Cantor Confusion
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From: mueckenh on 20 Nov 2006 11:14 Franziska Neugebauer schrieb: > > I cannot understand your explanation given there. If you say "The > > cardinality of omega is |omega| not omega", so you must have had in > > mind |omega| =/= omega, > > No Way! If you want to misapprehend me do so, but don't confuse your > misapprehensions with theorems of set theory. You wrote: "The cardinality of omega is |omega| not omega." Shall this sentence of yours express a difference between |omega| and omega or not? (Now I recognize why it is so difficult to convince the proponents sof set theory.) > > > Would it be meaningful to write "Y is X but not X"? > > This is not a "faithful" representation of my sentence you have objected > to because there is no character string which can be substituted for X > to get my original sentence back. If you meant |omega| = omega, then we have X = |omega| = omega. Regards, WM
From: Franziska Neugebauer on 20 Nov 2006 11:35 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > I cannot understand your explanation given there. If you say "The >> > cardinality of omega is |omega| not omega", so you must have had in >> > mind |omega| =/= omega, >> >> No Way! If you want to misapprehend me do so, but don't confuse your >> misapprehensions with theorems of set theory. > > You wrote: "The cardinality of omega is |omega| not omega." And what I meant was: "The cardinality of X is |X| not X". I have already clarified that my wording was misleading. > Shall this sentence of yours express a difference between |omega| and > omega or not? What exactly is so hard to understand? The cardinality of a set X is (written) |X| and -- in the case of omega -- equals under the common definition to omega. > (Now I recognize why it is so difficult to convince the proponents sof > set theory.) >> >> > Would it be meaningful to write "Y is X but not X"? >> >> This is not a "faithful" representation of my sentence you have >> objected to because there is no character string which can be >> substituted for X to get my original sentence back. > > If you meant |omega| = omega, then we have X = |omega| = omega. Let Y = "The cardinality of omega" and case 1: let X = "|omega|" "Y is X but not X" transforms into "The cardinality of omega is |omega| but not |omega|", case 2: let X = "omega" "Y is X but not X" transforms into "The cardinality of omega is omega but not omega". I did neither write nor mean what case 1 nor case 2 state. F. N. -- xyz
From: William Hughes on 20 Nov 2006 11:52 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". > > > > > > > > > > > > > > >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. In particluar, N, the set of all natural numbers, 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). > > 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. 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. > > 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. - William Hughes
From: William Hughes on 20 Nov 2006 12:01 mueckenh(a)rz.fh-augsburg.de wrote: < removal of the arg > William Hughes schrieb: > > > > > You are trying to argue > > > > The diagonal cannot be longer than every > > initial segment with a largest element because > > there is no set of lines without a largest line. > > No. The diagonal cannot be longer than every line because it consists > of line-ends only. > > > > There is no set of lines without a largest line > > because the diagonal cannot be longer than > > every initial segment with a largest element. > > > > This is circular. > > > No. Again the arguing is: The diagonal cannot be longer than every line > because it consists of line-ends only. Let's restore the bit you snipped. > Of course there are more conditions. They follow from the EIT. > Therefore I chose it. > 1 > 12 > 123 > ... > Every line has only different indexes (there are not two equal indexes > in one line). Yes > Line n has all indexes from 1 to n. Yes > Every initial segment of line ends is a subset of a line. Yes > Every initial segment of line ends is a subset of the diagonal. Yes > Every initial segment of the diagonal is a line. No. 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. Therefore you cannot say: The diagonal cannot be longer than every line because it consists of line-ends only. Do you intend to respond to this? - William Hughes
From: David Marcus on 20 Nov 2006 12:20
mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > Among other idiocies, WM keeps claiming that in his triangle there is a > > finite line as "long" as the infinite diagonal. > > Either you are a liar or too stupid to understand. I proved that there > is no infinite diagonal. (The diagonal is a subset of the line ends.) I believe you said the diagonal is not longer than every line. Consider this list: 1 2 2 3 3 3 .... where line n has n "n"'s. The "diagonal" is 1 2 3 ... Are you saying that this diagonal (i.e., 1 2 3 ...) is not longer than every line in the list? Are you saying that *you* think it is true that this diagonal is not longer than every line in the list? Are you saying that set theory implies that this diagonal is not longer than every line in the list? -- David Marcus |