Cantor Confusion
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From: mueckenh on 15 Jan 2007 09:32 Virgil schrieb: > In article <1168779834.276761.153130(a)a75g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > Hardly. Which digit marks an infinite initial segment? > If the diagonal construction did not produce an endless string > of digits, it would not produce a number known not to be in the list, > so finite initial segments don't cut it. If the union of finite trees (as defined) did not produce an infinite tree, then a natural number n would exist, which would not label a level in the union of trees. Such a number n does not exist. So the union of the trees is an infinite tree. An infinite tree has no finite path. Every infinite binary path is in this infinite binary tree. > > > Either it does not exist - or there are infinite paths in the union of > > finite trees. > > False dichotomy. Arguments? Regards, WM
From: mueckenh on 15 Jan 2007 09:42 Franziska Neugebauer schrieb: > >> Recall: Do get it out you must have put it in. > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > Either it does not exist - or there are infinite paths in the union of > > finite trees. > > False dilemma. It "consists" of only finite initial segments (not > finitely many therof, of course) _and_ it does exist _and_ there are > only _finite_ paths _as_ _members_ in the union of all finite trees. The union of all finite trees and the complete infinite tree are identical with respect to nodes and edges, but note 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 are represented by: r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > > Again: To get it out you must have put it in. Since noone put infinitely > deep trees into the union of trees it does not have such paths as > members. But as limits. I cannot believe that countable many paths can have uncountably many limits. > > >> [...] > >> >> 2. To "unary represent" every real in [0, 1] > >> > > >> > You cannot "unary represent" every real in [0, 1]. > >> > >> It's rather unimportant whether I "_can_" do so. > > > > It is important that it is in principle impossible. > > A mathematical argument is not that "I 'cannot' do this or that" That is clear, but I wanted to be polite. > but > "that no unary representation of every real in [0, 1]" exists. Why then did you say you could it? > > > > > Obviuosly yoou intermingle "unary" and "binary". > > Indeed I mean "binary represenation". To resume: If your set of binary > representations does not contain the representation of 1/3 your proof > is meaningless already before starting to write it down. The representation of 1/3 is in the union of all finite trees. It is the limit of a set of finite paths like N is the limit of all of its finite initial segments. Regars, WM
From: Dik T. Winter on 15 Jan 2007 10:52 In article <1168780521.340713.316740(a)51g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > A natural number n can be represented in a special unary notation: n = > > > 0.111...1 with n digits 1 (the leading 0. playing no role). Example: 1 > > > = 0.1, 2 = 0.11, 3 = 0.111, ... > > > In this notation the definition of the set of natural numbers, (1, 2, > > > 3, ...} = N, reads > > > > > > {0.1, 0.11, 0.111, ...} = 0.111.... (*) > > > > So you are now defining here 0.111... as the set of natural numbers. > > A pretty amzing notation, but let it stand. > > > > > Note that also the union of all finite initial segments of N, {1, 2, 3, > > > ..., n}, is N = {1, 2, 3, ...}. Therefore (*) can also be interpreted > > > as union of initial seqments of the real number 0.111.... > > > > But 0.111... is not a real number, you just defined it as the set of > > natural numbers! > > 0.111... and its initial segments are *representations* (according to > you, they cannot be numbers). They can represent natural numbers as > well as certain real numbers. Obviously both sets of numbers are > isomorphic. What is the isomorphism? And 0.111... does *not* represent a natural number in unary notation. Do you claim there is an isomorphism between the set of natural numbers and the set of real numbers: {1/2, 3/4, 7/8, ..., 1/9} That is false. There is a bijection, but that bijection does *not* work through the representation. > > But I think I understand what you are doing. You are > > creating sets of natural numbers based on binary expansions. And so > > 0.010101... would map to the set of even natural numbers. > > This is not required. Only initial segments of N and of 0.111... are > required. I am talking about the path 0.01010101... which you claim is also in the complete tree. > > I have two > > objections to this. > > (1) There is no difference between the mapping of 0.1 and 0.10. > > No problem. > > > (2) A single real number can map to two different sets of natural > > numbers. Consider 1/2 = 0.100000... = 0.0111... . > > Yes, some real numbers can do. Therefore there are more paths in the > tree than real numbers in [0, 1]. Well, there is a bijection between them, but *not* the natural bijection you get through the representtion. > > > Taking the uninion of all finite binary trees, we get the complete > > > infinite binary tree with all levels. All infinite paths representing > > > real numbers r of the real interval [0, 1] are in this union. We can > > > see this by the path always turning right, 0.111..., which is present > > > in the tree, according to (*). > > > > But the *set* of paths in the union is not the union of the *sets* of > > paths in the finite trees. > > When all nodes are there , then all paths are there. All paths are in the complete tree, but *not* in the union of the sets of paths in the finite trees, otherwise for each path in the complete tree there *must* be a finite tree that contains it. > Or the set of digits of Cantor's diagonal is not an infinite number. I see no connection. > > > Conclusion: Every finite binary tree contains a finite set of path. > > > > Right. > > > > > The > > > countable union of finite sets is countable. > > > > RIght again. > > > > > The set of paths is > > > countable. > > > > Wrong. The union of the set of paths of the finite trees is *not* the > > set of paths in the union of the finite trees. You can not get the > > union of a sequence of sets of paths by taking the union of individual > > paths. > > I take the union of all nodes. That guarantees the presence of all > paths. But in that case you are not talking about the union of finite sets of paths. That is my objection to your reasoning above. You still assume that the union of the sets of paths of the finite trees *is* the set of paths in the complete tree. That is *false*. The complete tree contains the path 0.010101... . If that path is not in *any* of the sets of paths in the finite trees, that path is also not in their union. So while the countable union of the sets of paths in finite trees is countable, that is *not* the set of paths in the complete tree. > Or the set of digits of Cantor's diagonal is not an infinite number. I see no connection. > > Consider the following finite trees (I see now that in finite trees you > > terminate paths with a node). Look: > > 0 0 > > / \ / \ > > / \ / \ > > 1 2 1 2 > > / \ / \ > > 3 4 5 6 > > > > On the left-hand side we have two paths, so the set of paths is: > > {0-1, 0-2} > > and on the right-hand side we have: > > {0-1-3, 0-1-4, 0-2-5, 0-2-6} > > the union of these two sets of paths is: > > {0-1, 0-2, 0-1-3, 0-1-4, 0-2-5, 0-2-6}. > > And so is the set of paths in the tree on the right-hand side not equal to > > the union of the sets of paths in both trees. > > But every path of the left tree is contained as an initial segment in a > path of the right tree. Yes. I see no problem. Do you agree that the *union* of the sets of paths is *not* the set of paths in the right-hand tree? You are talking about unions of sets of paths, segments have nothing to do with such. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Jan 2007 11:07 In article <1168869430.273702.199810(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Wrong. You can name the paths anything you like, that does not make it right. At level 2 the finite tree has only paths that go through two nodes. At level 3 the finite tree has only paths that go through three nodes. In the union of the sets of paths there are both paths that go through two nodes and paths that go through three nodes. They are distinct. > > > 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? See above. > What is the difference > between the union of all finite trees and the infinite tree in terms of > nodes? See above. And ultimately, in the union of the sets of paths of all finite trees, there are only paths that go through finitely many nodes. There is *no* path that goes through infinitely many 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. That is something different. Above we are talking about union. Here you are talking about limit. > > 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). Perhaps. But when you talk about unions of sets of paths you are not takling about union of paths. Those are different concepts. > > 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. Indeed. But the union is not the complete collection. > A set of sequencs cannot have more limits than there are sequences. Again switching to limit from union. The two concepts are different. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Jan 2007 11:02
In article <1168868712.421404.294530(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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, Sorry, you were talking about unions of sets of paths. Segments (initial or not) have nothing to do with that. Unions work only on elements, not on parts of elements, and the elements of the sets are paths, not segments. > > > 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? The diagonal number is not infinite. > > 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. The limit of a sequence is *not* the union of a sequence. Union only works for sets. The union of a collection of sets is precisely defined. An element is in the union if and only if there is *at least* one set in the collection that contains that element. In the union there is *no* limit involved. > > 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). Yes, but it is not in the set. And it is *not* the limit, at least, in mathematics it is in general not used as such. > Then also an infinite path with omega nodes can be created by a set of > finite segments of paths. As a limit, yes. But that does still tell us nothing about unions of sets of paths. When you create a union of two sets you take all elements from every set and remove the duplicates. > > > > 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. But I am not stating that. I am stating that "the union of the sets of paths in the finite trees is not the set of paths in the infinite tree". There is no consideration needed here about nodes, edges and whatever. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |