Cantor Confusion
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From: Dik T. Winter on 8 Jan 2007 19:29 In article <1168291438.929439.77360(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > > > Name a level or a node or and edge which are in one of the trees > > > > > only. > > > > > > > > Why should I do that? > > > > > > In order to prove that there is evidence for at least one path being > > > not in both trees. > > > > But for that I do not need to mention a node or edge, I only need to > > mention a path. > > Unless you can mention a node or edge, your path may be a ghost. What is a ghost path? > The union of all finite trees is also the union of all last levels. > That is the union of all levels. If some path is missing, then it must > stretch over more than all levels indexed by natural numbers. Ah, again about that diagonal. No, you are wrong. > > > > I can name a *path* that is in the complete infinite > > > > tree but not in the union of finite trees. > > > > > > Without different nodes that is an empty assertion. > > > > Well, I have shown it by proof. So it is not an empty assertion. > > That was not a proof. Why was it not a proof? > > The union of the sets of paths is an infinite set of paths. But there is > > no infinite element in there. > > The union of all finite initial segments (paths) is not an infinite > segment (path)? I am talking about the union of *sets* of paths, not the union of paths. Do you not see the difference? How do you define the union of paths? And how do you define the union of the *sets* of paths that are in the finite trees? > > > The infinite paths in the union of trees are the unions of the > > > respective finite paths. So much should be clear to every reader. > > > > What would be clear to every reader is that you are confusing issues. > > Unions are about sets. A tree has a set of nodes, a set of edges and > > a set of paths. When we do a union of trees we can do a union of those > > three sets for each tree. > > A tree is determined by its set of nodes as well as by its set of > edges. > Trees with same nodes and edges are identical with respect to paths. Ok, so you do *not* use the union of the sets of paths but something else. Indeed, you actually do not use those sets at all in forming your union. Nevertheless you want to draw conclusions about the cardinality of the *set* of paths in that union from the cardinalities of the *sets* of paths in the finite trees. > > Unless you define union completely different > > (and if so, pray give a definition) we can not use unions of elements > > of those sets. So you can not take the union of paths but only the > > union of sets of paths. On the other hand, once you define your union > > such that we also must consider unions of individual paths, I also want to > > consider unions of individual edges. Can I? If not, why not? > > The union is easily defined: Every finite tree (up to level n) contains > the union of all smaller trees (up to levels n - 1) and its own level > n. That is not a definition. How do you *define* the union of two trees? I saw a tree as consisting of three sets: E, N and P for edges, nodes and paths. And I defined the union of two trees T1 and T2 as: [ {E1, E2}, {N1, N2}, {P1, P2} ] (where those unions are sets, so duplicated elements can be elided). -- 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 8 Jan 2007 19:40 In article <1168291967.835156.121790(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > But in that case you have to properly define what you are doing. So pray > > tell, how do you *define* the union of trees. > > See my last posting. > > > > > You are dense. The well-ordering is given under the conditions of the > > > > axiom V=L. > > > > > > THEN GIVE IT PLEASE, HERE AND NOW ! Simply reproduce it. > > > > I have not looked into it, but the axiom V=L immediately gives a > > well-ordering, not only of R. > > Empty assertion! > > > > > If the axiom is false the well-ordering is not given. > > > > > > How could it disappear after having existed? Who would force us to > > > forget it, if it had been given? > > > > Did you look at that axiom? No, I think not. > > Of course I did not. The axiom may state the existence of a > well-ordering but does not construct or define it (because it cannot be > constructed or defined) No, that does AC (although the statement is indirectly). V=L is the axiom of constructibility and states that every set is constructible. Most mathematicians think it is false, but it can not be disproven and is "consistent" with ZF. It *gives* a construction. > > > What counts is only: There cannot be more unions than elements. > > > > Eh? As I understand it, there is only one union. But, what is the > > relevance? > > > There is a union of paths for every path. How do you define a union of paths? > There is a union of trees yielding the infinite tree. How do you define a union of trees? > Every finite tree reaches from the root to a level n. > The union contains every level which is a natural number. I have no idea, because I have not yet seen a proper definition. > Every node which is placed on a finite level (= every digit with a > finite index) is in the union of all finite trees. Therefore every path > containing nodes on a finite level (= every sequence of digits at > places with finite indexes) is in the union of finite trees. There is > nothing remaining! But that does *not* mean that the set of paths is the union of the sets of paths in the finite trees. > > > I have a union of all finite trees. The result is an infinite tree > > > (which need not be an element of the union). > > > > But you are talking about the union of sets of paths. What is the > > relevance? > > An infinite tree contains all possible paths. The uninon of all finite > trees is the infinite tree. Other infinite trees are not available. > There is no real number available which is not represented in the > union of all finte trees. You are again, *not* talking about the union of sets of paths. So what is the relevance? > Nevertheless, the representations in the union of all finite trees are > countable as the countable union of finite sets. A new term again. What are "the representations"? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Tony Orlow on 8 Jan 2007 20:07 Virgil wrote: > In article <45a25d13(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> >>> A number is not a digit string. The fact that there are no rationals >>> between two reals constitutes proof that those reals are identical. >>> See the reference above for an explanation of Dedekind cuts. >>> >>> >> What if one of those reals is a rational itself? Is there necessarily a >> rational between every rational and every real, or are there irrational >> reals "adjacent" to rational reals? > > Given any dense subset of the reals, the absence of a member of that set > between two supposedly different reals, regardless of the nature of > those reals, proves them equal. > > And in dense sets, such as the reals or rationals, there is no such > thing as "adjacency". So, there are spaces between the reals?
From: William Hughes on 8 Jan 2007 20:29 Tony Orlow wrote: > Virgil wrote: > > In article <45a25d13(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> > >>> A number is not a digit string. The fact that there are no rationals > >>> between two reals constitutes proof that those reals are identical. > >>> See the reference above for an explanation of Dedekind cuts. > >>> > >>> > >> What if one of those reals is a rational itself? Is there necessarily a > >> rational between every rational and every real, or are there irrational > >> reals "adjacent" to rational reals? > > > > Given any dense subset of the reals, the absence of a member of that set > > between two supposedly different reals, regardless of the nature of > > those reals, proves them equal. > > > > And in dense sets, such as the reals or rationals, there is no such > > thing as "adjacency". > > So, there are spaces between the reals? No. There is no last real that is less than 1.0 So there is no real that is adjacent to 1.0 Despite this, there is no interval [1.0-epsilon,1.0] that does not contain an infinite number of reals. - William Hughes
From: MoeBlee on 8 Jan 2007 20:45
Tony Orlow wrote: > Virgil wrote: > > And in dense sets, such as the reals or rationals, there is no such > > thing as "adjacency". > > So, there are spaces between the reals? In what theory? In Z set theory? If, in Z set theory, you define 'space between', then we can address the question of whether there is a space between any two real numbers or whatever question it is that you wish to ask in regard to 'space between'. MoeBlee |