Cantor Confusion
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From: mueckenh on 3 Nov 2006 09:34 William Hughes schrieb: > No. I believe that the length of the diagonal is the supremum of > the length of the lines. This supremum does not have to be the > length of any line. Call it omega. Then omega is the *supremum* > of the set of lines (equally the supremum of the set of columns). The set of lines has infinitely many elements, no line has an infinite number of columns. That is a difference. If you consider omega to be not the maximum but only the supremum of the set of lines, then we agree that actual infinty does not exist. But we both then disagree with ZF which states *there exits* the set of lines. They all are there. And there does not exist a set of infinitely many columns (= an infinite natural number). Regards, WM
From: Dik T. Winter on 3 Nov 2006 10:18 In article <1162557998.763485.221280(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > No, something quite different was argued there. Namely that the limit > > > > *also* is the number of edges in the infinite tree (or somesuch) > > > > requires transfinite induction. > > > > > > But is does not. > > > > Indeed, it does not. But I thought you were maintaining that it would be? > > Good heavens! I should require transfinite induction? Sorry, I misread. You insist that the limit *also* is the number of edges in the infinite tree. To prove that you need transfinite induction. > > You need transfinite induction to show that "the number of edges in the > > infinite tree" is equal to the limit of "the number of edges in the > > finite tree". > > No. The number of edges in a finite tree is without interest. We > consider only the infinite tree. The set of edges there can be > enumerated like the set of rational numbers, for instance. If a mapping > N --> Q is defined for every element of Q, then Q is countable. There was an interesting mapping in another thread from N to the positive rationals by David R Tribble, that is easily converted to something different: For any q in Q+ get the continued fraction [a0, a1, ..., an], define the following function: f(q) = 2^a0.(1 + 2^a1.(1 + ...(1 + 2^(an-1))...)) than f is a bijection between N and Q+. > The > mapping N --> {edges} has been established such hat every edge knows > its number. You are wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 3 Nov 2006 10:36 In article <1162559082.298879.296060(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Because it is unrelated with the question where 1/oo was defined. > > But to answer, I do not know who and when the definition of ordinal > > number was introduced. I know *why* it might have been introduced: > > to remove possible confusion. I have no idea what you mean with > > "contrary meaning". As your statement A n: n < omega does not > > indicate where n is coming from, I can not state whether it is > > correct or not. But if the natural numbers are meant, than it is > > certainly correct (if we assume the natural comparison between ordinal > > numbers and natural numbers). > > n e N was implied. If you can compare two entities in some dimension > like length or weight , then they are of the same sort with respect to > this dimension. Perhaps. But we do not talk here about length or weight. We are talking mathematics. You can compare two entities of a different kind if, and only if, a comparison function is defined. But as the natural numbers can be embedded isomorphically in the ordinal numbers, there is a natural comparison function. > > > If omega is a number > n, then 1/omega is a number < 1/n. For all n e > > > N. > > > > First, how do you *define* 1/omega? You know how arithmetic normally is > > defined? But of course, once you define it, some properties of arithmetic > > are lost. What is omega * (1/omega)? > > This property is lost. Still, how do you *define* 1/omega? You just state 1/omega = 0? Oh, well. Perhaps. That makes you indeed lose a few arithmetical properties, so you can not use them in general, but must use them with extreme care. > > > This definition has been conserved up to our days: Limesordinalzahl or > > > limit ordinal number. > > > > Yes. But "limit" has not really been defined. > > Grenze ist immer an sich etwas festes, unver?nderliches, daher kann > von den beiden Unendlichkeitsbegriffen nur das Transfinitum als seiend > und unter Umst?nden und in gewissem Sinne auch als feste Grenze > gedacht werden. Still, I see *no* definition of "limit". > > > In ZF. > > > > > > Karel Hrbacek and Thomas Jech: "Introduction to set theory" Marcel > > > Dekker Inc., New York, 1984, 2nd edition, p. 2: "So the only objects > > > with which we are concerned from now on are sets." > > > > Again, depends on the model. > > Of course. But the usual set theory without yet specifying any models > starts from this point. Not when I followed the courses on set theory. > > But even if we allow that, it can not > > be applied to the ball and vase problem. In that case there is a > > limit involved (in your models of that problem). And to use those > > limits (as you do) you have > > (1) to define the limit of a particular sequence of sets > > (2) to show that the cardinality of that limit is equal to the > > limit of the cardinalities. > > Wrong. The only thing to be shown is that omega is the limit of the > natural numbers. This is the fundamental assumption of set theory. > Multiplication by 9 is negligible. But you still have to define the limit of a particular sequence of sets and you have still to show that the cardinality of that limit is equal to the limit of the cardinalities. It is easy to construct examples where the latter is false (when we have properly defined the limit of the sets involved). Take, for instance, the set of zeros of the functions cos(pi.x/n) where n is an integer >= 1. Each set is infinite, nevertheless, in the limit the set becomes empty. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 3 Nov 2006 12:30 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > No. I believe that the length of the diagonal is the supremum of > > the length of the lines. This supremum does not have to be the > > length of any line. Call it omega. Then omega is the *supremum* > > of the set of lines (equally the supremum of the set of columns). > > The set of lines has infinitely many elements, no line has an infinite > number of columns. That is a difference. Correct (in ZFC). > If you consider omega to be not the maximum but only the supremum of > the set of lines, then we agree that actual infinty does not exist. Omega is the sup of the set of *lengths* of lines. > But we both then disagree with ZF which states *there exits* the set of > lines. They all are there. Why would William disagree that there is a set of lines? I'm sure he doesn't. > And there does not exist a set of infinitely > many columns (= an infinite natural number). What do you mean a "set of infinitely many columns"? We can consider the set of column numbers. Is that what you mean? -- David Marcus
From: David Marcus on 3 Nov 2006 12:35
mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > > > Let's make it simple. I'll give a statement and you say whether you > > > > think it is provable in ZFC. Is > > > > > > > > The set of natural numbers is infinite > > > > > > > > provable in ZFC? Please answer "Yes" or "No". > > > > I see you didn't answer my question. > > Did you really expect an answer? Strangely enough, if I ask a question, I do hope for an answer. I'm not always rewarded. > Yes, it is provable in ZFC. I don't > know anybody who would deny this. > > > > > > > > > As for the current definition of omega, Kunen's book is a good > > > > reference. According to Kunen, omega is not a natural number. > > > > > > That is out of any question. > > > > Sorry, but I don't understand. Please rephrase. > > It is somewhat tedious to correspond with you. It can be difficult to be forced to explain what you mean. > Every natural number is > a finite number by definition. This doesn't seem to clarify things. I said that Kunen says that omega is not a natural number. You said, "that is out of any question". I said I didn't know what you mean. And, you said that "every natural number is a finite number by definition". > > > > I'm > > > > guessing that by "whole number" you mean natural number, but I really > > > > don't know, since you seem to have your own language for everything and > > > > you never give definitions for any of the words that you use. > > > > > > "Whole number" is Cantor's name for his creation. > > > > > > My question is : Do you maintain omega > n for all n e N? > > > > Before I can answer the question, I need to know what you mean by the > > words/terms. So, please define "omega", "N", and ">". Also, by > > "maintain" do you mean that ZFC proves it? > > Do you keep on thinking that omega > n for all N, applying all > words/terms as Kunen would understand them. Using Kunen's definitions, then ZFC proves that if n is a natural number, then n < omega. Do you disagree that ZFC proves this? -- David Marcus |