Cantor Confusion
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From: mueckenh on 1 Feb 2007 07:23 On 1 Feb., 13:12, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 30 Jan., 17:18, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> >> 2. BECAUSE it is _undefined_ until the "sum of all natural > >> >> numbers" is defined. > > >> > It is defined if the set of all natural numbers s defined, > > > Here is the definition of the set N > > > 1 > > 2 > > 3 > > ... > > > Here is the definition of the sum of all elements of N > > > 1 > > 23 > > 456 > > ... > > Abstract art? > > F. N. > -- > xyz- Zitierten Text ausblenden - > > - Zitierten Text anzeigen - For people with less capabilities of abstract thinking: Imagine a rope and then thread the numbers: 1 32 456 .... Better now? Regards, WM
From: William Hughes on 1 Feb 2007 07:38 On Feb 1, 7:02 am, mueck...(a)rz.fh-augsburg.de wrote: > On 30 Jan., 17:43, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > On Jan 30, 8:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > You can try to utter again and agan this nonsense, but after a while > > > aI will cease to reply. Every such thing including only natural > > > numbers is covered by induction. All natural numbers are subject to > > > induction. > > > True, you can use induction to prove something about > > any natural number. However, the question is: "Can > > you use induction to prove something about a set > > of natural numbers?" > > This question has a trivial answer: > If the set of all natural numbers is nothing than all natural numbers, > then yes. > If the set of all natural numbers is more than all natural numbers, > then no. No there are two questions. i: Is X true for every element of the set A? ii: Is X true for the set A. The two questions are different. The answer to question ii can be No, even when the answer to question i is yes. Your trivial answer only applies to question i. > > > > > There are two types of sets of natural numbers > > > I: sets of natural numbers that are not (potentially) infinite > > > II: sets of natural numbers that are (potentially) infinite > > > Induction can only prove things about sets of type I. > > So you believe that type II sets have some esoteric supplement? No. The answer to question ii can be No, even if there is no "esoteric supplement" (i.e. the answer to question i is Yes). > > We can prove that every number is a number while a set of several > numbers is not a number. As you point out there is a difference between the elements of a set and the set. > There is no reason to distinguish between > finite and infinite sets. > > > > > By running induction longer and longer you can get more and more > > sets of type I. By running induction for an infinite time you can > > get all > > sets of type I. > > No. By two steps of induction, namely > P(1) > P(n) ==> P(n+1) > you can prove P for all natural numbers and for all sets of natural > number (except quantitative statements as I mentioned above). No you can prove things for all elements of all sets of natural numbers. There is a difference between proving that something is true of every element in a set A and proving that something is true of a set A. > > Your strong belief in the inaccesible infinite uttered above is your > (and some other people's) personal opinion but has nothing to do with > mathematics. > > > But you will never get a set of type II. (You may get > > a collection of sets whose union contains the same elements > > as a set of type II, but you will never get a single set of type > > II). > > Small wonder. There is no set of type II. No. We have already agreed that potentially infinite sets exist. Whether these sets "actually exist", that is whether every element in the set can be said to exist at once, is of no interest here. > > > > > The set of all natural numbers, the union of all natural numbers, N, > > is a set of type II. You cannot use induction to prove anything about > > the union of all natural numbers. > > Small wonder. This set does not exist, No, the potentially infinite set of natural numbers exists. We are not interested in the question of whether this set "actually exists". (Whether or not the set "actually exists", the question "Is the maximum element of N fixed (i.e. can it change if you change the subset of N that actually exists)?" has the answer No.) - William Hughes
From: Dik T. Winter on 1 Feb 2007 07:48 In article <1170327430.425220.185660(a)a34g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 31 Jan., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > The union of trees implies the union of outmost > > > left-hand side paths for example. This union is a path. > > > > Indeed, but it is *not* in the union of the sets of paths of finite trees. > > In the above example, that set does contain *all* the paths {0}, {0, 1}, > > {0, 1, 2}, {0, 1, 2, 3}, etc., but *not* the path {0, 1, 2, 3, ...}, > > because the last path is in *none* of the constituent sets. > > This is the same with the infinite tree. Why do you accept the > existence of the infinite tree? Why is it the same? The infinite tree *does* contain the path {0, 1, 2, 3, ...}. But that path is *not* in the union of the sets of finite paths. > The union of all finite paths of type p(r) establishes the infinite > path r. > The union of all finite paths of all finite trees establishes all > infinite paths. Yes. I do not contradict *that*. The "set of the union of paths" is *not* the "union of the sets of paths". -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Franziska Neugebauer on 1 Feb 2007 07:52 mueckenh(a)rz.fh-augsburg.de wrote: > For people with less capabilities of abstract thinking: [...] The three-eyed speaking to the two-eyed? F. N. -- xyz
From: Dik T. Winter on 1 Feb 2007 08:09
In article <1170328870.612961.224120(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 31 Jan., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170162837.477985.97...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > Continued: > > > > You are correct. > > > The union of all natural numbers is not a natural number. Therefore > > > you think it (the cardinality) can be infinite. > > > > Well, it is certainly not finite, so it must be infinite (translation: > > "not finite"). > > Not finite is correct. But you translate it as actually existing set, > i.e., finished infinity. That is incorrect. Not finite is identical to infinite. But the axiom of infinity tates that it exists. > > > The union of all paths is a path in many cases. In such a case it is > > > simultaneousy a union of natural numbers which is a natural number. > > > > The length of finite paths are natural numbers. The length of a > > finite union is a natural number. The length of an infinite union is > > not a natural number. Why do you think it is? > > Because there is nothing else in the union but natural numbers. > You cannot get an apple if you collect nuts. The union of all finite natural numbers is not a natural number. Why do you think it is? > Nevertheless, if you assume that the length of the union of all paths > (not the cardinal number !) is infinite, then you advocate infinite > number sizes are required for an infinite union of numbers. That is quite confusing as stated. But indeed, the infinite union of all natural numbers (considered as sets) is not a natural number. > > "There is no infinite union of finite paths unless there is an infinite > > path." > > Yes. And so what? There is an infinite number, more than one actually. > > But it is *not* a member of the set of finite numbers, as the infinite > > path is *not* a member of the set of finite paths. And natural numbers > > are (through the definitions) the numbers in the set of finite numbers, > > so there is no natural number that is infinite. And on the other hand, > > the infinite paths do not have a natural number as path-length. > > On the other hand many people assert that there is an infinite union > of finite pathlengths without containing or establishing an infinite > pathlength. Eh? Who? > > Yes. In what way does that contradict what I stated? The union of *sets* > > of paths does *not* contain union of paths, unless those unions are already > > present in one of the constituent *sets* of paths. > > Why do you accept then that the union of sets of nodes (finite trees) > establishes an infinite set of nodes (an infinite tree) without such > an infinite set (infinite tree) being contained in the union? Because that is something different. The union of sets of paths establishes an infinite set of paths. But that infinite set of paths does not contain an infinite path, because none of the constituent sets contains one. To translate to sets of nodes: the union of the sets of nodes establishes an infinite set of nodes that does not contain an infinite node. > > More formal, define > > a path p_k as {n_0, n_1, n_2, n_k-1}, where the n_i are nodes. The sets > > of paths from the finite trees contain p_k for all k. The union of the > > trees contains a path {n_0, n_1, n_2, ...}, but the union of the sets of > > paths does consist of: > > { {n_0, n_1, n_2, ..., n_k-1} | n_0 is a node, k in N} > > and so does not contain {n_0, n_1, n_2, ...}. So there is *no* infinite > > path in the union of the sets of paths from finite trees. > > How then does this infinite path manage to enter the union of the > finite trees? Because the set of paths of the union of finite trees is *not* contained in the union of the sets of finite paths. > > > Because I need the identity of path-lengths and numbers for my > > > argument. > > > > Well, it does not hold. If you measure the lengths as you did define, the > > union of all lengths is not a length, because it is not a natural number. > > If all natural numbers are finite, their union as defined for > pathlengths cannot be infinite. Why not? > Except you assert that an infinite union requires a number of infinite > size. It is not a natural number. > > > > > If T(oo) is constructed as the union of all finite trees T(k), > > > > > then every path in P is a path which is in the union > > > > > P(1) U P(2) U ... of the elements of P_C. > > > > > > > > Prove it. The union of P(i) contains finite paths only, *by the > > > > definition*. > > > > > > The infinite tree contains finite trees only. Is it finite on that > > > behalf? > > > > No. The union of finite sets can be infinite. But the union of sets > > with finite elements does not contain an infinite element. > > Finite trees are sets with finite elements, namely the finite number > of nodes. That number is not an element of a tree. The elements of the tree are *nodes* (by your own definition). The union of the finite trees contains finite elements only (i.e. nodes at a finite distance from the root). But there are infinite sequences of such elements. > > And, as > > your P(i) are sets with finite elements, their union does not contain > > an infinite element. On the other hand, their union *is* an infinite > > set. And so it is an infinite set of finite elements. > > The cardinal number is infinite, each pathlength is finite. Right. > The same you can require for the tree. A finite tree is nothing but a > sequence of nodes (if you count them as Cantor did count the rational > numbers: > > 0 > 12 > 6543 > 789...) > > What was your argument distinguishing infinite unions of finite trees > and finite paths, respectively? None. I distinguish unions of finite trees and unions of *sets* of finite paths. Above you are using P(1) U P(2) U ..., where each P(i) is a *set* of finite paths. > > Wrong. Infinite paths do not have a path-length that is a natural number. > > Pathlengths ARE natural numbers. For finite paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |