Cantor Confusion
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From: David Marcus on 8 Jan 2007 22:10 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? It is quite clear that you don't know any analysis or advanced calculus. You might wish to learn some. Math can be very interesting. Learning some would be a better use of your time than making clueless posts. -- David Marcus
From: Virgil on 9 Jan 2007 00:36 In article <1168291967.835156.121790(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > There is a union of paths for every path. There is no necessity that in two trees any node or edge or path be in both. > There is a union of trees yielding the infinite tree. Not without some serious assumptions about those trees which WM has not stated. In particular, one must at least presume that given any two trees in the union there is a tree containing both as subtrees. However there is on need in general for two trees to even have any nodes or edges in common. > Every finite tree reaches from the root to a level n. But may be totally disjoint from every other tree. > The union contains every level which is a natural number. As levels of disjoint trees. > > An infinite tree contains all possible paths. The uninon of all finite > trees is the infinite tree. Not necessarily. The union of all finite trees may be just a union of infinitely many totally disjoint trees. > Other infinite trees are not available. They are to me. > There is no real number available which is not represented in the > union of all finte trees. Actually, in the infinite union of disjoint finite trees, only terminating binaries are represented, that is not even all of the rationals in [0,1]. > > Nevertheless, the representations in the union of all finite trees are > countable as the countable union of finite sets. Not if that union is supposed to contain an infinite tree.
From: Virgil on 9 Jan 2007 00:41 In article <MPG.200ca367469c7d49989aae(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > In article <1168290342.751050.317770(a)51g2000cwl.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > No one, except possibly WM, is claiming that an infinite binary > > > > > > tree is > > > > > > the same as the union of all binary rational, and therefore finite, > > > > > > trees. > > > > > > > > > > > What else could it be? > > > > > > > > It could be, and is, the union of infinitely many finite trees. > > > > Since the union of two trees is not a tree, whyever should the union of > > > > more than two trees be a tree? > > > > > > The union of two trees is a tree. > > > > Then it will have nodes which are simultaneoulsy terminal nodes and not > > terminal nodes. One can embed one tree into a larger tree, but that > > operation is not union. > > Obviously, it is impossible to have a sensible discussion with WM. > However, I think the nonsense discussion will be easier for everyone to > follow if you and Dik assume that when WM says "union", he really means > to have all the finite trees be subsets of the infinite tree. If we do > it this way, then all the nodes and edges in the sequence of finite > trees are also in the infinite tree, but there are paths in the infinite > tree that do not appear in the finite trees. I would be glad to concede that the "union" of a suitably infinite set of finite subtrees of an infinite tree will form an infinite tree, but WM keeps denying that that is what he means. He seems to thing that all trees must be subtrees of some ur-tree, which is distinctly not the case. Disjoint trees are quite possible, and then their unions are not trees at all.
From: Virgil on 9 Jan 2007 00:45 In article <45a2ead2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> 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? The question is ambiguous. If TO is asking whether there is space between any two given reals for other reals to fit into, then yes. If TO is asking whether the entire set of reals has spaces for other objects to be fitted into by some repetition of the Dedekind cut or Cauchy sequence constructions, then no.
From: mueckenh on 9 Jan 2007 07:52
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. > 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. > > > > 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. > A line is not a potentially infinite set Correct. A line is or is not. But the system of lines is potentially infinite. > > > 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? That is a surprising aspect of existence. By showing that an element of the diagonal exists, you show that the line contaning it and every smaller element exists. Regards, WM |