Another AC anomaly?
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From: Marshall on 4 Dec 2009 09:46 On Dec 3, 10:31 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 3 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > So we can say that in "potential infinity" consists of a set of axioms? > > Here is, to my knowledge, the simplest possible explanation. Consider > the infinite binary tree: > > 0 > /\ > 0 1 > /\ /\ > 0 1 0 1 > ... http://www.flickr.com/photos/22984501(a)N06/2961175776/ Marshall
From: Virgil on 4 Dec 2009 17:18 In article <ce07a2f1-cf4c-4e53-b5d5-b1fcd2d80bb2(a)s20g2000yqd.googlegroups.com>, WM <mueckenh(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? Not even God knows. > > But as there is no such path. Hence there is no "all nodes". Or WM is wrong again about the need for one. > > Some matheologians claim that there are all nodes and that all nodes > are covere by all paths but that there is no path covering all nodes. Only WM claims that. Mathematicians claim that in ZFC there is a set of all nodes each represented by a finite. possibly empty, binary string of, say, zeroes and ones, and a set of all paths, each path being a function from N to {0,1}. Two such paths coincide on some initial string, the string of nodes preceding the first node at which they differ. Any two paths determine a > This would... But it is really useless to try to teach such ... It is not so much what WM does not know than hurts him, and his students, it's what he "knows" that ain't so, and then tries to teach to his poor captive students.
From: Carsten Schultz on 5 Dec 2009 06:41 WM schrieb: > On 4 Dez., 06:30, "K_h" <KHol...(a)SX729.com> wrote: >> "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message >> >> news:0c635f53-f5c4-4320-8825-de05f021a428(a)m3g2000yqf.googlegroups.com... >> On 2 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: >> >>> Same holds for the cylinder. Its contents is >>> {1}, {1}, {1}, {1} , ... >>> and that is not different from >>> {1}, {2}, {3}, {4} , ... >>> as you can see by renaming the elements. >> You need to provide a definition for your idea of a limit >> set. > > 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". > 2) N is the limit of the sequence a_n = ({1, 2, 3, ...,n}) > 3) N is the limit, i.,e. the infinite union of singletons {1} U {2} > U ... > And, one might add, you have to learn what a definition is. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
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. |