Cantor Confusion
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From: Virgil on 8 Feb 2007 04:27 In article <1170925367.150858.230020(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Feb., 20:12, Virgil <vir...(a)comcast.net> wrote: > > In article <1170852299.543791.43...(a)m58g2000cwm.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 6 Feb., 20:48, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1170754974.135681.22...(a)p10g2000cwp.googlegroups.com>, > > > > > > > > > The simplest reason is that omega - n = omega for all n in N. > > > > > > > > Where did you get that from? Reference? EB? > > > > > > > You could even figure it out by yourself. > > > > > > If one needs to adopt WM's crazy assumptions in order to come to WM's > > > > crazy conclusions, we are all better off without them. Both his > > > > assumptions and his conclusions > > > > > Excuse me, these were Cantor's assumptions and conclusions. > > > > Not the crazy ones I am referring to. > > We talked about: omega - n = omega for all n in N. > > > > For instance, one of WM's crazy assumptions is that a set of naturals > > which is not finite must contain a natural which is not finite. > > and cannot contain it! Why? All WM's supposed explanations of that claim are merely circular. If WM merely wishes to take as an axiom that there are no infinite sets, that would be one thing, but he wants to deduce it from something without assuming it, and he can't. > > > > Does WM claim that Cantor ever assumed or concluded that? > > He created this theroy. He was not and could not be aware of all of > its problems and implications. In other words, no.
From: Franziska Neugebauer on 8 Feb 2007 05:02 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 is not true. A tree *then* is the structure plus the set of paths and/or nodes. You persistently try to vaporize the essential constituents of a tree until it fits your preconception. > 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. 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. This can appropriately be expressed by T(n). >> >> > It contains paths as subsets. >> >> 0. In the usual definition of tree T = { v, e } this is obviously not >> true since neither of the non-trivial subsets >> {v}, {e}, {v, e} is a path at all. > > Forget this usual definition. We need not define the structure because > we talk about one structure only, a complete binary tree. Then you should talk about plain sets. You don't need to disguise them in the shape of trees. > If you are > too inflexible to swallow this then call my tree as you like it, but > do me a favour and retranslate your choice to tree when responding to > my letter. > >> 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. It leads again leads to the insight, that you posit what does not hold. > A path is an ordered set of nodes. In this picture a path is an ordered pair (S, <) of a set of nodes S and an order relation <. > But we need not use this order. To show that you are wrong even in this picture we *need* this order: As one easily sees T has only paths as elements, i.e. has ordered pair structure (S, <) but the elements of the path are S and <. Therefore there is no (non-trivial) p' which is a subset of a tree. Hence your conjecture "there are paths which are subsets of a tree" is wrong in this picture, too. > 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. You are performing the "fallacy of composition": ,----[ http://en.wikipedia.org/wiki/Fallacy_of_composition ] | A fallacy of composition arises when one infers that something is true | of the whole from the fact that it is true of some (or even every) | part of the whole. For example: "This fragment of metal cannot be | broken with a hammer, therefore the machine of which it is a part | cannot be broken with a hammer." This is clearly fallacious, because | many machines can be broken into their constituent parts without any | of those parts being so breakable. `---- >> >> Hence your conjecture "there are paths which are subsets of a tree" >> is wrong. > > See above. See above. F. N. -- xyz
From: Franziska Neugebauer on 8 Feb 2007 05:05 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? >> >> Let's call that numbers "named numbers". If the length of names >> "must" be finite one can easily prove, that any set of named numbers >> has cardinality <= card(omega). > > Of course. That is not new to me. It seems new to you that in real mathematics numbers do not necessarily have a name to exist. F. N. -- xyz
From: mueckenh on 8 Feb 2007 06:27 On 8 Feb., 11:02, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > This is not true. A tree *then* is the structure plus the set of paths > and/or nodes. You persistently try to vaporize the essential > constituents of a tree until it fits your preconception. What is wrong with my use of trees, except that it is inconvenient if one is a bit clumsy in adapting new ideas? This structure is given once and for all. Hence we do not need to mention it further. > > Froget this complicated picture. I does not lead ahead. > > It leads again leads to the insight, that you posit what does not hold. If you mention the structure in any step, the description gets complicated. The result does not differ from the case where the structure is given once and for all. > > > A path is an ordered set of nodes. > > In this picture a path is an ordered pair (S, <) of a set of nodes S and > an order relation <. The path also has some nodes. And we need only this aspect here. > > > But we need not use this order. > > To show that you are wrong even in this picture we *need* this order: But we don't need it for our conclusion. Therefore we do not use this picture. Show what is wrong with the conclusion that the nodes of a path are a subset of the set of nodes of the tree. > > > 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. > > You are performing the "fallacy of composition": If you find a path the identity of which depends on the representation as a sequence [0, 1, 6, 7] instead of a set {0, 1, 6, 7} then you may come up with your wisdom. For now I snip it. Regards, WM
From: mueckenh on 8 Feb 2007 06:31
On 8 Feb., 11:05, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(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, Mückenheim 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? > > >> Let's call that numbers "named numbers". If the length of names > >> "must" be finite one can easily prove, that any set of named numbers > >> has cardinality <= card(omega). > > > Of course. That is not new to me. > > It seems new to you that in real mathematics numbers do not necessarily > have a name to exist. What do those nameless numbers need to exist? How can we distinguish them from nonexisting numbers? By magic spell? Regards, WM |