Cantor Confusion
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From: David Marcus on 30 Jan 2007 17:50 MoeBlee wrote: > On Jan 30, 2:00 pm, Virgil <vir...(a)comcast.net> wrote: > > > If one can show a bijection between that interval and the set of all > > reals (which one can do quite easily) and one can also show that the > > "anti-diagonal" constructed must fall within the interval (which one can > > also do quite easily), the proof stands. > > Of course I don't disagree with that. Only, that just to show the > uncountability of the reals, we don't even need to show a bijection > between the interval and the set of reals. If a subset of the set of > reals is uncountable, then the set of reals is uncountable. If we take that subset to be [0,1) and then construct a real number not in the subset and not on the list, then we haven't proved anything. -- David Marcus
From: MoeBlee on 30 Jan 2007 18:11 On Jan 30, 2:48 pm, David Marcus <DavidMar...(a)alumdotmit.edu> wrote: > Can't say I've ever seen it applied to the entire set of reals, although > of course we could do it that way. In Suppes, for example. MoeBlee
From: MoeBlee on 30 Jan 2007 18:17 On Jan 30, 2:50 pm, David Marcus <DavidMar...(a)alumdotmit.edu> wrote: > MoeBlee wrote: > > On Jan 30, 2:00 pm, Virgil <vir...(a)comcast.net> wrote: > > > > If one can show a bijection between that interval and the set of all > > > reals (which one can do quite easily) and one can also show that the > > > "anti-diagonal" constructed must fall within the interval (which one can > > > also do quite easily), the proof stands. > > > Of course I don't disagree with that. Only, that just to show the > > uncountability of the reals, we don't even need to show a bijection > > between the interval and the set of reals. If a subset of the set of > > reals is uncountable, then the set of reals is uncountable. > > If we take that subset to be [0,1) and then construct a real number not > in the subset and not on the list, then we haven't proved anything. If we the range of a list is a subset of [0 1) and we show a real number not in the subset, then, just doing that doesn't prove the uncountability of the reals, I agree. But that doesn't contradict anything I've said. MoeBlee MoeBlee
From: Dik T. Winter on 30 Jan 2007 21:17 In article <1170162837.477985.97730(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 29 Jan., 03:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > If I consider the path length as the union of path length: = > > > {0,1,2,...,n-1}, I am right too. > > > > Yes, for this path. I never did have problems when you were uniting paths > > that you get infinite paths and all that that entails. > > But path-lengths are natural numbers! > > > But you are uniting > > *sets* of paths. In that case you do not unite paths at all. > > I unit finite trees. When you talk about the cardinality of the set of paths in the ultimate tree you are *using* the union of sets of paths. > This implies the union of sets of paths. This is wrong. > This > implies the union of paths, because the union of two or more paths > need not but can be a path (notice: the indexes are of the form > {0,1,2,...,n-1}). 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 extremely unclear. A path is a set of nodes (by your definition), > > its length is the cardinality of that set. Let's say that path p_n is > > the set of nodes {n_1, n_2, n_3, ..., n_n}, with path length (cardinality) > > n. The union of those paths is {n_1, n_2, n_3, ...} with cardinality > > aleph_0. But none of the paths in the union is infinite. If you do not > > agree with this, come up with a definition of path length when a path > > is a set of nodes. > > 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"). > 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? > 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. > > 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. 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. > > > 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 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. 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. > > > Therefore P is a subset of this union P(1) U P(2) U ... of elements of > > > P_C. > > > Is this correct? > > > > No. > > What is the origin of P when we unite finite trees? The subsets of the set of nodes in the union of the finite trees. > > > 0.010101... is not in the union of all T(n), n in N, as > > > you say. > > > > I do not say that. I explicitly state that it is in it. But it is *not* > > in the union of all P(n). > > Taking the union of all T(n) is analog to taking the union of all > P(n). Wrong. When you take the union of all T(n) you get as paths the (likely) unions of the paths in the T(n). But when you take the union of the P(n) you do not get those unions. > The union of all P(n) does contain elements > > like "0.0", "0.01", "0.010", "0.0101", but *not* "0.010101...". The > > union contains all finite initial segments of "0.010101...", but not > > that one itself. > > Then the path 0.010101... cannot be in the tree which is consructed by > taking all finite trees. By taking all finite trees you simultaneously > take all finite sets of finite paths. Yes, but in the infinite tree you do not *define* the paths as such. Rather, you state that some paths from the finite trees are not paths in the infinite tree. And actually *none* of the paths in the finite trees are also a path in the infinite tree. Or is {n_0, n_1} for some arbitrary connected nodes starting at the root a path in the infinite tree? You can answer two ways: (a) No. In that case, *none* of the elements of the sets of paths in finite trees are a path in the infinite trees. And so P is not contained in P(1) U P(2) U ... Worse, P(1) U P(2) != P(2). (b) Yes. In that case P(1) U P(2) U ... contains finite paths only, so the intersection of that set and P contains finite paths only. > > In a similar way, N contains all the finite initial > > segements of itself, but not N itself. > > N is not a natural number. But the union of all path-lengths is a path- > length. Wrong. Infinite paths do not have a path-length that is a natural number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Jan 2007 21:33
In article <1170163062.268110.161970(a)h3g2000cwc.googlegroups.com> mueckenh(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? What sets do 4 and 5 represent? > > 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. > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |