and again: > > p(oo) = p(0) U p(1) U p(2) U ... > > but that one is quite different from > > {p(0)} U {p(1)} U {p(2)} U ... > > 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. > > But I never denied that. But I now think I understand what you mean > > with that word "establish": > > A path is established if it is the union of a collection of a > > sequence of paths with the property that each path is a subset of > > its successor. > > Correct me if I am wrong, and if so, please state what you actually *do* > > mean with that word (not examples, a clear definition, please). > > I accept your definition: p(oo) is the union of all finite paths p(n). Good, let us keep that as a definition. > > 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). > 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)). 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 14 Feb 2007 10:51
On 13 Feb., 21:17, Virgil <vir...(a)comcast.net> wrote: > In article <1171364856.226197.135...(a)l53g2000cwa.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 12 Feb., 21:22, Virgil <vir...(a)comcast.net> wrote: > > > In article <1171283208.696664.25...(a)a75g2000cwd.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > On 12 Feb., 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1171205918.124082.214...(a)a75g2000cwd.googlegroups.com> > > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 11 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > ... > > > > > Well, I would concede that the above three things are representaions of > > > > > the > > > > > number three, using some convention. Anyhow, they are *not* the number > > > > > three. > > > > > What has "the number three" that is not expressed above? > > > > Universality. Just as no single man represents mankind, > > > I did not say that III expresses all numbers. It expresses all that > > the number 3 can express. > > Not to me. What is missing? > > > > > > no single > > > instance of a set with three objects represents all sets of three > > > objects, at least without common consent. > > > What about all existing sets with 3 objects, i.e., the fundamenal set > > of 3? > > What fundamental set of 3 does WM refer to? Olease read before writing. The set of all existing sets with 3 objects. > > In many set theories there cannot even exist such a "fundamental" set, Set theories which deny the existence of all existing sets should be abolished first. They are obviously nonsense, except in the eyes of every strongly loving beholder. Regards, WM |