Another AC anomaly?
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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/
From: Dik T. Winter on 7 Dec 2009 10:37 In article <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > > > > 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. > > Question. Isn't this simply a question of language? Not at all. When you define N as an infinite union there is no limit involved, there is even no sequence involved. N follows immediately from the axioms. > My > book on set theory defines omega, w, as follows: > > Define w to be the set N of natural numbers with its > usual order > < (given by membership in ZF). > > Now w is a limit ordinal so the ordered set N is, in the > ordinal sense, a limit. Of course w is not a member of N > becasuse then N would be a member of itself (not allowed by > foundation). Note here that N (the set of natural numbers) is *not* defined using a limit at all. That w is called a limit ordinal is a definition of the term "limit ordinal". It does not mean that the definition you use to define it actually uses a limit. (And if I remember right, a limit ordinal is an ordinal that has no predecessor, see, again no limit involved.) -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 7 Dec 2009 10:41
In article <5f8c7e7f-ec83-45ba-a584-b614759696f4(a)j4g2000yqe.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 3 Dez., 16:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > No. I asked you for a mathematical definition of "actual infinity" > > > > and you told me that it was "completed infinity". Next I asked you > > > > for a mathematical definition of "completed infinity" but you have > > > > not given an answer. So I still do not know what either "actual > > > > infinity" or "completed infinity" are. > > > > > > Both are nonsense. But both are asumed to make sense in set theory. > > > > No, set theory does not contain a definition of either of them. So in what way do they make sense in set theory? As I have not seen a definition of them at all, I can not see what that means. > > > The axiom of infinity is a definition of actual infinity. > > > "There *exists* a set such that ..." > > > Without that axiom there is only potential infinity, namely Peano > > > arithmetic. > > > > I see neither a definition of the words "actual infinity" neither > > a definition of "potential infinity". Or do you mean that "potential > > infinity" is Peano arithmetic (your words seem to imply that)? > > > > So we can say that in "potential infinity" consists of a set of axioms? > > Here is, to my knowledge, the simplest possible explanation. Darn, I ask for a definition, not for an explanation. > Consider > the infinite binary tree: > > 0 > /\ > 0 1 > /\ /\ > 0 1 0 1 > ... > > Paint all paths of the form > 0.111... > 0.0111... > 0.00111... > 0.000111... > and so on. > Potential infinity then says that every node and every edge on the > outmost left part of the tree gets painted. > Actual infinity says that there is a path 0.000... parts of which > remain unpainted. And that is wrong. So "potential infinity" and "actual infinity" are theories? But I do not know of any theory that states that there is any part of the path 0.000... that is unpainted. Can you show how any theory states that? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |