Cantor Confusion
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From: Dik T. Winter on 1 Feb 2007 08:15 In article <1170330236.064219.62780(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 31 Jan., 03:33, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170163062.268110.161...(a)h3g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > > In no basic way. To see the basics, you must continue IIII, IIIII, > > > > > etc. > > > > > > > > Ok. Pray define what sets those things do represent. > > > > > > > They represent the numbers 4 and 5. And: They are the numbers 4 and > > > 5. > > > > Circular reasoning. I did ask you in what way subsetting is the same as > > less than. Your proof was: 2 < 3 because II + I = III. So I am asking > > about sets (subsetting is a set operation), and you come back with > > decimal notation. So back again. How do you define subsetting such that > > 4 subset 5? > > If you have a set and you add something to this set (in unary > representation), then you have a super set of the former. The former > is a subset of the latter. This is not defining what I asked for. What are the elements of the sets you are talking about? > > > > Bizarre. So 3 is the set of all existing (what does that mean) sets > > > > with 3 elements, > > > > > > The number 3 is the set of all sets of 3 elements (unless they were > > > existing, they would not have 3 elements). But 3 is also here III. It > > > is every set of 3 elements. > > > > You failed to respond to what followed. If you *now* define it as every > > set of 3 elements, that is a circular definition. "3 is any set of 3 > > elements" looks quite circular to me. > > It is circular. But please do not forget that I do not *define* a > natural number! I leave that to Peano etc. I look for its *existence*. So it was not a definition at all. Note also that the Peano axioms to *not* talk about representation. > > > > I do not ask what possible answers there are, I only ask how you would > > > > like to propose it. And, pray extend to 111111111. > > > > > > This is a set of 9 distinguishable elements, it is a set representing > > > the number 9. And: It is the nunmber 9. Every set of 9 elements and > > > avery set of sets of 9 elements is the number 9. > > > > So you are now going to equivalence classes of sets with the same number > > of elements? Or what? As you write it your definition is still circular. > > Of course it is. You must start at some point and eventually you will > come back. Circular reasoning cannot be avoided. But that is not a > problem when existence is concerned. The definition of a number is > given on p. 3 and p 130 of my book. A definition that uses circular reasoning is not a definition. If a write: a foo is a foo I am not defining anything. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 1 Feb 2007 08:40 On 1 Feb., 13:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > 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. Yes, but it does not exist actually. Usually, you mistake one for the other. "The potentially infinite set of natural numbers exists" does not mean anything else than "there are natural numbers, and if n is a natural number, then n+1 is a natural number". All questions concerning this set can be answered by two steps of complete induction. Regards, WM
From: mueckenh on 1 Feb 2007 08:44 On 1 Feb., 13:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170327430.425220.185...(a)a34g2000cwb.googlegroups.com> mueck...(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. And that path is ot contained in the union of sets of finite trees. > > > 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". No, it is only a subset. Regards, WM
From: mueckenh on 1 Feb 2007 08:49 On 1 Feb., 14:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170328870.612961.224...(a)v33g2000cwv.googlegroups.com> mueck...(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 states that > it exists. Not finite is identical to infinite. The axiom of infinity, however, finishes infinity. > > > > > 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? The union of paths is a path. The projection of a finite number on a pathlength remains a finite number. The projection of all finite numbers on a pathlength remains a finite number. > > > 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. It is only unfamiliar to think about that. > But indeed, the infinite union of > all natural numbers (considered as sets) is not a natural number. Maybe. But the natural numbers consideed as paths remain natural numbers in any case. > > > > "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? Everybody who says that there is "an infinite set of finite pathlengths". Regards, WM
From: William Hughes on 1 Feb 2007 09:00
On Feb 1, 8:40 am, mueck...(a)rz.fh-augsburg.de wrote: > On 1 Feb., 13:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > 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. > > Yes, but it does not exist actually. Usually, you mistake one for the > other. > "The potentially infinite set of natural numbers exists" does > not mean anything else than "there are natural numbers, and if n is a > natural number, then n+1 is a natural number". > > All questions concerning this set can be answered by two steps of > complete induction. > No. The question Q: "Is the maximum value of N fixed (that is can the natural number referred to by the maximum value of N change if we change the subset of N that 'actually exists')?" is a question concerning N that cannot be answered by two steps of complete induction. However the two facts, "there are natural numbers" and "if n is a natural number, then n+1 is a natural number" can be used to show that the answer to question Q is No (it is not necessary to know that n+1 exists, only that n+1 can exist). - William Hughes |