Cantor Confusion
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From: Franziska Neugebauer on 9 Jan 2007 08:42 mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > >> > > A: L_D is a line, therefore L_D has a largest >> > > element. >> > > >> > > B: L_D contains any element that can be shown to exist >> > > in the diagonal, therefore L_D does not have a largest >> > > element. >> > > >> > > Contradiction. Therefore L_D does not exist. >> > >> > Therefore the union of line ends including the end of L_D does not >> > exist. >> >> The union of line ends is not a line. > > But it is supposed to exists as the complete, finished diagonal. Straw man. The diagonal is the ("potential infinite") diagonal. Every existing element of the diagonal is the end of a line. Nobody but you supposes that the diagonal "exists as complete". >> L_D does not exist >> so the union of line ends including the end of L_D does not exist. > > Therefre a complete, finished diagonal does not exist. As far as I can see is the diagonal "only potentially infinite". Nobody claimed that it is "complete". Nonetheless, the conclusion that a single line L_D does not exist remains valid. >> > Fine. There are no infinite sets. There is only potential >> > infinity. Every line including the diagonal >> >> The diagonal is not a line. The diagonal is a potentially infinite >> set. > > Correct. Therefore it does not contain all natural numbers (as > indexes) because then it would be actually infinite. Please pay attention to your wording! The diagonal _contains_ numbers (values). Not "as indexes", but as value of the diagonal _at_ a certain index. So your statement is rather meaningless. >> A line is not a potentially infinite set > > Correct. A line is or is not. But the system of lines is potentially > infinite. Since the system of lines is not a line your remark is meaningless. >> > has a greatest element >> > until the existence of a greater one is shown. >> >> However, and contrary to your repeated claim, there is no >> single line which contains every element that can be shown >> to be in the diagonal. > > Every element that can be shown to be in the diagonal does exist as > the end of a line. Yes? > > But not every end of a line does belong to a line? No one has claimed that. > That is a surprising aspect of existence. Your effort to not understand by twisting the words is amazing. > By showing that an element of the diagonal exists, you show that the > line contaning it and every smaller element exists. No one doubts this. It is obviously WM who has problems identifying the issue of this subthread. F. N. -- xyz
From: Franziska Neugebauer on 9 Jan 2007 08:49 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: >> In article <1168271417.069985.286650(a)51g2000cwl.googlegroups.com>, >> mueckenh(a)rz.fh-augsburg.de wrote: >> > William Hughes schrieb: >> > > >> > > That L_D does not exist. >> > > >> > > Assume a line that contains any element that can >> > > be shown to be in the diagonal exists. >> > >> > Otherwise the diagonal could not exist, because it is, by >> > definition, the union of line ends in the EIT. >> > >> > > Call >> > > this line L_D. >> > > >> > > A: L_D is a line, therefore L_D has a largest >> > > element. >> > > >> > > B: L_D contains any element that can be shown to exist >> > > in the diagonal, therefore L_D does not have a largest >> > > element. >> > > >> > > Contradiction. Therefore L_D does not exist. >> > >> > Therefore the union of line ends including the end of L_D does not >> > exist. >> >> It is any L_D which can not exist since it would be required to have >> both a last member and also the successsor to its last member. >> >> >> Fine. There are no infinite sets. There is only potential >> > infinity. Every line including the diagonal >> >> The diagonal, by not having a last member, is not a line, all of >> which, by definition, must have last members. > > But the diagonal consists of last members. This "argument" does not at all support a "refutation" of Virgil's statement. You have reversed the quantifiers again. > If there was no line for each member of the diagonal, which included > this and every smaller member, then the diagonal would not exist. Since there is a line for each member of the diagonal the ("potential infinite") existence of the diagonal is left untouched. > Therefore, the diagonal contains only such elements with indexes > {1,2,3,...n} which are simultaneously in a line - in one line. Is "n" denoting a bounded or an unbounded variable? > All together! Un pour tous, tous pour un? F. N. -- xyz
From: mueckenh on 9 Jan 2007 09:01 Dik T. Winter schrieb: > > What is a ghost path? A ghost path is a path which does not exist in a tree which contains the paths of all real numbers the bits of which can be indexed by natural numbers. > > > 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. All path the bits of which can be indexed by natural numbers are in the union of all finite trees. You know it. You cannot point to a bit of any number of [0, 1] which is missing. But you claim there were further numbers. Try to complete the union of all finite trees to the complete infinite tree. Which level, node or edge would you add? > > Why was it not a proof? Because you used crippled plants which are not under discussion in a forest of well grown trees. > > > > 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? Which edge of any infinite path is missing? Which bit could be added to one of the numbers represented? > > Ok, so you do *not* use the union of the sets of paths but something else. The material from which the paths are constructed. > 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. The completeness of U{n e N} {1,2,3,...n} = N is the same as the completeness of the union of all levels of the infinite tree, and that is the same as the union of all finite trees. And all these unions are counatble. > > 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. You can also unify the levels. > 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). It is sufficient to build the union of all finite trees, or of all levels, or of all edges, or of all nodes. This union is an infinite tree. An infinite tree cannot be surpassed (by another tree with naturally indexed levels). Therefore the union is *the* infinite tree and, hence, it contains all possible paths. Regards, WM
From: mueckenh on 9 Jan 2007 09:12 Dik T. Winter schrieb: > > > > > 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. But, alas, this construction cannot be communicated. It is and remains a top secret. Otherwise most mathematicians could easily be proved wrong by simply constructing a well-ordering of R. > > > > 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? For example: the infinite union of {0.01} and {0.010} and {0.0101} ... yields the path corresponding to 1/3. > > > 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. If the tree is complete with respect to edges, then it cannot be completed any further. > > > > > 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? If the tree is complete with respect to edges, then it cannot be completed any further. > > > 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"? The paths are representations of real numbers. Regards, WM
From: Franziska Neugebauer on 9 Jan 2007 09:14
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > The finite union of two or more finite trees is a finite tree. An >> > infinite union of finite trees is the infinite tree. >> >> What is an infinite union? (Please give a definition) > > An infinite union of trees is the union of all (or nearly all) finite > trees which reach from level 0 to level n where n eps N. 1. Please give a definition of "union of all finite trees". 2. Please give a definition of "union nearly all finite trees". F. N. -- xyz |