Another AC anomaly?
in [Logic]
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From: WM on 5 Dec 2009 06:58 On 4 Dez., 23:18, Virgil <Vir...(a)home.esc> wrote: > In article > <ce07a2f1-cf4c-4e53-b5d5-b1fcd2d80...(a)s20g2000yqd.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 4 Dez., 09:20, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <949e3b4b-04c5-4d6e-ba64-7f95d37db...(a)r24g2000yqd.googlegroups.com>, > > > > Actual mathematics would say that while every > > > node and every edge is included in the sequence, there is no member of > > > the sequence having all of them at once. > > > Everey member covers the first node. > > The n-the member covers all nodes including 1 and n. > > So logic forces us (i.e. those who can) to conclude that if all nodes > > are covered then there is a path covering all nodes. > > That WM's particular form of logic is rife with contradictions. > > For every 0 node there is a path in WM's set NOT containing it, > and for every path in WM's set there is a node NOT contained in it, > so why does WM insist that there is a path in your set containing every > node? That is potential infinity. For every node there is a path surpassing it For every path there is a node surpassing it. There is no "all nodes" and no "all paths". Actual infinity infinity or set theory, what is the same, claim that there is a path containing all nodes (namely the path 0.000...) and that the union of all paths 0.111... 0.0111... 0.00111... .... contains the path 0.000... but that there is no single path of the list containing all nodes of 0.000... That is simply impossible. The union of paths of the list contains exactly as many nodes of 0.000... as one of those paths to be unioned. If none of them contains all nodes of 0.000..., then even infinitely many will not accomplish that goal. The union of paths cannot result in more than any contributing path. The union can be potentially infinite. A union of finite paths cannot yield a path of fixed length the lengths of which surpasses every contributing path. Regards, WM Regards, WM
From: Virgil on 5 Dec 2009 14:06 In article <985b08f2-a385-48b6-9d94-0d45815bc066(a)f16g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Dez., 23:18, Virgil <Vir...(a)home.esc> wrote: > > In article > > <ce07a2f1-cf4c-4e53-b5d5-b1fcd2d80...(a)s20g2000yqd.googlegroups.com>, > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 4 Dez., 09:20, Virgil <Vir...(a)home.esc> wrote: > > > > In article > > > > <949e3b4b-04c5-4d6e-ba64-7f95d37db...(a)r24g2000yqd.googlegroups.com>, > > > > > > Actual mathematics would say that while every > > > > node and every edge is included in the sequence, there is no member of > > > > the sequence having all of them at once. > > > > > Everey member covers the first node. > > > The n-the member covers all nodes including 1 and n. > > > So logic forces us (i.e. those who can) to conclude that if all nodes > > > are covered then there is a path covering all nodes. > > > > That WM's particular form of logic is rife with contradictions. > > > > For every 0 node there is a path in WM's set NOT containing it, > > and for every path in WM's set there is a node NOT contained in it, > > so why does WM insist that there is a path in your set containing every > > node? > > That is potential infinity. > For every node there is a path surpassing it > For every path there is a node surpassing it. While I can guess what you mean by saying "For every node there is a path surpassing it", I have no idea what you might mean by "For every path there is a node surpassing it". > There is no "all nodes" and no "all paths". There is in ZF and in most set theories. If there is no "all nodes" your claim of "For every node" is impossible. If there is no "all paths" your claim of "For every path" is impossible. You can't have one without the other. > > Actual infinity infinity or set theory, what is the same, claim that > there is a path containing all nodes Only in unary trees. In binary trees no path contains all nodes. > (namely the path 0.000...) Which does not contain the node 0.1 > and > that the union of all paths > 0.111... > 0.0111... > 0.00111... > ... > contains the path 0.000... Not outside of Wolkenmuekenheim > but that there is no single path of the list containing all nodes of > 0.000... That path, 0.000..., certainly contains all nodes of that path!!!!! WM's already slim grasp of reality is fading fast. > > That is simply impossible. The union of paths of the list contains > exactly as many nodes of 0.000... as one of those paths to be unioned. WRONG!!! Each such path contains only finitely many nodes of 0.000..., but the union of all of them contains more than any finite number of those nodes. > If none of them contains all nodes of 0.000..., then even infinitely > many will not accomplish that goal. If each of infinitely many of them contains a node not contained in any of its predecessors, which is the case, then that union will contain all of them, at least outside of Wolkenmuekenheim > > The union of paths cannot result in more than any contributing path. The set of nodes of the union of even two paths results already in a proper superset of the nodes in any one path, at least outside of Wolkenmuekenheim. > The union can be potentially infinite. A union of finite paths cannot > yield a path of fixed length the lengths of which surpasses every > contributing path. In an infinite binary tree there are no finite paths, and unions of paths do not produce paths at all in any tree, at least outside of Wolkenmuekenheim.
From: WM on 5 Dec 2009 16:22 On 5 Dez., 20:06, Virgil <Vir...(a)home.esc> wrote: > > That is potential infinity. > > For every node there is a path surpassing it > > For every path there is a node surpassing it. > > While I can guess what you mean by saying "For every node there is a > path surpassing it", I have no idea what you might mean by "For every > path there is a node surpassing it". There is a symmetry: For every path of the form 0.000...000111... there is a node 0 at the outmost left side of the tree not covered by that path. And for every node 0 at the outmost left side of the tree, there is a path of the form 0.000...000111... covering it. > > > There is no "all nodes" and no "all paths". > > There is in ZF and in most set theories. > > If there is no "all nodes" your claim of "For every node" is impossible. > If there is no "all paths" your claim of "For every path" is impossible. > > You can't have one without the other. Correct. You can't have an infinite number of natural numbers without having also an infinite natural number (which is a self- contradiction). > > > > > Actual infinity infinity or set theory, what is the same, claim that > > there is a path containing all nodes > > Only in unary trees. In binary trees no path contains all nodes. > > > (namely the path 0.000...) > > Which does not contain the node 0.1 Which does contain all nodes of the outmost left side of the tree. > > > and > > that the union of all paths > > 0.111... > > 0.0111... > > 0.00111... > > ... > > contains the path 0.000... > > Not outside of Wolkenmuekenheim > > > but that there is no single path of the list containing all nodes of > > 0.000... > > That path, 0.000..., certainly contains all nodes of that path!!!!! But there are not all nodes of that form. > > > > That is simply impossible. The union of paths of the list contains > > exactly as many nodes of 0.000... as one of those paths to be unioned. > > WRONG!!! Each such path contains only finitely many nodes of 0.000..., > but the union of all of them contains more than any finite number of > those nodes. That is nonsense. Each path starts at the root and stretches until a finite number is reached. Even an infinity of finite paths does not contain a path which contains an infinity of nodes. The union of paths however is a path. Regards, WM
From: Virgil on 6 Dec 2009 00:45 In article <3b98d3f1-0ef1-4df0-bd0c-8020648df5a6(a)s19g2000vbm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 5 Dez., 20:06, Virgil <Vir...(a)home.esc> wrote: > > > > That is potential infinity. > > > For every node there is a path surpassing it > > > For every path there is a node surpassing it. > > > > While I can guess what you mean by saying "For every node there is a > > path surpassing it", I have no idea what you might mean by "For every > > path there is a node surpassing it". > > There is a symmetry: > For every path of the form 0.000...000111... there is a node 0 at the > outmost left side of the tree not covered by that path. > And for every node 0 at the outmost left side of the tree, there is a > path of the form 0.000...000111... covering it. Then why did you not say so instead of being deliberately ambiguous? And while you re, for once correct, your statements are trivial and of no relevance to the falsities you are trying to establish. > > > > > > There is no "all nodes" and no "all paths". > > > > There is in ZF and in most set theories. > > > > If there is no "all nodes" your claim of "For every node" is impossible. > > If there is no "all paths" your claim of "For every path" is impossible. > > > > You can't have one without the other. > > Correct. The why do you keep claiming one without the other? > You can't have an infinite number of natural numbers without > having also an infinite natural number (which is a self- > contradiction). Both a non sequitur,and a false statement!!! ZF has an infinite number of naturals but does not have any infinite naturals, at least according to any definition of "natural number" allowed in ZF. What goes on in WM's Wolkenmuekenheim is, of course, irrelevant to mathematics. > > > > > > > > > Actual infinity infinity or set theory, what is the same, claim that > > > there is a path containing all nodes > > > > Only in unary trees. In binary trees no path contains all nodes. > > > > > (namely the path 0.000...) > > > > Which does not contain the node 0.1 > > Which does contain all nodes of the outmost left side of the tree. That is an entirely different matter which, as you left it out of your original statement, makes you original statement false, as I previously said it was. > > > > > and > > > that the union of all paths > > > 0.111... > > > 0.0111... > > > 0.00111... > > > ... > > > contains the path 0.000... > > > > Not outside of Wolkenmuekenheim > > > > > but that there is no single path of the list containing all nodes of > > > 0.000... > > > > That path, 0.000..., �certainly contains all nodes of that path!!!!! > > But there are not all nodes of that form. If one is not embedded in Wolkenmuekenheim, there are. > > > > > > > That is simply impossible. The union of paths of the list contains > > > exactly as many nodes of 0.000... as one of those paths to be unioned. > > > > WRONG!!! Each such path contains only finitely many nodes of 0.000..., > > but the union of all �of them contains more than any finite number of > > those nodes. > > That is nonsense. What WM claims makes no sense within his Wolkenmuekenheim makes good sense everywhere else > Each path starts at the root and stretches until a > finite number is reached. In a standard infinite binary tree, none of the uncountably many infinite paths has such a last node. So whatever WM is talking about, it is not a standard infinite binary tree. > Even an infinity of finite paths does not > contain a path which contains an infinity of nodes. I do not know where WM is getting his paths, but he is being fed a very non-standard and inferior quality bunch of them, and, no doubt, being grossly overcharged into the bargain. > The union of paths > however is a path. REALLY non-standard and inferior quality!!!
From: Dik T. Winter on 7 Dec 2009 10:32
In article <7772c857-57b0-4422-b688-9a4c8b923467(a)h10g2000vbm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 3 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > The limit set is not necessarily "assumed" by the sequence of sets > > > > (if by that you mean that there is a set in the sequence that is > > > > equal to the limit set, if you mean something else I do not > > > > understand it at all). > > > > When you mean with your statement about N: > > > > N = union{n is natural} {n} > > > > then that is not a limit. Check the definitions about it. > > > > > > It is a limit. That is independent from any definition. > > > > It is not a limit. Nowhere in the definition of that union a limit is > > used or mentioned. > > 1) N is a set that follows (as omega, but that is not important) from > the axiom of infinity. You can take it "from the shelf". Actually: N is the smallest inductive set that starts with 1. The axiom of infinity states that that set does exist, but with that definition it is *not* defined as a limit. > 2) N is the limit of the sequence a_n = ({1, 2, 3, ...,n}) You have first to define the limit of a sequence of sets before you can state such. N is (in set theory) defined before even the concept of the limit of a sequence of sets is defined (if that concept is defined at all in the particular treatise). > 3) N is the limit, i.,e. the infinite union of singletons {1} U {2} > U ... No, the infinite union is *not* a limit. It is defined without even the presence of the concept of a limit of a sequence of sets. Rid yourself of all those ideas. > This is fact. It is not. > But if (3) is correct, then N must also be the limit of the process > described in my > http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#394,22,Folie 22 > without and *with* the intermediate cylinder. I see no process there, only a picture of a cylinder with the digits 1, 2, 3, 4 and 5, and an open cube. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |