Cantor Confusion
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From: Lester Zick on 21 Nov 2006 14:09 On Mon, 20 Nov 2006 17:24:44 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <MPG.1fcbf909c89f84c3989993(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > >> Lester Zick wrote: >> > On Mon, 20 Nov 2006 14:35:20 -0500, David Marcus >> > <DavidMarcus(a)alumdotmit.edu> wrote: >> > >Lester Zick wrote: >> > >> On Sun, 19 Nov 2006 17:53:12 -0500, David Marcus >> > >> <DavidMarcus(a)alumdotmit.edu> wrote: >> > >> >Lester Zick wrote: >> > >> >> On Sat, 18 Nov 2006 13:44:58 -0500, David Marcus >> > >> >> <DavidMarcus(a)alumdotmit.edu> wrote: >> > >> >> >Lester Zick wrote: >> > >> > >> > >> >> >> I'm saying that you don't understand what a mathematical >> > >> >> >> definition is >> > >> >> >> but nonetheless want to pretend you do. If a mathematical >> > >> >> >> definition >> > >> >> >> were "just" an abbreviation as you claim you wouldn't have any way >> > >> >> >> to >> > >> >> >> tell one mathematical definition from another. >> > >> >> > >> > >> >> >Why not? Suppose I make the following definitions. >> > >> >> > >> > >> >> > Let N denote the set of natural numbers. >> > >> >> > Let R denote the set of real numbers. >> > >> >> > >> > >> >> >Then I can tell N and R are different because their defintions are >> > >> >> >different. If I write >> > >> >> > >> > >> >> > 0.5 is not in N, >> > >> >> > >> > >> >> >then this means the same as >> > >> >> > >> > >> >> > 0.5 is not in the set of natural numbers. >> > >> >> > >> > >> >> >And, it means something different from >> > >> >> > >> > >> >> > 0.5 is not in R. >> > >> >> >> > >> >> But the problem, sport, is you claim mathematical definitions are >> > >> >> "only abbreviations". Granted I suppose even mathematikers can tell >> > >> >> the difference between N and R in typographical terms. I mean they >> > >> >> may >> > >> >> be too lazy and stupid to demonstrate the truth of what they say but >> > >> >> even they can see differences in typography. But in terms of >> > >> >> abbreviations alone we can't really say what the difference is >> > >> >> between >> > >> >> N and R because you insist their definitions are "only abbreviations" >> > >> >> and not their conceptual content. >> > >> > >> > >> >You said there was no way to tell two definitions apart. The >> > >> >typographical difference suffices to tell the definitions apart (as you >> > >> >just admitted). >> > >> >> > >> So what exactly is the difference between definitions N and R in >> > >> conceptual terms if definitions are "only abbreviations"? I mean you >> > >> say certain things about definitions which are mutually inconsistent. >> > >> If definitions were "only abbreviations" as you indicate then your >> > >> definitions for N and R would be restricted to those abbreviations N >> > >> and R. Instead you append certain properties to each and pretend that >> > >> they're part of the definitions for N and R which we'll all can see >> > >> are not part of their abbreviations such that your definition for >> > >> definitions is "only abbreviations which are not only abbreviations". >> > >> Obviously this kind of logic extends way beyond your doctoral thesis >> > >> in philosophy but is nonetheless true. >> > > >> > >That's impressive. Either you are trolling or you have completely >> > >misunderstood what people mean when they say "definitions are >> > >abbreviations". >> > >> > Or quite possibly you misunderstand what people mean by >> > "abbreviations". Not quite the same as the sloth and professional >> > turpitude of mathematical definitions you're used to I daresay. N and >> > R are "abbreviations". Whatever you imagine by what you attach to >> > abbreviations are not abbreviations. But please do go on and don't >> > allow me to distract you from your "modern mathematical" definition of >> > abbreviations which I'm quite confindent will afford us many a >> > pleasant evening of muffled laughter. >> >> And, we can let "U.S." stand for the United States. Is this >> abbreviation/definition/whatever true or false, in your opinion? > >There is no point in trying to communicate with Zick, as he is not the >least bit interested in communicating. Yes, but I must note that you're not the least bit interested in communicating what Zick is not interested in communicating. >Your best response to his blather is kill filing him. Yes well since Aatu called you pathetic have you also killfiled him? ~v~~
From: David Marcus on 21 Nov 2006 14:25 Franziska Neugebauer wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > Franziska Neugebauer schrieb: > >> 2. Since neither Virgil nor I did posit any claim which needs > >> support, not even by "naming one single element of the diagonal ...", > >> we don't have to. > > > > Virgil claimed the diagonal be longer than any line. That needs > > support. > > Of course. It is proven that for every single line l |l| < |diagonal|. > > > Therefore he has to name an element which is not member of a > > line. > > No Way. You reversed the quantifiers. WM does that a lot. When people catch him doing it, he denies it. -- David Marcus
From: Virgil on 21 Nov 2006 15:07 In article <1164105686.380911.138600(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > > The axiom of infinity says that the set N exists. Your iia > > > > > > > > says > > > > > > > > > > > > > > > > "we cannot recognize or treat all of its elements" > > > > > > > > > > > > > > This is no contradiction to the axiom. Compare the proof that the > > > > > > > real > > > > > > > numbers can be well-ordered. We cannot construct or define or > > > > > > > recognize a well ordering. > > > > > > > > > > > > The fact that we cannot construct a real ordering does not > > > > > > stop us from proving such an ordering exists > > > > > > > > > > LOL. I know. > > > > > > > > > > > and using > > > > > > the existence of such an ordering. > > > > > > > > > > The existence of a well-order does not guarantee the constructibility > > > > > or definability of a real ordering. > > > > > The existence of a set does not guarantee the existence or > > > > > definability > > > > > of a bijection or an identity mapping. > > > > > > > > No. > > > > > > > > To say that a set N exists is to say that all elements n of N exist. > > > > Thus the mapping n ->n for all elements n of N exists. > > > > > > That is exaggerated. There are models of ZFC (including countably many > > > elements a part of which could be interpreted as the set of natural > > > numbers) where no mapping on N exists. Why should it fail if you were > > > right? > > > > Piffle. Any defintion of mapping under which we can have > > an existing set of elements, but no existing identity map goes > > under the technical term of "stupid". > > That is a strong argument. > > You never heard of countable uncountable models? > > What about all contructible numbers or all words? They cannot be mapped > on N though they are a countable set. We are talking about one set mapping to ITSELF under the identity function x |--> x, not about one set being mapped to another. Such an identity mapping exists for any set because it is a set, at least in ZF or NBG. Does WM know what WM is talking about? > > > > Trichotomy is a property of ordered sets, not of sets. > > > > > > Only ordered sets have ordinal numbers / are ordinal numbers. > > > > > > > Yes. But not every set that has an ordering/satisfies trichotomy > > is the ordinals (e.g. real numbers with the usual ordering). > > These examples are called "order types" but not ordinals. The examples are *ordered sets* . Different ordered sets can exhibit different order types, but are not themselves order types. An order type applies to a set of ordered sets in which the order relations have in common some property over and above merely being an order relation. > > > > Note, however, cardinals are not ordered by definition. > > Note however that in order to attach a definite cardinal number to > every set, well-ordering must be possible for every set. That is why > Cantor insisted on well-ordering of all sets. That depends on one's definition of cardinal. If one is allowed to take an arbitrary representative of each cardinality , rather than the least ordinal of a given cardinality, then well-ordering of all sets is not necessary, though in that case, one does not have a provable trichotomy on cardinals without an axiom of choice. > > > > > If we put the natural ordering on the > > > > ordinals, then trichotomy holds for all ordinals (finite and infinite). > > > > > > Yes, why then do you dislike the name? > > > > I don't. Since trichotomy holds for both omega (for all ordinals) > > and for aleph_0 (at least within sets of natural numbers > > and within all sets if we assume AC ) > > there is no problem. I merely point out that it is not necessary > > to decide whether or not trichotomy holds in order to show > > that omega is an ordinal and aleph_0 is a cardinal. > > Neither the definition for ordinal, nor the definition for cardinal > > contains the term trichotomy or any term that means > > the same thing. > > Nevertheless we can recognize ordinals by trichotomy: If two numbers > both are ordinals, then they show trichotomy. If two numbers don't show > trichotomy, then they are not both ordinals. "Numbers" is ambiguous. What of the cases where "numbers" show trichotomy but are not ordinals, or are not even numbers in any usual sense? If "numbers" includes cardinals and our definition of cardinals does not identify them with ordinals and we do not have an axiom of choice, then it is quite possible to have two distinct cardinals neither of which is "smaller" than the other.
From: Virgil on 21 Nov 2006 15:34 In article <1164105950.440715.109720(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > > > > > Name one single element of the diagonal which is not contained in a > > > line (which contains this and all preceding elements). > > > > 1. Burden of proof weighs upon you who claimed "And there are lines as > > long as the diagonal is.". The question remains: Which lines? > > The diagonal D is a subset of the union U of lines L_n. Therefore D - > UL_n = empty set. > > > > 2. Since neither Virgil nor I did posit any claim which needs support, > > not even by "naming one single element of the diagonal ...", we don't > > have to. > > Virgil claimed the diagonal be longer than any line. That needs > support. Therefore he has to name an element which is not member of a > line. I have supported it many times, but apparently WM has a form of selective blindness which prevents him from seeing anything that conflicts with his beliefs. For ANY line n, d_{n+1,n+1} is not a member of that line. Thus for every n in N, the diagonal is longer than line n. To refute this proof, WM must find some line, and therefore some n in N numbering that line, containing every term of the diagonal. > > > This is what I call the one-eyedness of set theory: Its proponents see > > > that for every line, there is a diagonal element not contained in this > > > and all preceding lines. > > > > This is your wording not mine and very likely not Virgils, too. As you > > know from my matrix-view of your list every diagonal element d_nn is > > necessarily located in row n and column n. Where else? Outside of this > > matrix there are no diagonal elements. > > That's just my arguing. But with set theorists you never can be sure > that not the foolishest claims are supported. With anti-set-theorists, such as WM, claiming that the diagonal (a set that contains the successor of every element in it) can be contained within a proper subset ( a line, bounded initial segment with an element having no successor in the set), the are no in a position to call the kettle black. > > > > > But they don't see, or at least dispel it, that there is no element of > > > the diagonal which is outside of any line. > > > > The sequence of diagonal elements is "embedded" in the matrix. There are > > neither n of the index set which have no d_nn nor are there d_mm which > > are not of the index set of the matrix. > > > > > The first observation leads to the theorem: The diagonal is superset > > > of all lines. > > > > Only sets have supersets. Define "all lines" please. > > The diagonal is a sequence and as such an ordered set. Every coumn/line > is a sequence and as such an ordered set. I don't define "all". Look it > up in a dictionary of mathematical expressions. Regarding the diagonal and each line as merely sets, the diagonal is a superset of every line, and a proper superset of every line into the bargain. > > > > > The second observation (together with the fact that > > > every line is a superset of all preceding lines) leads to the theorem: > > > There is at least one line which is superset of the diagonal. > > > > Please prove. > > I leave it to you as an exercise. As we have exercised our disproofs of that non-theorem several times now ( see above for example) with no adequate response, we find that WM need exercise a good deal more than we do. > > Hint: If element d_nn of the diagonal is a member of line n, then all > elements d_mm wit m =< n of the diagonal are members of the same line > n. In other words: There is never more than one single line required to > establish a bijection with all the elements of the diagonal d_mm with m > =< n. As the diagonal has only finite indexes n, there is never more > than one line required to establish a bijection with all elements of > the diagonal. WM is being silly again by ignoring that there is no n in N for which d_nn is the last term but for every line there is such an n. Alternately: For every line L, there is an n in N for which there is no nth term in that line, no L_n, L_n does not exist,but d_nn does.
From: Virgil on 21 Nov 2006 15:59
In article <1164106170.999705.36100(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > There are two types of initial segments > > > > Initial segments with a largest element > > Initial segments without a largest element. > > > > There is an initial segment of the diagonal that does > > not have a largest element. > > (Recall: the diagonal has a largest element if and only > > if there is a last line. There is no last line.) > > Every line has a largest element. Thus there is an initial > > segment of the diagonal that is not a line. > > The diagonal has only finite indexes n. But infinitely many of them. > Therefore it consists of line > ends only. Every line end is the end of a line, no? So what? > > > > Therefore you cannot say: > > > > The diagonal cannot be longer than every line > > because it consists of line-ends only. > > Of course that must be true. It is equally obviously false. More obviously to those with their eyes open. > The diagonal has only finite indexes n. > Every line end is the end of a finite line. > As it cannot be infinite, the complete diagonal does not exist. WM wants to prove that N does not exist, so that he keeps assuming things which he cannot prove to get that result. In ZF: For each n in N, let N_n = {m in N: m <= N} Note that for every n in N, N_n is a PROPER subset of N. Let L_n : N_n --> N : m |--> m Let D: N --> N : m |--> m Then L_n is the nth line and D is the diagonal. For m <= n, L_n(m) = m is the m'th term of line n For m <= n, L_n(m) = D(m) = m is the m'th term of D For m > n, L_n(m) does not exist, but D(m) = m. For every n in M there is m > n , for example m = n+1, such that L_n(m) does not exist, but D(m) does. So for every n in N, D is "longer" than L_n. |