Cantor Confusion
in [Math]
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From: Virgil on 6 Mar 2007 14:44 In article <1173176849.548735.27970(a)n33g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 5 Mrz., 16:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173083672.999163.257...(a)s48g2000cws.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 5 Mrz., 03:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > Nonsense. The cross section of a tree is the number of nodes in its > > > > last level. So you now claim that the number of nodes in its last > > > > level > > > > of T(oo) is aleph-0? But there is no last level. > > > > > > Nevertheless, all last levels are there which make up the set of all > > > last levels of finite trees. If the set of all natural numbers exists > > > (e.g., as the set of all last elements n of initial segments > > > {1,2,3,...,n}), then the set of all last levels exists too. And we > > > know that this set of all last levels does not contain any element > > > which, as a set of nodes, has a cardinality of more than aleph_0. > > > > As a set of nodes, it indeed does have cardinality aleph_0. > > Of course. And we need not use the limit definition (of Hrbacek and > Jech or else) in order to get this result, because the whole tree > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > larger than countable. The set of binary places to the right of the radix point is countable, but the set of numbers representable by endless strings of binary places is not. Thus countably many levels of nodes must produce uncountably many strings. > > > > > Result: The union tree U(T(n)) is too narrow to contain more than > > > aleph_0 paths. > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > set of nodes, hence an element of the powerset of this set of nodes. > > We know that this powerset is not countable, so the set of paths can > > be (and actually is) uncountable. > > This argument should be recognizable as invalid. So WM wishes, but the facts refute him. > It could only be > applied if ALL combinations of nodes were paths. It only requires a choice of more than one node at each of a countable list of positions in the potential path, just like a choice of more than one digit at each position of an infinite binary string. And, unless WM claims that all these paths are finite in length, there must be a choice of two notes at every one of infinitely many positions. > But this is not the > case! And need not be the case. > Not every set of nodes is a path, as you well know. But enough are, as shown above! > In order to > count the paths (-bundles) which pass through a certain level of the > tree, it is sufficient to know the cross section of the tree. The complete infinite tree has no cross section, at least according to WM's definition of cross sections, as WM's definition requires a last level which such a tree does not have. Wm's notion of logic is that if his argument does not fit he only needs to get a bigger hammer to be able to pound it into place.
From: mueckenh on 6 Mar 2007 16:07 On 6 Mrz., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 5 Mrz., 16:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Nevertheless, all last levels are there which make up the set of all > > > > last levels of finite trees. If the set of all natural numbers exists > > > > (e.g., as the set of all last elements n of initial segments > > > > {1,2,3,...,n}), then the set of all last levels exists too. And we > > > > know that this set of all last levels does not contain any element > > > > which, as a set of nodes, has a cardinality of more than aleph_0. > > > > > > As a set of nodes, it indeed does have cardinality aleph_0. > > > > Of course. And we need not use the limit definition (of Hrbacek and > > Jech or else) in order to get this result, because the whole tree > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > larger than countable. > > The latter is an invalid conclusion. Did you ask the referee? Or are you the arbiter yourself? C(oo) is a part of T(oo). No, Dik. I know that the functions in set theory are not continuous and perform unpredictable jumps when leaving the finite and entering the infinite. One of these tricks is the set of all finite sets being countable and the set of all infinite sets being uncountable. Another trick is the famous observation: Forall n in N: |{2,4,6,...,2n}| < 2n while lim {n-->oo} |{2,4,6,...,2n}| > lim {n-->oo} 2n, and similar jokes. Therefore I devised the tree. Its nodes are connected by continuous paths. They prevent any discontinuity. If we have a cross section less than the number of nodes for any finite tree, then the cross section cannot surpass the number of nodes of the infinite tree. The structure of the tree forces the function C(n)/|T(n)| to remain continuous under any circumstances. > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > set of nodes, hence an element of the powerset of this set of nodes. Have you really been blinded for the fact that most of the subsets of nodes of the complete cannot serve as paths? > > > We know that this powerset is not countable, so the set of paths can > > > be (and actually is) uncountable. > > > > This argument should be recognizable as invalid. > > What is invalid? I may note that the parenthetical remark is *not* > part of the argument. Part of which argument? I cannot see any argument at all. You state (1) that Cantor's argument is valid (at this place I do no judge about it) and you state (2) that set theory is free of contradictions (I don't know who told you). From (1) and (2) you conclude that contrary to all mathematical evidence (I remind you of the ration of 2 nodes per path and the cross section of the tree which counts its paths) my result cannot be true. You are wrong. But similar to the case of Euclid's contradiction you can assert any illogic claim and you can assert that it is the best logic available. Nobody can convince you. > > > It could only be > > applied if ALL combinations of nodes were paths. But this is not the > > case! Not every set of nodes is a path, as you well know. In order to > > count the paths (-bundles) which pass through a certain level of the > > tree, it is sufficient to know the cross section of the tree. More > > than C(n) paths (-bundles) cannot fit into the tree. This proves > > without any doubt that the complete tree T(oo) cannot contain more > > than countably many paths (-bundles). > > This proves that in no way. Have you some argument which at least calms down your own doubts? Regards, WM
From: mueckenh on 6 Mar 2007 16:15 On 6 Mrz., 20:44, Virgil <vir...(a)comcast.net> wrote: > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 5 Mrz., 16:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1173083672.999163.257...(a)s48g2000cws.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 5 Mrz., 03:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > Nonsense. The cross section of a tree is the number of nodes in its > > > > > last level. So you now claim that the number of nodes in its last > > > > > level > > > > > of T(oo) is aleph-0? But there is no last level. > > > > > Nevertheless, all last levels are there which make up the set of all > > > > last levels of finite trees. If the set of all natural numbers exists > > > > (e.g., as the set of all last elements n of initial segments > > > > {1,2,3,...,n}), then the set of all last levels exists too. And we > > > > know that this set of all last levels does not contain any element > > > > which, as a set of nodes, has a cardinality of more than aleph_0. > > > > As a set of nodes, it indeed does have cardinality aleph_0. > > > Of course. And we need not use the limit definition (of Hrbacek and > > Jech or else) in order to get this result, because the whole tree > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > larger than countable. > > The set of binary places to the right of the radix point is countable, > but the set of numbers representable by endless strings of binary places > is not. > > Thus countably many levels of nodes must produce uncountably many > strings. > Then they must produce an uncountablke set of paths = cross section C(oo), which they do not! > > > > > > Result: The union tree U(T(n)) is too narrow to contain more than > > > > aleph_0 paths. > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > set of nodes, hence an element of the powerset of this set of nodes. > > > We know that this powerset is not countable, so the set of paths can > > > be (and actually is) uncountable. > > > This argument should be recognizable as invalid. > > So WM wishes, but the facts refute him. > > > It could only be > > applied if ALL combinations of nodes were paths. > > It only requires a choice of more than one node at each of a countable > list of positions in the potential path, just like a choice of more than > one digit at each position of an infinite binary string. Compare the proof of Oresme. There are only countably many unit fractions although their number dubles in infinity, i.e. for any finite n without ending. > > And, unless WM claims that all these paths are finite in length, there > must be a choice of two nodes at every one of infinitely many positions. This yields exactly |2^omega| = aleph_0 paths. > > Not every set of nodes is a path, as you well know. > > But enough are, as shown above! Never more than aleph_0 - for any finite n without ever coming to an end. > > > In order to > > count the paths (-bundles) which pass through a certain level of the > > tree, it is sufficient to know the cross section of the tree. > > The complete infinite tree has no cross section, at least according to > WM's definition of cross sections, as WM's definition requires a last > level which such a tree does not have. No it requires only a cross section continuosly growing with length n of the tree by 2^n. This is done and will never end. That's all about infinity. Regards, WM
From: Virgil on 6 Mar 2007 17:22 In article <1173215257.272668.121480(a)64g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 6 Mrz., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 5 Mrz., 16:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > Nevertheless, all last levels are there which make up the set of > > > > > all > > > > > last levels of finite trees. If the set of all natural numbers > > > > > exists > > > > > (e.g., as the set of all last elements n of initial segments > > > > > {1,2,3,...,n}), then the set of all last levels exists too. And we > > > > > know that this set of all last levels does not contain any element > > > > > which, as a set of nodes, has a cardinality of more than aleph_0. > > > > > > > > As a set of nodes, it indeed does have cardinality aleph_0. > > > > > > Of course. And we need not use the limit definition (of Hrbacek and > > > Jech or else) in order to get this result, because the whole tree > > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > > larger than countable. > > > > The latter is an invalid conclusion. > > Did you ask the referee? Did WM ask "the referee"? If he had he would have found himself penalized for multiple fouls. > C(oo) is a > part of T(oo). Then it cannot contain all paths of a complete infinite binary tree, as has been repeatedly proved here. The referees agree that WM is wrong. > > No, Dik. I know that the functions in set theory are not continuous > and perform unpredictable jumps when leaving the finite and entering > the infinite. One of these tricks is the set of all finite sets being > countable and the set of all infinite sets being uncountable. Another > trick is the famous observation: > Forall n in N: |{2,4,6,...,2n}| < 2n > while > lim {n-->oo} |{2,4,6,...,2n}| > lim {n-->oo} 2n, and similar jokes. That last may be true in wilds of WM_land, but it is not true in mathematics, as neither limit is defined in mathematics. > > Therefore I devised the tree. What "tree" is it that has such imposssible properties? > Its nodes are connected by continuous > paths. They prevent any discontinuity. Show us just how any edge is a "continuous" connection. By what definition of continuity? WM likes to throw big words around without regard for their meaning. But in the context of the edges connecting nodes, and edge is no more than an ordered pair of nodes, and there is nothing "continuous" about an ordered pair. If we have a cross section less > than the number of nodes for any finite tree, then the cross section > cannot surpass the number of nodes of the infinite tree. The structure > of the tree forces the function C(n)/|T(n)| to remain continuous under > any circumstances. How does WM define "continuity" of a function whose domain is the highly discontinuous set of finite naturals? > > > > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > > set of nodes, hence an element of the powerset of this set of nodes. > > Have you really been blinded for the fact that most of the subsets of > nodes of the complete cannot serve as paths? As there are uncountably many which CAN serve, the ones that cannot are irrelevant. > > > > > We know that this powerset is not countable, so the set of paths can > > > > be (and actually is) uncountable. > > > > > > This argument should be recognizable as invalid. > > > > What is invalid? I may note that the parenthetical remark is *not* > > part of the argument. > > Part of which argument? I cannot see any argument at all. Those who choose not to see can blind themselves to anything. > > > > > > It could only be > > > applied if ALL combinations of nodes were paths. It only requires that enough combinations be paths, and enough are. Every node in an infinite path in this infinite tee is parent to a choice of two child nodes, which gives an infinite sequence of binary choices, and Cantor long since proved the set of all such to be uncountable. despite WM's most strident objections. > > > case! Not every set of nodes is a path, as you well know. But uncountably many such sets are, as shown above. I This proves > > > without any doubt that the complete tree T(oo) cannot contain more > > > than countably many paths (-bundles). > > > > This proves that in no way. > > Have you some argument which at least calms down your own doubts? Does WM declare that an infinite path is not the result of an arbitrary infinite sequence of binary options?
From: Virgil on 6 Mar 2007 17:40
In article <1173215723.210725.37860(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 6 Mrz., 20:44, Virgil <vir...(a)comcast.net> wrote: > > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com>, > > > > > > > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 5 Mrz., 16:46, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1173083672.999163.257...(a)s48g2000cws.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 5 Mrz., 03:14, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > ... > > > > > > Nonsense. The cross section of a tree is the number of nodes in > > > > > > its > > > > > > last level. So you now claim that the number of nodes in its last > > > > > > level > > > > > > of T(oo) is aleph-0? But there is no last level. > > > > > > > Nevertheless, all last levels are there which make up the set of all > > > > > last levels of finite trees. If the set of all natural numbers > > > > > exists > > > > > (e.g., as the set of all last elements n of initial segments > > > > > {1,2,3,...,n}), then the set of all last levels exists too. And we > > > > > know that this set of all last levels does not contain any element > > > > > which, as a set of nodes, has a cardinality of more than aleph_0. > > > > > > As a set of nodes, it indeed does have cardinality aleph_0. > > > > > Of course. And we need not use the limit definition (of Hrbacek and > > > Jech or else) in order to get this result, because the whole tree > > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > > larger than countable. > > > > The set of binary places to the right of the radix point is countable, > > but the set of numbers representable by endless strings of binary places > > is not. > > > > Thus countably many levels of nodes must produce uncountably many > > strings. > > > Then they must produce an uncountablke set of paths = cross section > C(oo), which they do not! They do produce an uncountable set of paths in both ZFC and NBG, and WM has yet to come up with an axiom system in which he can show that they do not. Perhaps WM's perception of what C(oo) actually is is warped as badly as the rest of his mathematics. > > > > > > > > > Result: The union tree U(T(n)) is too narrow to contain more than > > > > > aleph_0 paths. > > > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > > set of nodes, hence an element of the powerset of this set of nodes. > > > > We know that this powerset is not countable, so the set of paths can > > > > be (and actually is) uncountable. > > > > > This argument should be recognizable as invalid. > > > > So WM wishes, but the facts refute him. > > > > > It could only be > > > applied if ALL combinations of nodes were paths. Not so. It only requires uncountable subset of the uncountable power set of the set of nodes. There are enough that we don't need all of them. All we need is a choice of two child nodes for each parent node, and we get enough choices to create uncountably many infinite paths. > > > > It only requires a choice of more than one node at each of a countable > > list of positions in the potential path, just like a choice of more than > > one digit at each position of an infinite binary string. > > Compare the proof of Oresme. There are only countably many unit > fractions although their number dubles in infinity, i.e. for any > finite n without ending. Compare the original Cantor diagonal proof for functions from N to {w,m}. > > > > And, unless WM claims that all these paths are finite in length, there > > must be a choice of two nodes at every one of infinitely many positions. > > This yields exactly |2^omega| = aleph_0 paths. Wrong, it is 2^|omega| > aleph_0. > > > > Not every set of nodes is a path, as you well know. > > > > But enough are, as shown above! > > Never more than aleph_0 - for any finite n without ever coming to an > end. Cantor proved otherwise. > > > > > In order to > > > count the paths (-bundles) which pass through a certain level of the > > > tree, it is sufficient to know the cross section of the tree. > > > > The complete infinite tree has no cross section, at least according to > > WM's definition of cross sections, as WM's definition requires a last > > level which such a tree does not have. > > No it requires only a cross section continuosly growing with length n > of the tree by 2^n. This is done and will never end. That's all about > infinity. WM keeps trying to ignore that the construction of an infinite binary path requires an infinite sequence of binary choices, which Cantor proved leads to more that countably many number of paths. That simple and unrefuted proof overwhelms all of WM's futile objections. |