Cantor Confusion
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From: Virgil on 8 Feb 2007 04:05 In article <1170923642.541518.218750(a)h3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Feb., 16:48, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 6 Feb., 13:47, Franziska Neugebauer <Franziska- > > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> mueck...(a)rz.fh-augsburg.de wrote: > > >> > It can be proven. {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > >> > contains {p(0), p(1), p(2), ... } > > > > >> What does contain mean? > > > > > The tree is a set of nodes. > > > > This is not true. A tree is "more" than a set of nodes. > > If the structure is defined separately, the only variable is n, the > number of levels. A tree is then a set of levels as well as a set of > paths as well as a set of nodes. This set of nodes can be enumerated > by a "chain" in the manner of Cantor's first diagonalization > > 0. > 12 > 6543 > 789... > > The tree is completely defined by one single natural number. Only after a lot of preliminary definitions, and absent those preliminaries, you have no trees at all. This > could be a number of the chain, but as we will not use trees with > different pathlengths, we will use only the number n of the last level > of the tree. Since the complete infinite binary tree does not have a last level, and no finite chain of nodes in it constitutes a path, that isn't good enough. > > Forget this usual definition. We need not define the structure because > we talk about one structure only, a complete binary tree. Which does not work at all in the the way WM's alleged finite trees work. > If you are > too inflexible to swallow this One would have to be more flexible than a snake in heat to swallow that garbage. > then call my tree as you like it, Can't, at least in public. > but > do me a favour and retranslate your choice to tree when responding to > my letter. Since your tree/trees are insensible, there is not translation possible from actual trees to WM-trees. > > > > 1. Only in Virgil's view of tree T it is true that a tree is a set of > > paths { p1, ... } . > > It is true in any case. Among others a tree can be viewed as a set of > paths. > > > > 2. A set s is a subset of a superset S iff > > > > A x (x e s -> x e S) > > > > 3. I interpret your conjecture "It contains paths as subsets." as > > "there is (at least) a path which is a subset of the tree", formally > > > > E p (p c T) (*) > > > > which means > > > > E p (A x (x e p -> x e T)) > > > > Assume now that there is some p = p' for which (*) holds. We have then > > > > A x(x e p' -> x e T) (**) > > > > Now the definition of path comes into play. As explained below the > > elements of p' are ordered pairs, i.e. have structure (number, node) or > > (number, edge). As one easily sees T only has paths as elements, i.e. > > has sequence structure { ..., (number, node), ... }. Therefore there is > > no (non-trivial) p' which is a subset of a tree. > > Froget this complicated picture. I does not lead ahead. WM would have us forget everything that he does not like merely because he does not like it. > > A path is an ordered set of nodes. But we need not use this order. If > we apply the chain sketched above, then there is a path [0, 1, 6, 7]. > This path is well defined by its nodes. The order is unimportant. > Therefore we can denote by {0, 1, 6, 7} and consider it a set of > nodes, a subset of the tree. If you do not agree with this simple > picture, please say where it becomes undefined. Once the set of edges and set of nodes are complete, one can derive other things. But by what rule is {0, 1, 6, 7} a path in any tree without a rule for deciding parent and child node of each edge? > > > > > The special meaning of "contain" is irrelevant as long as we refer to > > > a unique meaning, i.e., > > > either: something which is contained in the tree is really in the > > > tree, > > > or: something which is contained in the tree does belong to the tree > > > as a supremum. > > > > It is absolutely necessary to carefully distinguish between membership > > (element-relation "e") and the subset-relation ("c"). > > If the paths p(1), p(2) , p(3) ... (all outmost left hand side) are in > the union of all finite trees as members then the union of all finite > trees contains the path p(oo) (if this path exists). Except that Wm has said that no such unions can exist, for it they did, N would have to be a completed infinity without any infinite elements. > > X is not finite -> there must be an x in X which is infinite Delusion upon delusion! > > > > You do persistently refer to that axiom but you don't grasp that you do. > > I start off with the observation that *for even positive numbers* |{2, > 4, 6, ..., 2n}| < n. Further I assume that the set of all even > positive numbers contains nothing but even positive numbers. There is > no axiom involved up to now. The conclusion is that aleph0 is not a > number larger than every even positive number. It is a "conclusion" that, as usual, does not follow from your premises.
From: Virgil on 8 Feb 2007 04:09 In article <1170923985.245438.56170(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Feb., 16:59, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 6 Feb., 23:07, Virgil <vir...(a)comcast.net> wrote: > > > > >> > The basic way to establish IV c V is to use the numbers in their > > >> > basic > > >> > form IIII c IIIII. (Numbers *are* their representations.) > > > > >> Numerals are no more numbers than names are people. > > > > > Wrong! People can exist and do exist without names. Numbers cannot. > > > > Who has told you that? Whatever, M�ckenheim axiom 2: > > > > "Numbers cannot exist without name" > > If you take all names (in the widest sense, including symbols and > notations and defining equations) from a number and attach it to > another number, what remains with the first? If you remove from a person absolutely everything that makes him/her a person, is whatever remains still a person? Is a corpse a person?
From: mueckenh on 8 Feb 2007 04:15 On 7 Feb., 20:22, Virgil <vir...(a)comcast.net> wrote: > In article <1170852533.563009.176...(a)j27g2000cwj.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 6 Feb., 21:15, Virgil <vir...(a)comcast.net> wrote: > > > In article <1170756166.580698.67...(a)p10g2000cwp.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > On 4 Feb., 20:28, Virgil <vir...(a)comcast.net> wrote: > > > > > > > > Not so. Induction can only prove every set of naturals which is > > > > > > > bounded > > > > > > > above by a natural is finite, but that does not, according to the > > > > > > > axiom > > > > > > > of infinity, exhaust the possible sets of naturals. > > > > > > > There is no natural number in this set which is not covered by > > > > > > induction. So, which number is missing to exhaust N? > > > > It is not numbers but sets of naturals of which I spoke. > > > The empty set of natural numbers? I am only interested in sets of > > natural numbers which contain at least one natural number and contain > > nothing but natural numbers. They are covered by induction. > > One can show by induction that any set of naturals is a subset of "the > set of all naturals", but only if one allows that "the set of all > naturals" exists in the first place. > > And one cannot, by induction or any other method, show that every set of > natural numbers is finite without assuming it. Wrong. If a set is subject to complete induction, i.e., if n implies the existence of n+1, then the set of these numbers n is infinite. That is the definition of infinity. If you deny this then you should give an upper bound. You cannot. But you will not respond to this paragraph. No, the problem is that set modern theory is not satisfied with this simple form of infinity (because it would turn out self- contradictive). Therefore you need some other form, a more infinite infinity, even for the "smallest" infinity. > So that assuming what he wants to be able to prove has been WM's > technique from the start. > > Intuitionalists are at least open about their assumptions. WM is not My assumptions are that all natural numbers are finite and that for all even positive numbers |{2,4,6,...,2n}| < n and that all subsets of natural numbers contain natural numbers only so that every set of natural numbers can be exhausted by induction. There are no further assumptions. Which should be dropped? Regards, WM
From: mueckenh on 8 Feb 2007 04:16 On 7 Feb., 20:24, Virgil <vir...(a)comcast.net> wrote: > In article <1170852711.217577.234...(a)a75g2000cwd.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 6 Feb., 23:07, Virgil <vir...(a)comcast.net> wrote: > > > > > The basic way to establish IV c V is to use the numbers in their basic > > > > form IIII c IIIII. (Numbers *are* their representations.) > > > > Numerals are no more numbers than names are people. > > > Wrong! People can exist and do exist without names. Numbers cannot. > > It is, as usual WM who is Wrong! Most numbers do not have names. Which number exists without any name? Regards, WM
From: mueckenh on 8 Feb 2007 04:25
On 7 Feb., 21:24, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > On Feb 7, 4:46 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > There is an > > > > axiom which requires the existence of the non existing and seems to > > > > make some people happy > > > > If there is such an axiom anywhere, it is only an axiom in WM's system, > > > not in anyone else's. > > > > > (like the axiom which requires the finity of > > > > the infinite). > > > > Any such axiom exists only in WM's system, and not in anyone else's. > > > Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: > > "Foundations of Set Theory", 2nd edn., North Holland, Amsterdam > > (1984): "Intuitionists reject the very notion of an arbitrary > > sequence of integers, as denoting something finished and definite as > > illegitimate. Such a sequence is considered to be a growing object > > only and not a finished one." > > > Who considers it a finished one? > > Which particular formal intuitionistic axiomatization do you have in > mind? (There are intuitionistic formal systems, but I don't know in > particular which one you have in mind so that we can evaluate whatever > it is you think holds in or about them.) > The question here is not at all what these or those intuitionists think, but what "others" think, according to the opinion of Frenkel et al. Intuitionists reject N as something finished. Who considers N as something finished? Above you could find the outset: > > > > (like the axiom which requires the finity of > > > > the infinite). > > > > Any such axiom exists only in WM's system, and not in anyone else's. Regards, WM |