Cantor Confusion
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From: mueckenh on 15 Jan 2007 08:45 Dik T. Winter schrieb: > > > 1 1 > > > / \ / \ > > > / \ / \ > > > / \ / \ > > > 2 3 2 3 > > > / \ / \ / \ / \ > > > 4 5 6 7 > > > / \ / \ / \ / \ > But with this interpretation the union of the sets of paths is *not* the > set of paths in the tree on the right. The set of paths in the tree on > the right is missing the path 1-2-L for example, which should be in the > union. Actually the set of paths on the right misses *all* paths from > the tree on the left. So when you state that there are only 4 paths in > the tree on the left and 8 in the tree on the right, I retract my > remark. In that case the union of the sets of paths has 12 elements. In the finite trees there are ony initial segments of the infinite paths of the infinite tree, because every path representing a real number is the limit of a sequence of finite paths. In the infinite union of all finite trees there are all sequences - countably many. If there are not more limits than sequences, then the limts are countable too. > > > > > If you agree, there is however one problem with the complete infinite tree. > > > *None* of the constituent threes contains an infinite path, and so the > > > union of the sets of paths also does not contain an infinite path. > > > > None of the finite initial segments of N contains N. Is there a problem > > with N? > > If no element of the constituent sets in a union has property A, also there > is no element in the union has property A. > No path in the sets of paths is infinite, so also no path in the union is > infinite. No number in the initial segments is infinite, so also no number > in the union is infinite. No initial segment of Cantor's diagonal number is infinite. Neverthelss there is an infinite diagonal number? > > Pray revise how union is defined. I think I quoted the definition quite > proper. It *is* required. If an element a is in the union of a collection > of sets, it *must* be in at least one set from the collection. The limit of a sequence is not an element of the sequence. > > > Are you stating that the Peano axioms are nonsense because there is no > set that satisfies the axioms? Or what else? If there actually is a set satisfying these axioms, then its number omega of members is not in the set as an element. An infinite number is created by a set of finite numbers or segments (as the limit). Then also an infinite path with omega nodes can be created by a set of finite segments of paths. > > > > Simply because there is no successor of the largest element. Now remove > > > that largest element. If there is still a largest element, the Peano > > > axioms can not be satisfied. However, the axiom of infinity states that > > > there is at least one set that satisfies the Peano axioms, and that N is > > > the smallest of those sets. You are claiming that there is no such set. > > > > You claim it for the set of paths. > > You are in confusion again. I do not claim that there is a largest set of > paths. > > > I think I was talking about sets of paths. > > > > representing finite and infinite sequences of numerals of real numbers. > > What is the relevance? The relevance is: If the union of all nodes of a path is not an infinite path, then the union of all digits of Cantor's diagonal is not a real number. Regards, WM
From: mueckenh on 15 Jan 2007 08:57 Dik T. Winter schrieb: > > See my recent posting. You need to assume that the complete union of > > levels of the binary tree does not contain all the paths representing > > real numbers. Therefoer something must be added, in order to get the > > complete set of paths. But there is nothing to be added - but ghosts. > > Oh. But depends on what is done. But even if you unite two finite levels, > the set of paths in the union is not the union of the sets of paths of > the separate levels. Depends on definition of "finite tree". If all finite paths are continued with sequences of 000... or 111..., then the union of paths of two finite trees is the set of paths of the larger tree. > > > > > All path the bits of which can be indexed by natural numbers are in the > > > > union of all finite trees. > > > > > > But *not* in the union of the sets of paths in the finite tree. > > > > Yes. The union of all finite trees is not an (actually) infinite tree. > > See my due explanations of the union of all lines of the EIT: > > And again you *fail* to argue with the union of sets of paths in mind. If you unite all finite trees (finirte paths continued by 000... and 111...), then you unite all possible nodes, you unite all possible edges, but you don't unite all possible paths? What is the difference between the union of all finite trees and the infinite tree in terms of nodes? If there is a difference, then it can exist also in Cantor's list. As the digit a_nn has always a finite distance from the first line it has also always a finite distance from the first column. You never cover the complete list. > > There are infinite paths in the union of the trees. There is *no* infinite > path in the union of the sets of paths. Those two things are different. The infinite paths are the unions (limits). > > > > > > Ok, so you do *not* use the union of the sets of paths but > > > > > something else. > > > > > > > > The material from which the paths are constructed. > > > > > > Makes no sense. > > > > But is fact. > > How than can you draw conclusions about the cardinality of the sets of > paths in the union from the cardinalities of the sets of paths in the > finite trees? A union cannot have more elements than are united. A set of sequencs cannot have more limits than there are sequences. Regards, WM
From: mueckenh on 15 Jan 2007 09:10 Dik T. Winter schrieb: > In article <1168533543.556066.104940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > But you do not see this small difference. You > > > > claim that the union of all finite numbers or of all finite initial > > > > segments {1,2,3,...,n} is the infinite set {1,2,3,...}. > > > > > > Elaborate on the difference. > > > > Why do you see it different in case of the union of all finite trees or > > paths? > > I am not talking about unions of paths. I am talking about unions of > *sets* of paths. That is something completely different. But it is quite irrelevant, because every path of the set of infinit paths is a union of finite segments of paths. There are only countably many sequences. > > > > > Therefore you > > > > must add, in addition to all finite segments of that very infinite > > > > path, something which you cannot describe. Something, however, which is > > > > certainly not a finite segment of that path. > > > > > > I think I was talking about sets of paths. > > > > representing finite and infinite sequences of numerals of real numbers. > > But the sets of the paths in the finite trees do *not* contain infinite > sequences, and so the union of those sets can not contain infinite > sequences. The union of all finite trees and the complete infinite tree are identical with respect to nodes and edges, but differ with respect to paths? Could you please specify the asserted difference concerning the paths p in terms of their definition? The paths are: p = Sequence (a_n) with n in N and a_n in {0, 1} while the real numbers in general are represented by: r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. Identical representations imply identical numbers. Regards, WM
From: mueckenh on 15 Jan 2007 09:20 Franziska Neugebauer schrieb: > >>> An infinite union of finite sets is an infinite object. > > > >> But that object does *not* have an infinite _member_. The union of all finite trees contains all terms of the sequence (a_n) with n in N and a_n in {0, 1} representing t real number r. The infinite path is the limit. Are there more limits than sequences? > > > > So your question "Why should it?" is easily answered: It should not, but > as Virgil has pointed out: "you invalidly assume an infinite [...] > union of finite sets must contain an infinite object". The union of all finite trees and the complete infinite tree are identical with respect to nodes and edges, but differ with respect to paths? Could you please specify the asserted difference concerning the paths p in terms of their definition? The paths are: p = Sequence (a_n) with n in N and a_n in {0, 1} while the real numbers in general are represented by: r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. Identical representations imply identical numbers. Regards, WM
From: mueckenh on 15 Jan 2007 09:25
Virgil schrieb: > In article <1168779322.952374.232460(a)q2g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > Which ones of those specifically "rooted finite paths" does WM claim > > > contain infinitely many nodes? > > > > Which segment {1,2,3,...,n} of N contains infinitely many elements? > > Is N the union of all initial segments? > > If one makes a claim about finite initial segments of N, that claim need > not be true for anything that is not a finite initial segment of N. > > And N is not a finite initial segment of N. And an infinite path is not a finite path. > > > > > > > Each of the sets {1}, {1,2}, {1,2,3}, etc., is both a member and a > > > subset of N, but N is not. > > > > And, in addition, it is not green. Yes. Relevance? > > So that WM claims every property common to all finite initial segments > of N must also be a property of N? How about finiteness? Why then should the finity of all the paths in the finite trees exclude the infinity of the paths of the union of all paths / trees? > > > > > > Even the diagonal of the EIT is no longer infinite, if it is > > > > inconvenient for set theory? > > > > > > As properly described, that diagonal is actually infinite however > > > inconvenient that may be for WM's version of set theory. > > > > This diagonal (infinite path) is the union of all digits (nodes) of the > > initial segments and the limit of the sequence of the initial segments. > > The union of all finite initial segments of N is not a finite initial > segment of N. The union of all finite trees is not a finite tree. An infinite tree, however, does not contain finite paths! Regards, WM |