Cantor Confusion
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From: mueckenh on 6 Feb 2007 06:54 On 5 Feb., 06:04, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170606249.100767.104...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 3 Feb., 04:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1170413742.648825.136...(a)l53g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > So you say: > > > > The union of finite trees contains an infinite tree. > > > > The union of finite chains contains an infinite chain. > > > > The union of finite paths contains an infinite path. > > > > > > Indeed. > > > > > How can a union of finite elements contain an infinite element? > > Where in the above is there a union of finite elements? In all three statements. > Unions are > about *sets*, not about elements. Every element is a set. There are only sets in ZFC. > > > > > So you say the union of finite paths > > > > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > > > contains the infinite path p(oo) = {0,0,0,...}. > > > > > > > > But the union of finite sets of finite paths > > > > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > > > > does not contain the union of finite paths > > > > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > > > > > Again, indeed. > > > > But that is incorrect! > > > > The union of finite sets of finite paths > > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > > is the set of all finite paths > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > Indeed. > > > which obviously contains the union of finite paths of the special form > > p(0), p(1), p(2), .... > > as a subset. > > Wrong. There is no obviously here. It can be proven. {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains {p(0), p(1), p(2), ... } > The union of sets of elements does > *not* contain the union of elements. The tree contains it. > This is abundantly clear from the > first eight chapters of your book. They report the history, not the truth. (The latter you will find in Chapter 9 and 10). > > > If you say, as you do above, that the union of finite paths contains > > an infinite path, > > p(0) U p(1) U p(2) U ... contains the infinite path p(oo) > > then this is also true for this subset. > > Let's get back to the finite. A = {{a, b}, {a, c}, {a, d}} and > B = {{a, d}, {b, d}, {c, d}}. A U B consists of: > {{a, b}, {a, c}, {a, d}, {b, d}, {c, d}}. > It does *not* contain unions of the sets that are elements of A or B. Why don't you construct an example which is represented in the tree? {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains {p(0), p(1), p(2), ... } > > It is impossible that you loose p(oo) because the set of all finite > > paths > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > contains some more paths than he union of special finite paths > > p(0), p(1), p(2), .... > > In what way does > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > contain: > p(0) U p(1) U p(2) U ...? In the way the tree contains them as sets of nodes: p(0) U p(1) U p(2) U ... = {p(0), p(1), p(2), ... } = p(oo) is a subset of the set of paths contained in the tree. > > > > > I would say: The union of all finite sets of finite path contains all > > > > finite paths. > > > > > > Right. As I have stated all the time. And so the set of paths in the > > > complete tree is *not* a subset of the union of all finite sets of > > > finite paths, something you have stated repeatedly. > > > > The union of all finite paths of a special kind contains the due > > infinite path. > > The union of all finite paths contains all infinite paths. > > *Sets* of paths. Remember? > > > > Right. But you *use* that the union of all finite sets of finite paths > > > *contains* the union of these paths. > > > > If the union p(0), p(1), p(2), ....contains p(oo) then the union > > p(0), p(1), q(1), p(2), q(2), r(2),s(2), ... > > cannot miss it. > > What *sets* of paths are you using? The set of paths contained in the finite tree T(0), the set of paths contained in the finite tree T(1), ..., the set of paths contained in the finite tree T(n), and so on. > > > > Then apply it twice. First you get the single set of all paths, second > > you get the set of elements. > > Which is a set of nodes. Correct. And a set of ndes can be a path. > So you are not taking the union but the union of > the union. Let us analyse. A set of paths is a set of sets of nodes. So > let us consider the basic two trees (level 1 and level 2). The sets of paths > are: L(0) = {{0}} > L1: {{0, 1}, {0, 2}} > L2: {{0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. > Two sets. Unite them once: > {{0, 1}, {0, 2}, {0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. > Unite this one, and we get: > {0, 1, 2, 3, 4, 5, 6}. > I do not see a path there. Every path is a subsets of the union, for example {0, 2, 6} c {0, 1, 2, 3, 4, 5, 6}. Which path do you miss? > > > > > If there is an infinite pathlength, then there must be an infinite > > > > number as well as an infinite segment of natural numbers. > > > > > > Why must there be an infinite number? > > > > Because a union of finite pathlengths cannot lead to an infinite > > pathength. > > > > The infinite union of pathlengths > > 1,1,1,... > > 2,2,2,... > > 3,3,3 > > 4,4 > > 5 > > > > yields a finite pathlength, namely 5. > > Yes. Infinite unions can give finite things. But that is not a > contradiction. It shows clearly that not the number of elements is decisive but their sizes. > > > > But rest assured, there may be an > > > infinite segment (depending on how you define segment). > > > > It is necessary. As an infinite natural number would be necessary to > > obtain an actually infinite set of numbers. > > You think so, but provide no proof. Look at the EIT. 1 11 111 .... If every element of the diagonal exists, then every line must exist. If the diagonal is longer than every natural number then a line must be longer too. > > > There cannot be an infinite segment {1,2,3,...n} with a last element > > unless the last element is infinite. > > Sure. But again that is not in contradiction to anything I state. It contradicts actual infinity. > > > > The union of the P(i) is the set of all paths. It includes the set of > > all paths of the form p(n). Why should it not contain p(oo) if the set > > of all paths of the form p(n) alone contains p(oo)? > > I think we are talking at cross purposes. And are misunderstanding each > other about what the meaning is of the term "the set of all paths p(n)". I think P(i) is the set of all paths of the finite tree with i levels. p(n) is a path like 0.000...0 with n zeros behind the point. The union of all p(n), i.e., of all nodes 0, is p(oo) = 0.000... > > > > And that is a single set containing every element which is in A or in > > B or in both. > > The union of some sets of paths is one set of paths. > > The union of all paths is one set of elements. > > Right. So the union of all paths is a set of nodes. But I do not think > that set is a path. That set is a tree (by your definition). That set is a tree but contains the paths as subsets. > > The union of {three paths} and {12 paths} and {17 paths} is a set of > > paths. These paths are unioned in the tree to give a set of nodes, > > perhaps including even one or more paths. > > This does not make sense at all. The union of the two paths in the tree T(1) and the 4 paths in the tree T(2) and the 8 paths in the tree T(3) is a set of 14 paths (a set of 14 sets of nodes). The union of these 14 sets is one set of nodes (the tree T(3)). This set of nodes contains all the 14 paths as subsets. Regards, WM
From: Franziska Neugebauer on 6 Feb 2007 07:21 mueckenh(a)rz.fh-augsburg.de wrote: > On 5 Feb., 05:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: [...] > Of course the requirement to decide whether n is in N is more circular > than the statement that 1 is in N. Contemporary set theories do not have these kind of problems. [...] >> it is impossible to talk about >> it, at least mathematically. > > Present mathematics has nothing in common with existence. In present mathematics "existence" does not mean *physical* existence. [...] >> Existence is a mathematical thing when you can establish it by axioms >> or through theorems based on axioms. Anything else is merely >> phylosophy. > > There is something called reality and another thing called matheology. The latter is taught in Augsburg. F. N. -- xyz
From: William Hughes on 6 Feb 2007 07:28 On Feb 6, 5:18 am, mueck...(a)rz.fh-augsburg.de wrote: > On 4 Feb., 20:33, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 4, 1:01 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 4 Feb., 15:31, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > > Similarly one can prove by induction: Every set of even positive > > > > > numbers contains numbers which are larger than the cardinal number of > > > > > the set. > > > > > No. You make your usual mistake. > > > > No, you make your usual mistake by thinking that a potentially > > > infinite set can have an infinite cardinal number. > > > I make no assumptions whatsoever about the cardinal > > number of a potentially infinite set. (I do not even assume > > that one exists). > > You said: "E contains numbers which are larger than the *cardinal > number of E*." See below. > I said that this statement is false. If E has a cardinal number this statement is false. If E does not have a cardinal number this statement is false. Saying that the statement is false does not require making an assumption about whether of not E has a cardinal number. > > > > > > Let E be the potentailly infinite set of even numbers. > > > > Two statements > > > > > i: For every element e of E, the set > > > > F(e) = {2,4,6,...,e} contains cardinal numbers which are > > > > larger > > > > than the cardinal number of F(e) > > > > > ii E contains numbers which are larger than the cardinal > > > > number of E > > > > > Statement i is true and can be shown by induction. Statement > > > > ii is false. > > > > Of course, > > > My claim is that statment ii is false. Your reply > > is "of course". We both agree that statement ii > > is false. > > Yes, because there is no cardinal number of E. > > > > > > namely because there is no cardinal number for potentially > > > infinite sets. > > > > > The fact that statment i is true does not mean > > > > that statement ii is true. > > > > The fact that statement I is true means that there are no elements of > > > the set which could increase its "number of elements" without > > > increasing the sizes of elements in the set. > > > As statement ii does not say or imply > > anything about increasing the "number of elements" in E, > > totally irrelevent. > > Statement ii is false. every statement beginning with "As statement ii > does " therefore is irrelevant. Not to questions of whether statement ii is true of false. > > > There is no element that can be shown to > > be in E that is not in some F(e), but you need more than one > > F(e) . There is no single F(e) that contains every element > > that can be shown to be in E. > > Wong. There is no element in E which is outside of every F(e). Hence > there are two possibilities: > Either all elements of E exist and there is necessarily one F(E) > or not all elements of E exist, then here is no F(E). > No If all elemens of E exist, then E is actually infinite. No F(e) is infinite so no F(e) contains E, and there is no one F(e). If not all elments of E exist then E is potentially infinite. No F(e) is potentially infinite so no F(e) contains every element of E that can be shown to exist. In neither case is there a single F(e). In neither case is there an element of E that is not in one of the F(e), but different elements of E may require different F(e). > > > > The fact that > > there is no element in E that is not in some set > > F(e), does not mean that E has the same properties > > as the sets F(e). > > > The fact that statement i is true > > does not imply that statement ii is true. > > Statement i is true, statement ii is false > > (note your "of course" above). > > The statement "Every set of even positive > > numbers contains numbers which are larger than the cardinal number of > > the set", is false. > > Potential infinity is in this case the sequence of sets F(e) - there > is no further set E other than this sequence. Let E be the sequence of the sets F(e_i). There are two statements. i. For each of the sets F(e_i) in the sequence E, there is an element of F(e_i) which is greater than the cardinal number of F(e_i) ii There an element of one of the F(e) that is greater than the cardinal number of E. i is true, ii is false, E is a counterexample. - William Hughes
From: Franziska Neugebauer on 6 Feb 2007 07:47 mueckenh(a)rz.fh-augsburg.de wrote: > It can be proven. {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > contains {p(0), p(1), p(2), ... } What does contain mean? ( ) contain as element ( ) contain as subset >> The union of sets of elements does >> *not* contain the union of elements. > > The tree contains it. Depends on the definition of tree in terms of sets and of the definition of union. Which definitions do you refer to? [...] >> > If you say, as you do above, that the union of finite paths >> > contains an infinite path, >> > p(0) U p(1) U p(2) U ... contains the infinite path p(oo) >> > then this is also true for this subset. >> >> Let's get back to the finite. A = {{a, b}, {a, c}, {a, d}} and >> B = {{a, d}, {b, d}, {c, d}}. A U B consists of: >> {{a, b}, {a, c}, {a, d}, {b, d}, {c, d}}. >> It does *not* contain unions of the sets that are elements of A or B. > > Why don't you construct an example which is represented in the tree? > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains {p(0), p(1), > p(2), ... } What does contain mean? ( ) contain as element ( ) contain as subset [...] >> In what way does >> {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } >> contain: >> p(0) U p(1) U p(2) U ...? > > In the way the tree contains them as sets of nodes: > p(0) U p(1) U p(2) U ... = {p(0), p(1), p(2), ... } 1. The usual set theoretic definition of a U B is U{a, b} which is { x | x e a v x e b }. 2. If your '=' means equality your formula is a statement form (Aussageform). Assume p(n) = {} A n >= 0. Then your formula states p(0) = { p(0), {} } which is wrong for all p(0). Hence your statement form is not universally valid, it is universally invalid. 3. If your '=' means ':='/'def=' (definition) you redefine the common meaning of the union symbol. (equivocation). [...] >> > Then apply it twice. First you get the single set of all paths, >> > second you get the set of elements. >> >> Which is a set of nodes. > > Correct. And a set of ndes can be a path. No. A paths *is* a) a sequence of nodes, or b) a sequence of edges. [...] >> > There cannot be an infinite segment {1,2,3,...n} with a last >> > element unless the last element is infinite. >> >> Sure. But again that is not in contradiction to anything I state. > > It contradicts actual infinity. You have simply stated the M�ckenheim axiom X is not finite -> there must be an x in X which is infinite which is alas not part of ZFC or any other modern set theory. F. N. -- xyz
From: Dik T. Winter on 6 Feb 2007 09:20
In article <1170757722.722299.113560(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 5 Feb., 04:33, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > Having read chapter 8 now completely, there are two errors there. > > Please let me know, and also that one in chapter 7. I would like to > correct these errors in the second edition. Will be in the review that is coming shortly. > > But > > it remains an excellent review about the history of thinking. I just > > started chapter 9, and, well, I will not say anymore now. I need a few > > days for the remaining chapters. > > Attention. Dangerous contents! Yes, that I do see. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |