Cantor Confusion
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From: Virgil on 21 Nov 2006 04:18 In article <4562B7D6.6000806(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 11/13/2006 10:21 PM, Virgil wrote: > > mueckenh(a)rz.fh-augsburg.de wrote: > >> Franziska Neugebauer schrieb: > >> > mueckenh(a)rz.fh-augsburg.de wrote: > > >> > > Yes. But, sorry to see, it is the fundament of modern mathematics. > >> > > >> > "Finished infinities" is your wording. > >> > >> Precisely describing the fundament of modern mathematics. > > > > It may describe WM's fundament, but need not describe anyone elses'. > > The style of discussion seems to correspond to the subject: > Cantor Confusion. Cantor seems to confuse anti-mathematicians like EB and WM. But he does not confuse mathematicians. > WM is absolutely correct. Isn't he? Not hardly.
From: Eckard Blumschein on 21 Nov 2006 04:42 On 11/21/2006 10:18 AM, Virgil wrote: > >> WM is absolutely correct. Isn't he? > > Not hardly. Nobody is always absolutely correct as he is also not always absolutely wrong. It depends. Serious people ask on what it depends. False Virgil's don't.
From: Virgil on 21 Nov 2006 04:46 In article <4562CA1F.9070708(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 11/21/2006 10:18 AM, Virgil wrote: > > > > >> WM is absolutely correct. Isn't he? > > > > Not hardly. > > Nobody is always absolutely correct as he is also not always absolutely > wrong. It depends. Serious people ask on what it depends. False Virgil's > don't. Since EB agrees with me that nobody, presumably including WM, is absolutely correct, what is his problem?
From: mueckenh on 21 Nov 2006 05:41 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. > > > > > > > > > > > > > > > > > > > >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. > > > > > > > > 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. > > > 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. Regards, WM
From: mueckenh on 21 Nov 2006 05:45
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. > > > 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. > > > 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. 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. > Regards, WM |