An uncountable countable set
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From: Virgil on 11 Aug 2006 16:35 In article <1155314772.096651.203180(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I never said that it is a natural. But according to Cantor it is a > number which stands in trichotomy with other numbers, in particular > with the natural numbers. So do the rationals and reals "stand in trichotomy with the naturals" (but not the compexes).
From: Virgil on 11 Aug 2006 16:39 In article <1155315288.866092.161030(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I mean that a natural number can be attributed to every digit position. And this can be done in uncountably many ways, and can still leave enough naturals unutilized to index every digit in countably many more infinite strings of digits.
From: Dik T. Winter on 11 Aug 2006 20:02 In article <1155314988.698343.137140(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1155067333.259873.193700(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > The v. Neumann model of natural numbers is just the model of segments. > > > n = {0, 1, 2, ..., n-1} > > > Every natural number is finite. > > > Hence every union of finite segments is a finite segment. > > > > Every finite union of finite segments is a finite segment. Your hence > > does not follow. Again, a misinterpretation. > > Every union of finite segments is a finite segment. "finite" does not > apply here because there are *infinitely* many finite numbers. Just because there are infinitely many finite numbers, the word "finite" *does* apply. An infinite union of finite segments is *not* necessarily finite. Where do you find that an infinite union of finite segments is finite? Prove it. > > > > Can you provide me with either a proof of that statement or a site > > > > where such a proof is given? > > > > > > Every natural is finite, by definition. > > > > I never argued that. That is not a proof. Pray provide a *proof* that > > every union of finite segments is a finite segment, or, alternatively, > > that every union of finite segments is a natural number. > > According the v. Neumann model it is. (According to the Zermelo model > too.) False. > You know that there are not only finitely many numbers, but infinitely > many. Hence we can unify infinitely many sequences. Nevertheless every > natural number is finite. Yes. But if we unify infinitely many sequences the result is *not* necessarily finite and *not* necessarily a finite number. Consider the sequence of finite sets {}, {0}, {0,1}, {0,1,2}, ... ; there are infinitely many of them. There union contains every number that is in at least one of the sets. As each natural number is in at least one of the sets, the union contains all natural numbers, and there are infinitely many. Also, there is no non-natural number that is in the union as there is none in any of the sequence of sets. This is just basic set theory. > > > The v. Neumann model of natural numbers is just the model of segments. > > > n = {0, 1, 2, ..., n-1} > > > > Again, you refuse to provide a response. > > Please look it up in text books. > The v. Neumann model is established. > The knowledge that here are infinitely many finite numbers is > established too. The knowledge the an arbitrary union of finite sets is also finite is *not* esatblished. But that was not at all what I asked. I asked something you stated about sets, and you "responded" by providing a model, where what you stated is also not true. > I do not want t discuss basics of set theory. I understand. Apparently you do not want to discuss how you arrive at your statement that an infinite union of finite sets is finite. But as that is not *in* set theory, but a conclusion from you outside set theory, you should at least tell the world how you arrive at that staggering conclusion. > > > > The last line of the triangle will have equal width as the length > > > > of the triangle. If we get at an infinitely long triangle, there > > > > is no last line. > > > > > > And there is no definite number aleph_0 of lines. > > > > Why not? > > The line number is infinite by the axiom of infinity. The number of lines is infinite by the axiom of infinity. There is a quite subtle distintion that you do not want to see. > The line number is infinite by the axiom of infinity. There is not an infinite line number by the axiom infinity. Pray stop misinterpreting what set theory tells. -- 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 11 Aug 2006 20:23 In article <1155315217.378759.114420(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > There is no natural number oo. Therefore {k = 1 .. oo} is nonsense. > > > There is, as you say, a definition for (*+){k = 1 .. n} Ak. More is > > > not possible. > > > > More is possible, when you define what the notion means. > > No more is possible if naturals are required. I have no idea what you are meaning here. > > > Because: index = natural number = "number that can be indexed" > > > > Again you are using a term in a single sentence to mean two different > > things. > > No. It is exactly the same. *No*. Darn. In your parlance "number that can be indexed" means a "number for which every digit position can be indexed". They are not the same. If they are the same show a proof. > > > > You state so, and remain stating so, without any proof. > > > > > > It is a definition. > > > > *What* definition? There is, as far as I know, no definition that states: > > A number can be indexed if all its digit positions can be indexed. > > so what definition are you alluding to. Pray, for once, be clear. > > Indexing is equivalent to covering. A number can be covered if all its > digit positions can be covered. Depends how you define that (you have upto now still failed to give a proper definition). > You state "all digit positions can be covered but the whole number > cannot be covered". That is nonsense. Why? > > > > It is impossible, because index number = number which can be indexed. > > > > Again, you fail to provide a definition for "can be indexed". In normal > > terminology, a number can be indexed if we can assign a natural number > > with it in the indexing process. > > That would be expressed by "enumerating" this number. I chose the word > "indexing" in order to focus on the digit positions. > > > You appear to be meaning something > > completely different, namely: a number can be indexed if all its digit > > positions can be assigned a natural number to it. > > Correct. Good. So now we have your definitions: enumerating: assigning a natural number to the elements of a set indexing: assigning a natural number to all its digit positions covering: there is a natural number such that all digit positions can be indexed by natural numbers smaller than or equal to that number. But in that case some of the logic you stated in this thread is false, by your own admission. Namely you stated: (every digit of n can be indexed) ==> (n is in the list) and you said that was by definition of the term list. But with your current words, the members of a list are enumerated and not indexed. Also you state that indexing is the same as covering. This is the same as telling: for all p there is an n such that ... is equivalent to: there is an n such that for all p ... which is false. Back again to quantifier dyslexia. -- 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 11 Aug 2006 20:31
In article <1155315288.866092.161030(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1155067888.355942.225340(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > You are claiming that K is in the list of indices. That makes no > > > > > > sense at all. > > > > > > > > > > I am claiming that every index is in the list of indexes. This list is > > > > > infinite. There are infinitely many indexes. And, yes, 0.111... is not > > > > > among them and therefore cannot be indexed. > > > > > > > > No proof, yet. K = 0.111... can also be written as "for every n in N, > > > > K[n] = 1". > > > > > > Every number which can be written in this manner is contained in the > > > list, because there is every n and every digit which can be indexed. > > > > Makes no sense at all. Proof please. > > > > > EVERY INDEX IS THERE. Every index position is there. It is the set of > > > all indexes and o all index positions. But 0.111... is not there. > > > Because this number cannot be indexed. > > > > But when I state "for every n in N, K[n] = 1" can the number K not be > > indexed? > > Every n in N is in the list. K is not in the list. That is not an answer. Can the number K be indexed or not? > > To be clear, what do you mean with "can be indexed". You > > are using two different, conflicting, definitions in your articles. > > I mean that a natural number can be attributed to every digit position. Yes, so can K be indexed or not? And if not, why not? Please use your definition of indexing. I specifically stated that K was that representation where each natural number points to a 1, and there are no other digit positions. So by your *own* definition it can be indexed. But indeed, it is not in the list (as you admit). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |