Cantor Confusion
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From: Dik T. Winter on 16 Jan 2007 06:52 In article <JBxywy.12K(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <MPG.20160647a0a74784989b3d(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > > Dik T. Winter wrote: > > > In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: > > > ... > > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > > > > this would imply that f(0) = 0? > > > > > > No (assuming any eta > 0). It would only imply that lim{x->0} f(x) = 0. > > > > ?? > > It appears to be clear to me. When the eta are != 0, you can say > nothing about f(0). Argh. I must haev been asleep. Of course it tell nothing about the limit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 16 Jan 2007 08:36 Dik T. Winter schrieb: > > 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. Neither does omega. An isomorphism between two ordered sets (P, <1) and (Q, <2) is a one-to-one function h with domain P and range Q such that for all p1, p2 in P: p1 < p2 if and only if h(p1) < h(p2). We have h(1) = 0.1 h(2) = 0.11 .... h(n) = 0.111...1 with n numerals "1". > > I am talking about the path 0.01010101... which you claim is also in the > complete tree. I every node is in the tree, then every node of 0.01010101... is in the tree too. Or is this assertion false? > > > > > When all nodes are there , then all paths are there. > > All paths are in the complete tree, But the union of all finite trees is not 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. Cantor's diagonal can be represented as a (diagonal) path in an infinite matrix. This matrix is the union of finite matrices. It is the same as with the trees. > > > > > 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. If there are all nodes then there are all paths. If there are all combinations of digits (or bits) then there is nothing remaining for a separate number 0.010101... > > > > 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. I am talking about the union of all nodes. The presence of all paths in the union of all nodes is obvious. You see it if you try to find out which node of the path 0.010101... may be missing in the union of trees but may be present in the complete tree. Regards, WM
From: mueckenh on 16 Jan 2007 08:47 Dik T. Winter schrieb: > 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. I meant the following: Accordng to my original definition, the tree from level 0 to level 1: 0. 0 1 has the paths 0.0000..., 0.0111..., 0.1000..., and 0.1111. (the tails 000... and 111... are always appended.) This set of four paths is also in the tree from level 0 to level 2 and in all greater trees. But this is irrelevant. We do not unite sets of paths. We unite sets of nodes. The completeness of a tree is proved by the complete set of nodes. If there are all nodes, then there are all paths. > > 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. In the union of all finite squares (11) (11), (12) (21), (22) etc. there is no infinite diagonal. > > > 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. Why do you refuse to talk about the limit in case of paths? > > > > 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. The union of finite trees contains all (initial) sequences (of paths). Their number is countable. > > > A set of sequencs cannot have more limits than there are sequences. > > Again switching to limit from union. The two concepts are different. Feel free to take what you think is necessary to abolish set theory. Regards, WM
From: mueckenh on 16 Jan 2007 08:57 Dik T. Winter schrieb: > > No initial segment of Cantor's diagonal number is infinite. Neverthelss > > there is an infinite diagonal number? > > The diagonal number is not infinite. > Not wih respect to the hight of its digits and not with respect to the value of the number, but it is infinite with respect to its number of digits. > > > 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. Yes, for the sets of nodes for instance. And for the sets of paths if each path is appended by 000... and 111... as I described in my last constribution. > > 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. We an do that with nodes and with paths (as originally defined). > > > > > > 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". Please use my original definition yielding infinite paths in finite trees. > There is no consideration needed here about nodes, edges and whatever. Paths cannot exist where no nodes are. Therefore paths cannot exist (partially) outside of the tree containing all nodes. Regards, WM
From: mueckenh on 16 Jan 2007 09:02
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > 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? > > Given an infinite (in your terms potentially infinite) sequence of > digits > (e.g Cantor's diagonal), there is one initial segment that is > not "marked by a digit". This is the sequence (in your terms > potentially infinite sequence) of all digits. Given the sequence of initial segments of binary 1/3: 0.0 0.01 0.010 0.0101 .... there is one element not marked by a digit, i.e., not contained in a finite binary tree but contained in the union of all finite binary trees. > Thus one cannot say > Cantor's diagonal "consists only of finite initial segments". One must say so. It is similar to saying N consists of finite numbers only. Regards, WM |