Cantor Confusion
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From: William Hughes on 21 Nov 2006 07:38 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. Therefore it consists of line > ends only. Every line end is the end of a line, no? A classic case of Person 1: P_1 is true. <reasons why P_1 is true> P_1 implies P_2 Therefore P_2 is true Person 2: No P_1 is true However, P_1 does not imply P_2 <reasons why P_1 does not imply P_2> P_2 is false Person 1: Clearly you do not understand. Here is another reason why P_1 must be true. Yes. P_1: The diagonal consists of the ends of lines. No. The fact that P_1: the diagonal consists of the ends of lines does not imply that P_2: the diagonal is not longer than every line. > > > > 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. 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. > > > > 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. Yes. However, since there is always an element (n+1), no single line can establish a bijection with *all* the elements of the diagonal (both those <=n and those > 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. No. No line with finite index n can establish a bijection with all elements of the diagonal. Every line has a finite index. Therefore no line can establish a bijection with all elements of the diagonal - William Hughes.
From: William Hughes on 21 Nov 2006 08:00 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. Absolute piffle. The question is whether the set N can be mapped to the set N, not whether some other set can be mapped to the set N. [Although it is interesting to note that both the constructable numbers and the set of all words can be mapped to the set N (please check the definiton of countable set)] > > > > > > > > > > > > > > > > > > > > > > > > >And even if we could, the > > > > > > > axiom of infinity does not prove that infinite ordinal and cardinal > > > > > > > numbers are numbers, i.e., that they stand in trichotomy with each > > > > > > > other. > > > > > > > > > > > > Check the definitions of ordinal and cardinal below. Look for > > > > > > the term trichotomy. Fail to find the term trichotomy. Draw > > > > > > the obvious conclusion. > > > > > > > > > > My conclusion is: You don't know this term. > > > > > > > > No, this is not the obvious conclusion (it is also wrong). > > > > The obvious conclusion is that the > > > > fact that the term tricotomy is not used when defining > > > > either the ordinals or the cardinals, means that we > > > > do not need to show trichotomy to show that > > > > something is an ordinal or a cardinal. > > > > > > It may be called by another name but trichotomy is implied. In fact all > > > ordinals are asserted to stand in trichotomy with each other. > > > > If you want to include the natural ordering as part of the ordering > > of the ordinals, yes. ALL ordinals, including infinite ordinals, stand > > in trichotomy with each other. > > Correct. that's what I said. And since the set of all natural numbers is an ordinal, it stands in trichotomy with all other ordinals. > > > > > > > > > > > Nevertheless, you will > > > > > agree if you learn what trichotomy means: a < b or a = b or a > b for > > > > > any two numbers a and b. > > > > > > > > > > > > > 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. > > > > > > We need to put > > > > an ordering on the ordinals or the cardinals before we can say > > > > whether trichotomy holds. > > > > > > Ordinals are ordered sets by definition. We need not put anything. > > > > > If you insist. > > Even if I wouldn't insist. It is the case by definition. > > > I would define the set of ordinals first, then put an > > order on it, but, you can do both simultaneously if you want. > > > > 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. Piffle. Well-ordering has nothing to do with the existence or not of at least one bijection. > > > > > 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. Correct. And since all ordinals show trichotomy, showing that something is an ordinal proves trichotomy. You do not need to demonstrate trichotomy directly. So. The fact that the set of natural numbers is an ordinal by definition (a definition that does not mention trichotomy) proves that the set of natural numbers shows trichotomy. You do not need to demonstrate trichotomy directly. - William Hughes
From: Franziska Neugebauer on 21 Nov 2006 08:16 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. Do I miscomprehend your sentence "And there are lines as long as the diagonal is."? Condider the following scenario: state 1: Given a parking space with *two* cars having their wheels mounted. (each car has 4 wheels) state 2: Unmount all the wheels of both cars and put them on one stack of wheels. (stack of wheels has 8 wheels) Proposition 1: "There are cars in state 1 which have as many wheels as the stack of wheels in state 2 has." Proposition 2: "There is *a* car in state 1 which has as many wheels as the stack of wheels in state 2 has." Now my interpretation: Claim 1 states that there are at least two (due to the plural form "cars" and due to the "are" instead of "is") cars each of which have the property of having as many wheels as the stack of wheels has (8). That is wrong in my view. Claim 2 states that there is (at least) one car which has as many wheels as the stack of wheels has (8). Which is wrong, too. Question 1: Do you agree with my interpretation? We frequently use formulas like Ex(p(x)) which is translated as "There is an x having property p". Question 2: Is it correct that you mean by writing "And there are lines as long as the diagonal is." you mean "There is a line as long as the diagonal" is? I. e. in the sense of Ex (p (x)) with x for line and p(x) "lenght of line equal length of diagonal"? >> 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. >> > 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. > >> > 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. I did not ask you to define "all" but "all lines". Do you mean "A line (p(line))" or do you mean "p(set of all lines)"? >> > 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. I see no superset and no proof. > 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. That is not true in the sense that the double index mm with m != n is not a location in line n but in line m. It may be true that the value of d_mm is equal to d_nm. But this is an issue of occupancy. > 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. I don't understand what you mean. > As the diagonal has only finite indexes n, Finite valued indexes n e omega, but not finitely many of them. > there is never more than one line required to establish a bijection > with all elements of the diagonal. Do you want to posit, that there is some line identical to the diagonal? That is not (yet) proven. F. N. -- xyz
From: Eckard Blumschein on 21 Nov 2006 08:24 On 10/30/2006 12:07 AM, William Hughes wrote: > But note, to define an integer, the definition must be > communicated. We have only a finite number of bits > to use for this communication (this includes comunicating > the details of any compression scheme). So there are > only a limited number of integers that it is possible to define. > Among these is the largest integer that it is possible to > define. > > If we claim that the only integers that exist are the ones > that have been or will be defined, > then there is a largest possible integer. > > - William Hughes Having looked around and realized a lot of impolite and opiniated persons, I consider you, William Hughes, like a perhaps rare positive exception. Hopefully you will not mistake me if I do no longer hesitate to nonetheless comment on your reasoning. While I do not question your last sentence being a correct conclusion if the premise holds, I cannot not see any possibility to reasonably and concisely quantify a largest possible number. Do we really have to define numbers? I tend to be satisfied when I have an executable rule which may create as many natural numbers as I like, as does Archimede's axiom. Executable means effectively about the same as communicable. It excludes postulated a priori existence. Do not get me wrong. I consider the properties executable and communicable like belonging to the abstract mathematical idea, not to any physical limitation to it. The process of counting indefinitely is never completely executable. Accordingly all created numbers are finite, and there is no principial reason for a largest possible number to be found. Eckard Blumschein
From: Eckard Blumschein on 21 Nov 2006 08:41
On 11/9/2006 10:32 PM, Virgil wrote: > > WM again uses "number" where we use "set". > > We have a SET which contains all of the /finite/ naturals or all of the > /finite/ ordinals. Yes, modern set-mathematicians have their Gruenderzeit belief. Gruenderzeit was the time after a won war against France when Germany was reunited for the first time and much money out of French colonies together with growing industry boosted anything including genuine science as well as Cantorian fiction. |