Cantor Confusion
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From: Dave Seaman on 30 Jan 2007 22:53 On 30 Jan 2007 13:46:28 -0800, MoeBlee wrote: > On Jan 30, 1:17 pm, Dave Seaman <dsea...(a)no.such.host> wrote: >> I disagree. The version of the diagonal proof that I have in mind is >> applied to [0,1), not to R. When correctly presented, the proof shows >> that there is a number in [0,1) that is not in the given list. The >> conclusion is that [0,1) is uncountable, and therefore R is uncountable. >> Andy's mistake was in using an incorrect version of the diagonal proof. > I understand your point, but I am noting that there is also a mistake > in logic even more fundamental. I also understand your point, but I feel that you are the one who is overlooking something. > The diagonal proof can be used to apply to certain intervals or to the > entire set of real numbers, as the argument is variously formulated. > To prove that the set of reals is uncountable, it is only required to > show that for any list of reals, there is a real not in the list. That > we can get the stronger result that that real not in the list is in a > particular interval is fine, but it is not required for merely proving > that the set of real numbers is uncountable (unless you set up the > diagonal argument in such a way as to require that the real not on the > list is in a particular interval). But that's the way the proof is normally presented. In fact, I have never seen it done otherwise. The standard method is to consider a mapping f: N -> [0,1) and show that f is not a surjection. The conclusion is that [0,1) is uncountable. I feel confident that Andy Smith had this proof in mind, but he misunderstood it to the extent that he thought the proof could be carried out in base 2 instead of base 10. When he found that this version of the proof was flawed, he wrongly concluded that the flaw was in the diagonal argument rather than in his base-2 version of it. It could not be a mistake to apply the proof to [0,1), because that was how the correct proof was done in the first place; his mistake lay entirely elsewhere, in changing the base and in failing to recognize the significance of the dual representation problem. > Meanwhile, even IF someone had, say, fooled us into thinking that > there is no diagonal method to construct a number in [0 1) but not in > the list he gave us, then that would STILL be irrelevent in the sense > I am mentioning, since we only have to keep in mind that to show the > uncountability of the reals does not require (even if it can be done) > showing that for every interval there is a diagonal method to > construct a real not in that interval. Yes, I understand your point, but it does not apply to the comparison of the two proofs in question. When a correct proof is converted into an incorrect proof, the mistake is in the part of the proof that was changed, not in some other part of the proof that was perfectly correct in the first place. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 1 Feb 2007 05:57 On 31 Jan., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > If you have a union of sets of paths, there is nothing that allows you > to unite paths. A path is in the union of sets of paths if and only > if it is an element of at least one of the constituent sets. It does > not matter whether it can be formed using unions of paths. In your > situation, the paths {0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3}, etc., can > be united to form the path {0, 1, 2, 3, ...}. On the other hand, the > sets that contain the paths {0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3}, etc., > can *not* be united to a set that contains the path {0, 1, 2, 3, ...}. > > > 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? Consider only such paths which are members of an ordering, like all paths with only nodes 1. (The ordering is by number of nodes which may be called length of the path. Such paths are initial segments of at least one path of the infinite tree, i.e., one finite path is a subset of the other finite path.) If we consider only paths r_n, r_m (with m,n in N) which are initial segments of at least one infinite path r of the infinite tree, then we can state: The union of two finite paths of type r_n, r_m is a path of type r_k with k sup (m,n). The union of all finite paths of type r_n is or contains or leads to the infinite path r. In order to find a correct expression let's say. The union of all finite paths of type r_n establishes the infinite path r. This is completely analog to: The union of two finite trees is a finite tree. The union of all finite trees is or contains the infinite tree. Conclusion: 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. Regards, WM
From: mueckenh on 1 Feb 2007 06:21 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. > > > 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. 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. > > > Therefore your and Cantors distinction between numbers and sets is > > wrong. There is no infinite union of finite numbers unless there is an > > infinite number. (Translate number by path, then you will see it.) > > "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. > > > > Yes, each path in T(oo) is the union of a set of finite paths. And each > > > path in T(oo) is not in the union of the sets of finite paths. So I did > > > apply the knowledge you agreed to. What now? > > > > Unions of paths (for instance the outmost left-hand sided) are paths! > > 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? > 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? > > > > > In case of path lengths measured as I defined, the union of all > > > > lengths is a length. > > > > > > We were talking about paths here, methink, not about lengths? So I wonder > > > why you are stating this here. > > > > 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. Except you assert that an infinite union requires a number of infinite size. > > > > > 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. > 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. 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? > > > Wrong. Infinite paths do not have a path-length that is a natural number. Pathlengths ARE natural numbers. If you accept an infinite pathlengths then you accept an infinite finite number. Regards, WM
From: mueckenh on 1 Feb 2007 06:43 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: > > > On 29 Jan., 04:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > And in what way is > > > > > III c IV c V > > > > > ? > > > > > > > > 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. > > > > 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*. (Cf. p. 130 of my book). > > > > 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. http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 The important last chapter which can be down loaded free of charge from my web site: http://www.fh-augsburg.de/~mueckenh/MR/P3 Kap 10.pdf is discussing the *existence* of numbers. Regards, WM
From: mueckenh on 1 Feb 2007 06:49
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 .... It is not difficult to sum all squares, cubes, etc. Regards, WM |