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From: WM on 7 Dec 2009 11:11 On 7 Dez., 16:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 1) N is a set that follows (as omega, but that is not important) from > > the axiom of infinity. You can take it "from the shelf". > > Actually: N is the smallest inductive set that starts with 1. The axiom > of infinity states that that set does exist, but with that definition it > is *not* defined as a limit. I said you can take it from the shelf. It is not defined as a limit (if you like so) although amazingly omega is called a limit ordinal. > > > 2) N is the limit of the sequence a_n = ({1, 2, 3, ...,n}) > > You have first to define the limit of a sequence of sets before you can > state such. No. > N is (in set theory) defined before even the concept of the > limit of a sequence of sets is defined (if that concept is defined at > all in the particular treatise). N is a concept of mathematics. That's enough. > > > 3) N is the limit, i.,e. the infinite union of singletons {1} U {2} > > U ... > > No, the infinite union is *not* a limit. It is defined without even the > presence of the concept of a limit of a sequence of sets. The infinite union is a limit. Why did you argue that limits of cardinality and sets are different, if there are no limits at all? > > But if (3) is correct, then N must also be the limit of the process > > described in my > >http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#394,22,Folie22 > > without and *with* the intermediate cylinder. > > I see no process there, only a picture of a cylinder with the digits 1, 2, 3, > 4 and 5, and an open cube. It is a good tactics to play possum or play stupid. Regards, WM
From: WM on 7 Dec 2009 11:20 On 7 Dez., 16:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <5f8c7e7f-ec83-45ba-a584-b61475969...(a)j4g2000yqe.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 3 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > No. I asked you for a mathematical definition of "actual infinity" > > > > > and you told me that it was "completed infinity". Next I asked you > > > > > for a mathematical definition of "completed infinity" but you have > > > > > not given an answer. So I still do not know what either "actual > > > > > infinity" or "completed infinity" are. > > > > > > > > Both are nonsense. But both are asumed to make sense in set theory. > > > > > > No, set theory does not contain a definition of either of them. > > So in what way do they make sense in set theory? As I have not seen a > definition of them at all, I can not see what that means. You have seen the axiom of infinity. It say that an infinite set exists and that implies that infinitely many elements of that set exist. That is actual infinity. > > > > > The axiom of infinity is a definition of actual infinity. > > > > "There *exists* a set such that ..." > > > > Without that axiom there is only potential infinity, namely Peano > > > > arithmetic. > > > > > > I see neither a definition of the words "actual infinity" neither > > > a definition of "potential infinity". Or do you mean that "potential > > > infinity" is Peano arithmetic (your words seem to imply that)? > > > > > > So we can say that in "potential infinity" consists of a set of axioms? > > > > Here is, to my knowledge, the simplest possible explanation. > > Darn, I ask for a definition, not for an explanation. The definition of an actually infinite set is given in set theory by the axiom of infinity. You should know it or know where to find it. (You can look it up in my book.) The definition of a potentially infinite set is given by 1 in N n in N then n+1 in N. > > > Consider > > the infinite binary tree: > > > > 0 > > /\ > > 0 1 > > /\ /\ > > 0 1 0 1 > > ... > > > > Paint all paths of the form > > 0.111... > > 0.0111... > > 0.00111... > > 0.000111... > > and so on. > > Potential infinity then says that every node and every edge on the > > outmost left part of the tree gets painted. > > Actual infinity says that there is a path 0.000... parts of which > > remain unpainted. And that is wrong. > > So "potential infinity" and "actual infinity" are theories? But I do not > know of any theory that states that there is any part of the path 0.000.... > that is unpainted. That is the inconsistency of set theory. The complete infinite binary tree can be constructed using countably many finite paths (each one connecting a node to the root node), such that every node is there and no node is missing and every finite path is there and no finite path is missing. Nevertheless set theory says that there is something missing in a tree thus constructed. What do you think is missing? (If nothing is missing, there are only countably many paths.) Regards, WM
From: WM on 7 Dec 2009 11:34 On 7 Dez., 16:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dn...(a)giganews.com> "K_h" <KHol...(a)SX729.com> writes: > > ... > > > > > When you mean with your statement about N: > > > > > N = union{n is natural} {n} > > > > > then that is not a limit. Check the definitions about > > > > > it. > > > > > > > > It is a limit. That is independent from any definition. > > > > > > It is not a limit. Nowhere in the definition of that > > > union a limit is used > > > or mentioned. > > > > Question. Isn't this simply a question of language? > > Not at all. When you define N as an infinite union there is no limit > involved, there is even no sequence involved. N follows immediately > from the axioms. Nevertheless it is a limit ordinal. > > > My > > book on set theory defines omega, w, as follows: > > > > Define w to be the set N of natural numbers with its > > usual order > > < (given by membership in ZF). > > > > Now w is a limit ordinal so the ordered set N is, in the > > ordinal sense, a limit. Of course w is not a member of N > > becasuse then N would be a member of itself (not allowed by > > foundation). > > Note here that N (the set of natural numbers) is *not* defined using a > limit at all. That w is called a limit ordinal is a definition of the > term "limit ordinal". It does not mean that the definition you use to > define it actually uses a limit. (And if I remember right, a limit > ordinal is an ordinal that has no predecessor, see, again no limit > involved.) Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" Marcel Dekker Inc., New York, 1984, 2nd edition, p. 134 As a starting point, we use the fact hat each natural number is identified with the set of all smaller natural numbers: n = {m in N : m < n}. Thus we let w, the least transfinite number, to be the set N of all natural numbers: w = N = {0, 1, 2, 3, }. It is easy to continue the process after this limit step is made: The operation of successor can be used to produce numbers following w in the same way we used it to produce numbers following 0: Regards, WM
From: Virgil on 7 Dec 2009 15:21 In article <KuAGqH.FrI(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dnZ2d(a)giganews.com> "K_h" > <KHolmes(a)SX729.com> writes: > ... > > > > > When you mean with your statement about N: > > > > > N = union{n is natural} {n} > > > > > then that is not a limit. Check the definitions about > > > > > it. > > > > > > > > It is a limit. That is independent from any definition. > > > > > > It is not a limit. Nowhere in the definition of that > > > union a limit is used > > > or mentioned. > > > > Question. Isn't this simply a question of language? > > Not at all. When you define N as an infinite union there is no limit > involved, there is even no sequence involved. N follows immediately > from the axioms. > > > My > > book on set theory defines omega, w, as follows: > > > > Define w to be the set N of natural numbers with its > > usual order > > < (given by membership in ZF). > > > > Now w is a limit ordinal so the ordered set N is, in the > > ordinal sense, a limit. Of course w is not a member of N > > becasuse then N would be a member of itself (not allowed by > > foundation). > > Note here that N (the set of natural numbers) is *not* defined using a > limit at all. That w is called a limit ordinal is a definition of the > term "limit ordinal". It does not mean that the definition you use to > define it actually uses a limit. (And if I remember right, a limit > ordinal is an ordinal that has no predecessor, see, again no limit > involved.) Other than the first ordinal, though that restriction is not really relevant here.
From: Virgil on 7 Dec 2009 15:23
In article <KuAGIy.FE1(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article > <7772c857-57b0-4422-b688-9a4c8b923467(a)h10g2000vbm.googlegroups.com> WM > <mueckenh(a)rz.fh-augsburg.de> writes: > > On 3 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > The limit set is not necessarily "assumed" by the sequence of sets > > > > > (if by that you mean that there is a set in the sequence that is > > > > > equal to the limit set, if you mean something else I do not > > > > > understand it at all). > > > > > When you mean with your statement about N: > > > > > N = union{n is natural} {n} > > > > > then that is not a limit. Check the definitions about it. > > > > > > > > It is a limit. That is independent from any definition. > > > > > > It is not a limit. Nowhere in the definition of that union a limit is > > > used or mentioned. > > > > 1) N is a set that follows (as omega, but that is not important) from > > the axiom of infinity. You can take it "from the shelf". > > Actually: N is the smallest inductive set that starts with 1. The axiom > of infinity states that that set does exist, but with that definition it > is *not* defined as a limit. For some people, it starts with 0. > > > 2) N is the limit of the sequence a_n = ({1, 2, 3, ...,n}) > > You have first to define the limit of a sequence of sets before you can > state such. N is (in set theory) defined before even the concept of the > limit of a sequence of sets is defined (if that concept is defined at > all in the particular treatise). > > > 3) N is the limit, i.,e. the infinite union of singletons {1} U {2} > > U ... > > No, the infinite union is *not* a limit. It is defined without even the > presence of the concept of a limit of a sequence of sets. > > Rid yourself of all those ideas. > > > This is fact. > > It is not. > > > But if (3) is correct, then N must also be the limit of the process > > described in my > > http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#394,22,Folie 22 > > without and *with* the intermediate cylinder. > > I see no process there, only a picture of a cylinder with the digits 1, 2, 3, > 4 and 5, and an open cube. |