Cantor Confusion
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From: Franziska Neugebauer on 20 Nov 2006 04:26 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> This _existence_ is "assumed" rightly. The function >> >> B := { <n, n> | n e omega } >> >> is the desired bijection. > > Only if it includes all natural numbers. Do you want to posit, that B does not "include" every natural number? If so which is not? F. N. -- xyz
From: stephen on 20 Nov 2006 07:42 David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > imaginatorium(a)despammed.com wrote: >> Lester Zick wrote: >> > On Sat, 18 Nov 2006 18:31:05 +0000 (UTC), stephen(a)nomail.com wrote: >> > >David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >> > >> Lester Zick wrote: >> > >>> On Thu, 16 Nov 2006 02:02:49 -0500, David Marcus >> > >>> <DavidMarcus(a)alumdotmit.edu> wrote: >> > >>> >Please give a specific example of something that you think is absurd or >> > >>> >a contradiction. I don't know what you mean by "containment of sets and >> > >>> >subsets". >> > >>> >> > >>> Well as I recollect Stephen seems to think infinite sets are proper >> > >>> subsets of themselves. >> > > >> > >> Are you sure that is what Stephen thinks? >> > > >> > >I see Lester has resorted to lying. >> >> That was Stephen... I must say I've always been puzzled by accusations >> of "lying" in this group. In particular, how can a claim to "seem to >> recollect.." be lying? > Depends on whether the person's memory is really hazy or they are being > disingenuous. And given the fact that google exists, there is no excuse for a hazy memory. Stephen
From: William Hughes on 20 Nov 2006 08:09 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. > > > > > > >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. > 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. We need to put an ordering on the ordinals or the cardinals before we can say whether trichotomy holds. If we put the natural ordering on the ordinals, then trichotomy holds for all ordinals (finite and infinite). If we put a partial ordering on the cardinals by "a<=b if there exists an injection from a to b", trichotomy will hold for all sets if we assume AC. If we do not assume AC, trichotomy may not hold for all sets, but it does hold for all sets of natural numbers (including the set of all natural numbers). > > > There are even models without a bijection to the natural numbers. > > > > Piffle. If the natural numbers exist, then the identity map is > > a bijection. > > If it exists, of course. > Piffle. If the set of natural numbers exists then the identiy map on the natural numbers exists. - William Hughes
From: mueckenh on 20 Nov 2006 08:38 Franziska Neugebauer schrieb: > >> A theory of _sets_ is not a theory of _columns_. > > > > Experience has shown that practically all notions used in contemporary > > mathematics can be defined, and their mathematical properties derived, > > in ZFC. In this sense, the axiomatic set theory serves as a > > satisfactory foundation for al[l] other branches of mathematics. > > It can describe every mathematical notion --- with the exception of > > what a column is? > > I do not speculate about what is describable and what not. If you want > to posit a definition do so! "Practically all" notions! If you again refuse to learn from me, look here: Experience has shown that practically all notions used in contemporary mathematics can be defined, and their mathematical properties derived, in this axiomatic system. In this sense, the axiomatic set theory serves as a satisfactory foundation for the other branches of mathematics. [Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" Marcel Dekker Inc., New York, 1984, 2nd edition, p. 3] > > It is curious that you obviously don't like to give precise definitions > of certain notions even when you have explicitly been asked for. I defined the EIT. A column is a vertical row. The first "column" of it is the first vertical row. "First" is counted from the left hand side. Perhaps I should add that "first" means really first, not zeroth. > > [...] > > >> > Sometimes it is necessary to quote. In particular if you are > >> > uninformed but nevertheless refuse to take advice from me. > >> > >> I don't need any advice from you. > > > > You don't know it. That' s why you would need to learn a lot. Look, > > you have learned from me meanwhile, even against your furious > > opposition, that omega = |omega| in modern set theory. > > You do no longer persue your plan to prove a contradiction in modern set > theory? Perhaps I will find another proof, but I think those delivered are sufficient. > > > Doesn't this case make you wonder whether there are other things which > > you do not yet know but which you could learn from me? > > > >> My opinion is > >> > >> A (n e omega & |{0, 1, 2, ..., n}| < |omega|) > > > > Seems an empty opinion. What is the symbol A refers to? > > A n (n e omega & |{0, 1, 2, ..., n}| < |omega|) That is correct. All (initial) segments of natural numbers (which consist only of natural numbers!) have a finite number of members. > > What I say is: The cardinality of the sequence (d_nn) is the cardinality > of omega. And the ordinality of (d_nn) is omega too, like the ordinality of the first column. Therefore it is clear that all segments of natural numbers (which all are subsets of the numbers contained in the lines of the EIT) are finite. Regards, WM
From: mueckenh on 20 Nov 2006 08:48
William Hughes schrieb: > > The diagonal consists of line indexes, i.e., of the line ends. > > Therefore it is a subset of the line indexes. > > > > Yes, the diagonal is the set of line indexes (a subset, but > not a proper subset). Every line is a subset of the diagonal. And the diagonal is a subset of the line ends. > > We need more than this. Consider the set > X={1,2} and the two finite sequences A and B > > A= {1,2,1,2} > > B={1,2} > > A and B both consist of elements of the set X, but A is > longer than B. So to say that both the diagonal and the > lines are subsets of the line indexes is not enough > to show that A cannot be longer than B. > > So you must have something more in mind when > you claim that the fact that every element of the diagonal > is a line index means that the diagonal cannot > be longer than every line. What is this? 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). Line n has all indexes from 1 to n. Every initial segment of line ends is a subset of a line. Every initial segment of line ends is a subset of the diagonal. Every initial segment of the diagonal is a line. Regards, WM |