Cantor Confusion
in [Math]
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From: mueckenh on 15 Feb 2007 16:42 On 14 Feb., 18:24, "William Hughes" <wpihug...(a)hotmail.com> wrote: > 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. It is true for every form of existence E can assume. But in order to avoid any misunderstanding: Every set of finite even numbers contains numbers which are larger than the cardinal number of the set, if the set has a cardinal 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. > > > 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. Could you please look up the proof that the actually infnite set N has a cardinal number larger than every natural? Usually it is done by induction, and it starts: Certainly 0 < |N|. This is the fundamental error. No, certainly 0 is not less than |N|. No axiom and no proof lead to this statement. >of > >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. The sparrow is fixed. It is a songbird and it sings: The number of elements of my set is not 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? Because vectors and tensors are closely related to complex numbers, Quaternions and Cayley numbers, but are not called numbers. In my opinion, all numbers should be ordered by size. > > > > > > > > > > > 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. Correct. But as all existing elements trivially exist, and as all elements are in trichotomy, there must be a greatest element. Of course it cannot be fixed. Hence there is a not fixed largest element. > > > > > > > > > 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?!? There is only one type of convergence. Convergence means coming together, not staying apart. (Of course, in set theory convergence means 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| > > 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). Anyhow, it is something which is not larger than some other thing. > > > > > > > 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. best denoted by oo. > > > > > > 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. Good luck. Proof? Regards, WM
From: mueckenh on 15 Feb 2007 16:44 On 15 Feb., 14:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171470343.811036.193...(a)k78g2000cwa.googlegroups.com> mueck...(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. {p(0)} U {p(1)} U ... is a subset of the union of P(0) U P(1) U .... > > 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 ... . p(0) U p(1) U ... is a subset of P(0) U P(1) U ... . This subset is unioned by the tree. It yields p(oo). P(oo) is composed of the sets p(oo), q(oo), r(oo). > > > > > > 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. It does have bearing on the presence of p(oo) and all other infinite paths in the union of finite sets of paths, i.e., in the union of finite trees. You cannot avoid this by formal retreat fights. > > > > > 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. No. If p(oo) is established as the union of p(i), then it is established by the P(i). This holds for every infinite path q(oo), r(oo), ... which s element of P(oo). Hence, P(oo) is established by the union of finite trees. > > > > 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. UP(i) = {p(0), p(1), p(2), ..., and other finite paths} The tree contains a path which *is* the union of the p(i), namely p(oo). The tree, i.e., the union of finite trees is or establishes the paths q(oo), r(oo), ... too. Therefore it establishes all elements of the set P(oo). Regards, WM
From: Carsten Schultz on 15 Feb 2007 16:55 mueckenh(a)rz.fh-augsburg.de schrieb: > Correct. But as all existing elements trivially exist, and as all > elements are in trichotomy, there must be a greatest element. Of > course it cannot be fixed. Hence there is a not fixed largest element. A floating maximum! -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: MoeBlee on 15 Feb 2007 17:23 On Feb 15, 1:42 pm, mueck...(a)rz.fh-augsburg.de wrote: > Every set of finite even numbers contains numbers which are larger > than the cardinal > number of the set, if the set has a cardinal number. Since even numbers are natural numbers, thus finite, I suppose that what you actually mean is: Every finite set of even numbers has a member greater than the cardinality of the set. But that's not even true. {0 2} is a finite set of even numbers and no member is greater than the cardinality of the set. > if the set has a cardinal number. In ZFC, every set has a cardinal number. Even in just ZF (using Scott's method), every set has a cardinal number. And, no matter, every set of natural numbers has a cardinal number. And, no matter, every finite set has a finite cardinal number. > Could you please look up the proof that the actually infnite set N > has a cardinal number larger than every natural? Usually it is done by > induction, and it starts: Certainly 0 < |N|. This is the fundamental > error. No, certainly 0 is not less than |N|. No axiom and no proof > lead to this statement. You're nuts. Of course it is trivial to prove that there is an injection from 0 into N and no bijection between 0 and N. Anyway, it definitely is a theorem of ZFC that every natural number has cardinality strictly less than N, by application of the pigeonhole principle, of which you can find a proof in just about any set theory textbook and many textbooks in other subjects in undergraduate mathematics. > There is only one type of convergence. Convergence means coming > together, not staying apart. (Of course, in set theory convergence > means divergence.) What theorem of set theory can be understand to say that "convergence means divergence"? MoeBlee
From: Virgil on 15 Feb 2007 17:32
In article <1171575734.891375.54010(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 14 Feb., 18:24, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > 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. > > It is true for every form of existence E can assume. But in order to > avoid any misunderstanding: > > Every set of finite even numbers contains numbers which are larger > than the cardinal > number of the set, if the set has a cardinal number. That requires that only naturals can be cardinals, which is neither desireable nor necessary in a consistent set theory. > > > > > 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. > > Could you please look up the proof that the actually infnite set N > has a cardinal number larger than every natural? E certainly has a sparrow at least as large as the sparrow of the successor of any natural number, so until WM can show us a natural whose sparrow is as large as the sparrow of its successor, Sparrow(E) > sparrow(n) for all n.. > Usually it is done by > induction, and it starts: Certainly 0 < |N|. This is the fundamental > error. No, certainly 0 is not less than |N|. No axiom and no proof > lead to this statement. The definitions of cardinality and the order relation on cardinals, however, do lead to precisely the more relevant statement |0| < |N|. At least until WM can show us a surjection from {} to N. > > > >of > > >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. > > The sparrow is fixed. It is a songbird and it sings: The number of > elements of my set is not fixed. Then it is not a set at all, since sets in proper set theories are not allowed WM's sort of vagueness. > > > > > > > > > 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? > > Because vectors and tensors are closely related to complex numbers, > Quaternions and Cayley numbers, but are not called numbers. In my > opinion, all numbers should be ordered by size. WM's opinions on matters mathematical have repeatedly shown themselves to be not worth a tinker's dam. > > 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. > > Correct. But as all existing elements trivially exist, and as all > elements are in trichotomy, there must be a greatest element. Where does that foolishness come from, WM? Trichotomy may imply order but does not imply boundedness, much less a extrema. > Of > course it cannot be fixed. Hence there is a not fixed largest element. Hence no largest element at all. > > > > > > > > > > > > > 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?!? > > There is only one type of convergence. To someone whole mind is too small to contain more than one thing at a time there may be only one sort of convergence, but to the rest of us it is only WM's limitations speaking. Convergence means coming > together, not staying apart. (Of course, in set theory convergence > means divergence.) Then WM is now acknowledging at least two types of convergence, which means he was lying to claim only one. > > > > 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| Is that misspelling a Freudian slip? > > > > 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). > > Anyhow, it is something which is not larger than some other thing. It is not larger that itself nor larger that the sparrow of a power set of E, but is certainly provably larger than the sparrow of any natural. > > > > > > > > > > > > > which is > > > fixed as a growing number. Oxymoron! With WM being the latter part. > > > > > 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. > > best denoted by oo. Most of us prefer other representations. > > > > > > > > > > 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. > > Good luck. Proof? Since WM is so clearly unable to provide proofs of any of his own claims, one wonders why he would bother to ask for proofs of anyone else's. |