Cantor Confusion
in [Math]
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From: mueckenh on 22 Mar 2007 07:27 On 21 Mrz., 16:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174481309.873623.92...(a)e1g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > > What portion of the first node > > > is assigned to your path? > > > > How much does the last term of the geometric series contribute to the > > value o the series? This is the value which the first node contributes > > to he path. > > As there is no last term of the geometric series this makes no sense. Nevertheless, it makes sense to calculate the sum of the series. > > > > > Can we agree upon the following facts? > > > > (1) The infinite paths 0.000... and 0.111... are completely contained > > > > in the complete tree T(oo), i.e., every node belonging to one of > > > > these paths is a node belonging to the tree T(oo). > > > > > > Indeed. > > > > > > > (2) The set of nodes of the tree T(oo) is countable. In particular, > > > > the number of nodes belonging to one level L(n) is countable for > > > > every n. > > > > > > Indeed. > > > > > > > (3) There is no node in the tree (including every node of the paths > > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > > many paths arrive at or start off from or cross through. > > > > > > No. I would state that at the root node of the tree there start > > > uncountably many paths. Why do you think that number is not uncountable? > > > (Remember, paths are non-terminating.) > > > > (3, clarified) There is no node in the tree (including every node of > > the paths > > 0.000... and 0.111...) which belongs to a level where uncountably > > many paths have separated. > > At every node in the tree uncountably many paths go to the left and > uncountably many paths go to the right. I do now know what you are > meaning here. At no node in the tree there exists a single path. This means: There are no single paths. > > > > > But it is connected to the root node like every other path. The > > > > branching offs are countable. > > > > > > Yes, I never contested that the number of branching offs is countable. So > > > what? > > > > In the whole tree there can be no more separated paths than nodes. > > Pray give a *proof*, not just a statement. The number of separated paths up to level n is given by the number of nodes of level n. No level of the tree has an uncountable number of nodes. > > > > > > You state it again. at no finite distance from the root. But you are > > > > > wrong. At each finite distance from the root there are uncountably > > > > > many paths that go through that level. That number is the same for > > > > > each level. > > > > > > > > That number is not visible. Let us take what can be proved: The number > > > > of separated paths *is* countable because, within the tree, there > > > > *cannot be more than countably many separated paths - at no finite > > > > distance from the root! > > > > > > And so what, what does that prove about the complete tree? Indeed, at no > > > finite distance from the root. What this *means* is that the number of > > > terminating paths is countable. > > > > No. It means that the number of paths which consist only of nodes with > > natural indexes is countable. > > Not at all. Pray provide a *proof*. The number of separated paths up to level n is given by the number of nodes of level n. No level of the tree has an uncountable number of nodes. > > > > But there is no finite distance from > > > the root where all non-terminating paths are separated from each other, > > > so the reasoning does not apply to them. > > > > The "distance" is measured by the index of the due node, i.e., by the > > number n of the corresponding level. "At no finite distance from the > > root" means at no node which can be enumerated by a natural number. > > Right. There is *no* node where all non-terminating paths are separated > from each other. There is no point in the tree where more than countably many paths are separated, although the tree is infinite. Outside of the tree there is no mathematics of real numbers. Regards, WM
From: mueckenh on 22 Mar 2007 07:34 On 21 Mrz., 16:16, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174481667.234070.21...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Yes. But the path-lengths are natural numbers, and there is no > > > > infinite natural number. > > > > > > The path-length of a non-terminating path is *not* a natural number. > > > > How then can it be contained in a set which contains only sets of > > finite paths? > > It is *not* an element of a set which contains only sets of finite paths. > Pray get your statements correct. A set that contains only sets of finite > paths has as elements sets of finite paths. Not paths. We can form the union of the sets. Then we get a set of finite paths. > Moreover, it is > also not an element of a set of all finite paths, because all elements are > finite. So an infinite path is not an element of it. And I would prefer > if you refrain from the use of the ambiguous word "contains". That can > mean different things. Whatever it means. The set of sets of finite paths does not contain any infinite path, also after the union of the sets has been taken. > > > > > The axiom of infinity grants the existence of all natural numbers, but > > > > not the existence of an infinite natural number. > > > > > > Indeed. > > > > why then do you believe that an infinite path (number) is contained in > > the union of sets of finite paths (numbers). > > It is not an element of the union of sets of finite paths. Why do you think > that I do think so? The union of sets of finite paths is a set of finite > paths, so there is no infinite path in it as element. Therefore, after having the union of all paths which are elements of finite trees, there is no infinite path. > > > > > > Now an > > > > > infinite paths is, for instance, the set {(i, 1) | i in N}, which is > > > > > a subset of U(T(n)). > > > > > > > > You try to get an infinite element out of a union of finite elements. > > > > That does not work. > > > > > > Where is the infinite element? > > > > You said: "The path-length of a non-terminating path is *not* a > > natural number." And you find it in a set of sets of finite paths. > > Eh? N is the union of the sets of initial segments, the cardinal number > of N is not a natural number. Where is the difference? The path-length > is the cardinal number of the set that represents a path (actually one > less if the path is finite). No. You are in error. The cardinal number of an infinite set of finite paths is infinite. But that does not produce any path of infinite length. > > > > > (3) There is no node in the tree (including every node of the paths > > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > > many paths arrive at or start off from or cross through. > > > > > > No. From the root node of the tree there start uncountably many paths. > > > > But they do never separate into uncountably many paths. > > At each node, uncountably many paths go to the left and uncountably many > paths go to the right. And everything which belongs to the mathematics of real numbers of [0, 1] happens within the tree. A path which is not isolated within the infinite tree does not exist as isolated path. Regards, WM
From: mueckenh on 22 Mar 2007 07:40 On 21 Mrz., 19:09, Virgil <vir...(a)comcast.net> wrote: > In article <1174481667.234070.21...(a)e65g2000hsc.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > The path-length of a non-terminating path is *not* a natural number. > > > How then can it be contained in a set which contains only sets of > > finite paths? > > If a nested set of finite paths, each representable by its leaf node, is > taken to represent the infinite path through each of those leaf nodes, > one has a fair representation of an infinite path. Then you have an infinite set of finite paths. That is not an infinite path. > > How else but as a set of nodes would WM represent an infinite path? In the union tree U(T(n)) there is no infinite path. In the EIT there is no infinite line. Yopu may take the diagonal, i.e., the leaf nde of every line, to et an infinite "path". But this s not a path of the tree but only a measure for the infinite number of paths. Regards, WM >
From: mueckenh on 22 Mar 2007 07:45 On 21 Mrz., 19:28, Virgil <vir...(a)comcast.net> wrote: > In article <1174482399.743448.268...(a)p15g2000hsd.googlegroups.com>, > > > > and a negative answer would confirms WM's misrepresentation > > > of the issue. > > > > > And at no point of > > > > the tree more than countably many infinite paths have separated. > > At every "point" of the tree, if that means node, uncountably many paths > through the left branch separate from uncountably many through the right > branch. > > WM is again applying methods only suitable to finite trees to the CIBT > Why should the determination of uncountably many paths by this inductive method be restricted to finite trees, if this determination was possible? > > Let us consider only the points (or better nodes) which in fact do > > exist (because they can be enumerated or indexed by natural numbers): > > > Do you think that the set of those paths, which consist only of > > existing nodes, is countable? > > In what tree? In a finite tree in which every path has n edges, the set > of all paths has cardinality 2^n > > In a tree in which every path has Aleph_0 edges , the set of all paths > has cardinality 2^Aleph_0 >Aleph_0 > At which point within the tree are these many paths existing? > > > > And since there is no end of "points", that no more proves the > > > countability of the set of all paths than the finiteness of each natural > > > proves the finiteness of the set of all naturals. > > > But you believe that the complete path 0.000... as well as the > > complete path 0.010101... etc. including their missing ends are > > contained in the complete tree? > > What "missing ends"? > When a circle has no end, does it necessarily have a "missing end"? > When an infinite sequence has no end, is there a "missing end". > > Do I believe that there exists a CIBT? Yes! If so, then there exists a level with uncountably many paths crossing, i.e., with uncountably many nodes. Other, it does not exist as* complete* binary tree. Regards, WM
From: Dik T. Winter on 22 Mar 2007 11:29
In article <1174562837.696384.225170(a)b75g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 21 Mrz., 16:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174481309.873623.92...(a)e1g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > > > What portion of the first node > > > > is assigned to your path? > > > > > > How much does the last term of the geometric series contribute to the > > > value o the series? This is the value which the first node contributes > > > to he path. > > > > As there is no last term of the geometric series this makes no sense. > > Nevertheless, it makes sense to calculate the sum of the series. You can calculate the sum of the series. But what the relation is to the contribution of the nodes to the paths is extremely unclear. From your statement I derive that the first node contributes nothing to the path. > > > > No. I would state that at the root node of the tree there start > > > > uncountably many paths. Why do you think that number is not > > > > uncountable? > > > > (Remember, paths are non-terminating.) > > > > > > (3, clarified) There is no node in the tree (including every node of > > > the paths > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > many paths have separated. > > > > At every node in the tree uncountably many paths go to the left and > > uncountably many paths go to the right. I do now know what you are > > meaning here. > > At no node in the tree there exists a single path. This means: There > are no single paths. Still unclear. Through each node go uncountably many paths. But what you mean with "there are no single paths" is unclear. Every two individual paths diverge at some node from each other. > > > > > But it is connected to the root node like every other path. The > > > > > branching offs are countable. > > > > > > > > Yes, I never contested that the number of branching offs is > > > > countable. So what? > > > > > > In the whole tree there can be no more separated paths than nodes. > > > > Pray give a *proof*, not just a statement. > > The number of separated paths up to level n is given by the number of > nodes of level n. No level of the tree has an uncountable number of > nodes. Makes no sense at all and is no proof. At each level n there are 2^(n-1) separated groups of paths, where each group contains uncountably many paths. > > > > And so what, what does that prove about the complete tree? Indeed, > > > > at no finite distance from the root. What this *means* is that the > > > > number of terminating paths is countable. > > > > > > No. It means that the number of paths which consist only of nodes with > > > natural indexes is countable. > > > > Not at all. Pray provide a *proof*. > > The number of separated paths up to level n is given by the number of > nodes of level n. No level of the tree has an uncountable number of > nodes. Yes, so what? That is no proof of your statement. When we talk about the set of paths, we are *not* talking about any finite level. What happens at finite levels is irrelevant to the total result. > > > The "distance" is measured by the index of the due node, i.e., by the > > > number n of the corresponding level. "At no finite distance from the > > > root" means at no node which can be enumerated by a natural number. > > > > Right. There is *no* node where all non-terminating paths are separated > > from each other. > > There is no point in the tree where more than countably many paths are > separated, although the tree is infinite. Outside of the tree there is > no mathematics of real numbers. As at each point in the tree there are uncountably many paths going through it, I wonder what you mean with "countably many paths are separated". -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |