Another AC anomaly?
in [Logic]
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From: WM on 8 Dec 2009 10:53 On 8 Dez., 15:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 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. > Yes, it is called a limit ordinal because by definition each ordinal that > has no predecessor is called a limit ordinal (that is the definition of the > term "limit ordinal"). It has in itself nothing to do with limits. No, that is not the reason. The reason is that omega is a limit without axiom of infinity, and omega is older than that axiom. > > N is a concept of mathematics. That's enough. > > Yes, and it is a concept of mathematics because it is defined within > mathematics, and it is not defined as a limit. It is a concept of mathematics without any being defined. > > The infinite union is a limit. > > I do not think you have looked at the definition of an infinite union, if > you had done so you would find that (in your words) such a union is found > on the shelf and does not involve limits. Try to start doing mathematics > and rid yourself of the idea that an infinite union is a limit. An infinite union *is* not at all. But if it were, it was a limit. > > > Why did you argue that limits of > > cardinality and sets are different, if there are no limits at all? > > I have explicitly defined the limit of a sequence of sets. With that > definition (and the common definition of limits of sequences of natural > numbers) I found that the cardinality of the limit is not necessarily > equal to the limit of the cardinalities. That means that you are wrong. Regards, WM
From: WM on 8 Dec 2009 11:02 On 8 Dez., 16:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 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. > > Oh, so actual infinity means that a set with infinitely many elements exists? Yes. > In that case you should reject the axiom of infinity. You are allowed to do > that, and you will get different mathematics. But you can not claim that > mathematics with the axiom of infinity is nonsense just because you do not > like it. But go ahead without the axiom of infinity, I think you have to > redo quite a bit of mathematics. Before 1908 there was quite a lot of mathematics possible. There was quite a lot of possible mathematics. > > The definition of an actually infinite set is given in set theory by > > the axiom of infinity. > > You are wrong, the axiom of infinity says nothing about "actually infinite > set". Actually the axiom of infinity does not define anything. It just > states that a particular set with a particular property does exist. That is just the definition of actual infinity. > > > The definition of a potentially infinite set is given by > > 1 in N > > n in N then n+1 in N. > > That does not make sense. Without the axiom of infinity the set N does not > necessarily exist, so stating 1 in N is wrong unless you can prove that N > does exist or have some other means to have the existence of N, but that > would be equivalent to the axiom of infinity. N need not exist as a set. If n is a natural number, then n + 1 is a natural numbers too. Why should sets be needed? > > 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. > > Right. > > > 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.) > > And here again you are wrong. There are countably many finite paths. There > are not countably many infinite paths, and although you have tried many > times you never did show that there were countably many infinite paths. There is not even one single infinite path! But there is every path which you believe to be an infinite path!! Which one is missing in your opinion? Do you see that 1/3 is there? What node of pi is missing in the tree constructed by a countable number of finite paths (not even as a limit but by the axiom of infinity)? Regards, WM
From: WM on 8 Dec 2009 11:10 On 8 Dez., 16:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Nevertheless it is a limit ordinal. > > Yes, that does not mean that necessarily a limit is involved. It is a limit > in the sense that you do not get there by continuously getting at the > successor, in that case it is a limiting process. But when you define > N as an infinite union you do *not* go there by continuously getting at > a successor. The union of a collection (finite, countably infinite or > some other infinity) is defined whithout resorting to successor operations. > Moreover, they would even not make sens if the collection is infinite but > not countably infinite. That does not make sense in either respect, so or so. > > > 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}. > > Note that here N is apparently already defined, without using a limit. Natural numbers can be defined without using a set. > > > 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. > > Yes. So what? That you can define things using a limit does *not* > imply that it is necessarily defined as a limit. O I see. That's like cardinality. The limit cardinality is not the cardinality of the limit (because the limit is not a limit). Regards, WM
From: A on 8 Dec 2009 13:36 On Dec 8, 11:10 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Dez., 16:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Nevertheless it is a limit ordinal. > > > Yes, that does not mean that necessarily a limit is involved. It is a limit > > in the sense that you do not get there by continuously getting at the > > successor, in that case it is a limiting process. But when you define > > N as an infinite union you do *not* go there by continuously getting at > > a successor. The union of a collection (finite, countably infinite or > > some other infinity) is defined whithout resorting to successor operations. > > Moreover, they would even not make sens if the collection is infinite but > > not countably infinite. > > That does not make sense in either respect, so or so. > > > > > > 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}. > > > Note that here N is apparently already defined, without using a limit. > > Natural numbers can be defined without using a set. > > > > > > 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. > > > Yes. So what? That you can define things using a limit does *not* > > imply that it is necessarily defined as a limit. > > O I see. That's like cardinality. The limit cardinality is not the > cardinality of the limit (because the limit is not a limit). > > Regards, WM A function f is said to be continuous at a point x in its domain if the limit of f(a), as a approaches x, is equal to f(x); in others words, the limit of the values of f is equal to the value of f at the limit, speaking loosely. Of course, not every function is continuous at every point in its domain, and some functions are not even continuous at any point in their domains at all. The situation for sets and cardinality is no more mysterious than that. The cardinality of a limit of subsets of the integers is not guaranteed to be the limit of the cardinalities of those subsets. You don't expect an arbitrary function to always be continuous, so perhaps it's unreasonable to expect the cardinality "function," defined on subsets of the integers, to be continuous.
From: Virgil on 8 Dec 2009 15:57
In article <69368271-d841-4c3e-9f73-57259312f585(a)g12g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 8 Dez., 15:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > �> 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. > > > Yes, it is called a limit ordinal because by definition each ordinal that > > has no predecessor is called a limit ordinal (that is the definition of the > > term "limit ordinal"). �It has in itself nothing to do with limits. > > No, that is not the reason. Yes it is!!! > The reason is that omega is a limit > without axiom of infinity By what definition of "limit"? > > > �> N is a concept of mathematics. That's enough. > > > > Yes, and it is a concept of mathematics because it is defined within > > mathematics, and it is not defined as a limit. > > It is a concept of mathematics without any being defined. Not outside of Wolkenmuekenheim. > > > �> The infinite union is a limit. > > > > I do not think you have looked at the definition of an infinite union, if > > you had done so you would find that (in your words) such a union is found > > on the shelf and does not involve limits. �Try to start doing mathematics > > and rid yourself of the idea that an infinite union is a limit. > > An infinite union *is* not at all. But if it were, it was a limit. In ZF, any union of the sets which are members of a set is "defined" by the axiom of union, and in ZF there is no other form of union at all. So that what WM is saying about unions is false in ZF. > > > > �> � � � � � � � � � � � � � � � �Why did you argue that limits of > > �> cardinality and sets are different, if there are no limits at all? > > > > I have explicitly defined the limit of a sequence of sets. �With that > > definition (and the common definition of limits of sequences of natural > > numbers) I found that the cardinality of the limit is not necessarily > > equal to the limit of the cardinalities. > > That means that you are wrong. It shows that Dik is right and that WM is wrong. As usual! > > Regards, WM |