Cantor Confusion
in [Math]
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From: mueckenh on 14 Feb 2007 11:15 On 14 Feb., 02:21, "William Hughes" <wpihug...(a)hotmail.com> wrote: > 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. > No, it is true. How can you conclude it was wrong? > > 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. Absolutely wrong. Correct is: The equivalence class |E| is different from any equivalence class containing a set with cardinality a (fixed) finite number. This is trivially true because |E| is not a fixed 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? !!! > > 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. Every element of E is finite. Show me a part of E which is outside of every finite initial segment. > > > > > 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! We are talking about convergence. This: |n - omega| = omega is not convergence but divergence. > > > 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. Of curse, but it is not a number. The sparrow of E is |E| which is fixed as a growing number. > > > 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. Yes. We can fix: Eery potentially infinite set has cardinality |E|, i.e., a growing number. > We can call the sparrow of E a number greater than > any natural number Do you also call John Wayne or salty greater than any natural number? > There is no problem is calling > an equivalence class of sets a number. Except that ist is easily proven wrong, for example by | {2,4,6,...,2n}| = n. Regards, WM
From: mueckenh on 14 Feb 2007 11:19 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. By the way, above is only number 2 given. Regards, WM
From: mueckenh on 14 Feb 2007 11:25 On 14 Feb., 02:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171366968.518530.122...(a)v33g2000cwv.googlegroups.com> mueck...(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. 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 ... > > > > 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). That is a formal result. Take he union of this union, then you have p(oo). This union is done in the tree. > > > 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). > 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). Regards, WM
From: Ralf Bader on 14 Feb 2007 12:24 mueckenh(a)rz.fh-augsburg.de wrote: > 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 The set of, say, 1000-element-subsets of a set of 3000 elements doesn't exist in your idiotic "Matherealism". Let alone the set of all 1000-element-sets. Or the set of all 3-element-sets (somewhere in the incredulously foolisg nonsense you put on the arxiv you say that no set could have more than 10^100 or so elements) Nevertheless, "Matherealism" is not among the theories which should be abolished according to your criterion, because this stupid piece of junk doesn't qualify as a theory.
From: William Hughes on 14 Feb 2007 12:24
On Feb 14, 11:15 am, mueck...(a)rz.fh-augsburg.de wrote: > On 14 Feb., 02:21, "William Hughes" <wpihug...(a)hotmail.com> wrote:> 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. > > No, it is true. How can you conclude it was wrong? Because it does not imply that the set of even finite numbers has a cardinality, it is not true for E, and E is a set of finite even numbers. > > > > 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. > > Absolutely wrong. > > Correct is: The equivalence class |E| |E| is your notation for the cardinality of E. We are not talking about the cardinality of E (your definition) but the sparrow of E. >is different from any > equivalence class containing a set with cardinality a (fixed) finite > number. This is trivially true because |E| is not a fixed number. |E| is not the sparrow of E. The sparrow of E, as you have agreed, is fixed. > > > > 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? > > !!! > Yes, why not? > > > > > 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. > > Every element of E is finite. Show me a part of E which is outside of > every finite initial segment. My second statement was "No element of E is outside of every initial segment." The fact that no element of E is outside of every initial segment does not mean that E has a fixed maximum. > > > > > > > > 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! > > We are talking about convergence. This: |n - omega| = omega is not > convergence but divergence. PIFFLE!!! Do you really not know that there is more than one type of convergence?!? > > > > > 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. > > Of curse, but it is not a number. The sparrow of E is |E| No |E| is your notation for your definition of the cardinality of E. Your definition of the cardinality of E (something that changes) is not the same as the sparrow of E (something which does not change). > which is > fixed as a growing number. > > > > > > 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. > > Yes. We can fix: Eery potentially infinite set has cardinality |E|, > i.e., a growing number. Every potentially infinite set of natural numbers has the same sparrow which is something which is not |E| (your definition) but something which does not change. > > > We can call the sparrow of E a number greater than > > any natural number > > Do you also call John Wayne or salty greater than any natural number? No, but there is nothing to stop me. - William Hughes |