Cantor Confusion
in [Math]
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From: Dik T. Winter on 22 Mar 2007 11:45 In article <1174563273.137227.294160(a)y66g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Yes, we can do that. And then we get a set of finite paths that does *not* contain a non-terminating path as an element, and also not the path-length of such a path as an element. So what now? > > 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. So you prefer to use ambiguous words? Why? To confuse everybody? > The set of sets of finite paths does not contain > any infinite path, also after the union of the sets has been taken. Again "contain". But, so what? I fail to see the relevance to the path-length of non-terminating paths. > > > 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. Wrong. In the first place, paths are *not* elements of trees according to your definitions. Paths are *subsets* of trees according to them. Pray use the correct wording. Second, the union of all paths etc., is *not* a path, it is just an infinite set of nodes, the same as the union of all trees. The paths are in it as subsets, as are the infinite paths. > > > > 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. But I do not state that it does. The infinite paths are *unions* of finite paths, where each finite path is an initial segment of the infinite path. I am not talking about the cardinality of a set of path, but of the cardinality of a path (the number of nodes in it). That does not mean that an infinite set of paths contains an infinite path, that would be false 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. 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. Pray define "isolated". -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 22 Mar 2007 14:53 In article <1174563273.137227.294160(a)y66g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. But forming the set theoretic union of the nested finite trees to form a supertree does not automatically require forming the set theoretic unions of the appropriate nested sets of paths to form the requisite superpaths. That secondary unioning would have to be done separately. Unless one wishes to define the paths of a union of a nested set of trees as a maximal nested sequence of paths (nested in the sense that each path (as a finite set of nodes) is a subset of all the longer paths (as larger finite sets of nodes). > > > 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. A union of trees is not a tree unless it has paths in it. And the paths of members of a union of trees are not paths of the union. So that WM's "union" of all finite trees is not a tree at all. > > > > > > > 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. The union is no tree either. And in order to make a tree out a union of trees, one must also make paths out of of suitable unions of paths. And any such making of a union suitably nested finite trees into a tree makes the union of suitably nested sets of paths into a path. And this process applied to all binary trees produces an infintie tree with countably many nodes and "levels" and uncountably many paths. > > > > > > > > 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. If each of those denumerably many paths, as a set of nodes, is a subset of each larger one, then their union, like N, contains a denumerable number of nodes, and is a path in the WM's psuedo-union tree. > > 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. One can define an order relation on the paths in any tree such that for distinct paths p1 and p2, p1 < p2 iff at the first branching at which they differ, p1 branches left and p2 branches right. Under this ordering, every finite tree is discretely ordered, meaning any but the largest has a next larger and any but the smallest has a next smaller. But every CIBT is densely ordered, meaning between any two paths there is another (actually, between any two, there will be uncountably many others). So that for CIBTs none of those uncountably many paths is isolated. Thus,to say that there are no isolated paths is a triviality, but leaves the actuality that there are uncountably many un-isolated paths between any two distinct paths. And WM has provided 0.000... and 0.111... as two such distinct paths.
From: Virgil on 22 Mar 2007 14:58 In article <1174563615.043961.305440(a)y66g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. If your "union" of trees really is a tree, then suitable unions of paths will be its paths. > > > > 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. Then there are no paths at all and it is not a tree at all. WM wants to have an infinite binary tree without any paths, since as soon as he allows it to have any paths, it has more than he wants.
From: Virgil on 22 Mar 2007 15:11 In article <1174563931.158002.44140(a)p15g2000hsd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? Whoever said finite trees have infinitely many anything? Wm wants to take an infinite sequence of finite binary trees where each path in each tree, as a set of nodes, is a proper subset of some path in each subsequent tree. Then WM claims the union of those nested trees is somehow a tree but the union of corresponding nested paths is not a path. > > > 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? When one has no last point. > > > > > > 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. That is a false claim. The number of nodes at each finite level n of the tree is 2^n, regardless of whether the tree has a finite or infinite number of levels, but that n only limits the number of paths if the tree ends at that nth level. Other, it does not exist as* > complete* binary tree. The CIBT has uncountably many paths, regardless of WM's many machinations to establish otherwise. > > Regards, WM
From: mueckenh on 23 Mar 2007 08:58
On 22 Mrz., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174562837.696384.225...(a)b75g2000hsg.googlegroups.com> mueck...(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. the first node contributes exactly as much as the last term of the geometric series contributes to the sum 2 of the series. As this term does not exist, it does not contribute anything. The next one before the last one also does not exist and not contribute. But somehow some nodes manage to contribute enough to obtain the result 2 after all. If we exchange infinitely many terms of the geometric series, its sum remains 2, because it is absolutely converging. As the geometric series contains only the smallest possible infinity of terms, it should not cause problems to read it from behind. > > > > > > 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. Consider all of them simultaneously. You like to do so when Cantor's diagonal proof is concerned. Every exchanged digit is followed by infinitely many digits which have to be exchanged. Nevertheless you say, it is possible to exchange all of them at one time. Consider the paths o he tree - all at the same time. This is the big mistake of set theory. The unreasonable allowance or prohibition of instantaneously possible actions. Cantor: yes. Tree: no. Why? > > > > > > > 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. That is wrong, because none of these paths ever gets isolated. Is it similar to the quarks? Path-confinement? We can safely state that at no level there are uncountably many separated paths. As every node is in the tree ad no node is outside, there are at no place in the world uncountably many numbers. > > > > > > 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. Wrong. Every number has digits only at finite distance from the decimal point. In the tree there is no level "infinite", but all the infinitely many levels are there - each one in a finite distance from the root node. > > > > > 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". At no point of the tree (and of the universe) more than countably many path can be distinguished. Regards, WM |