Cantor Confusion
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From: davidmarcus on 27 Jan 2007 10:00 On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > Induction can possibly > > prove that all the members of V* have some property, but can prove > > nothing about V* itself. > We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. > Therefore all the path-*lengths* in the union are natural numbers. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, then at least > one of the paths in the union must be infinite. > > Is this so? The way to answer questions in mathematics is to provide a proof. Can you prove the claim or its negation?
From: Franziska Neugebauer on 27 Jan 2007 10:52 mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > >> Induction can possibly >> prove that all the members of V* have some property, but can prove >> nothing about V* itself. > > We can boil down the discussion about trees to the following simple > question, considering only one path, Please define path: [X] Sequence of nodes. [ ] Sequence of edges. [ ] Set of edges. [ ] other, define: ___________________________________ > for instance the path p on the outmost left hand side of the tree. In the sequence-of-nodes picture this is the path p = { (0, 2^i) | i e omega } refering to the numbering of G given in <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net>. > This path p (in terms of nodes) is the union of all paths of finite > trees with length n, n in N. Define "in the union": [ ] member (element) [ ] subset We call p_n n e omega the finite left-path of length n + 1: p_n := { (0, 2^0), ..., (n, 2^n) } Let P_l be the set of all finite left-paths: P_l := { p_n | n e omega } This set only exists if omega exists. (If we assumed omega not to exist P_l and the set of all finite left-paths would not provably exist). Since we assume that omega exists (we do so in contemporary set theory) P_l exists. According to the axiom of union even the union U P_l exists. As one can easily show it is true that U P_l = p > Therefore all the path-*lengths* in the union are natural numbers. 1. The union of paths is itself is not "many paths" but (at least) is single path. *In* the union there is no single paths as member. The union is equal to the path p. The length of path p is (by definition) not a natural number. Its length is card(p) = card(omega) +_card 1 = card(omega). 2. P_l contains paths each of which has finite length (depth). Namely depth(p_n) = card(p_n) + 1 = n + 1 n e omega. > Notwithstanding the question whether there are infinitely many paths > in the union The union U P_l = p does not have any sequence as member (element) at all. It equals p. So the union is a path. Every p_n is a subset of U P_l = p. > or not: If the union path p is infinite, True. U P_l = p has infinite cardinality. > then at least one of the paths in the union must be infinite. The union does not contain any single path as member (element!). P_l does contain only finite paths p_n n e omega as members. > Is this so? You proved that silly questions do really exist. 1. Define "in the union": a) there is no member (element) in U P_l = p which is a path b) there are subsets of U P_l which are paths. Namely every p_n n e omega and p itself are subsets of U P_l. (p is no a proper subset though). 2. You probably mean be a member of P_l (the set of all finite left-paths). Here indeed every p_n n e omega is element of P_l. But p is not an element of P_l, since P_l only contains finite paths. Probably you confuse P_l and U P_l. All in all you presented today next version of your 2003's theme X is not finite -> there must be an x in X which is infinite cast into paths. What comes next? F. N. -- xyz
From: G. Frege on 27 Jan 2007 11:31 On Sat, 27 Jan 2007 16:52:31 +0100, Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > > All in all you presented today next version of your 2003's theme > > X is not finite -> there must be an x in X which is infinite > > cast into paths. [...] > I guess, you hit the nail on the head! Quote (from a recent post): "Unless there is an infinite number the number of [natural] numbers [...] cannot be infinite." (W. M�ckenheim) F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 27 Jan 2007 11:37 On 26 Jan., 23:15, Virgil <vir...(a)comcast.net> wrote: > In article <1169803740.861911.258...(a)m58g2000cwm.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > I am not interested in any property of what you call the set N. >Then WM disclaims any interest in most of ordinary mathematics, which is > build on the set N. It is built on natural numbers. > > Purset nonsense. > WM doesn't even spell his falsehoods correctly. I saw it before posting. But it looked nice. Pure set nonsense sounds well for your identical nonidentical trees. >> II is a subset of III. > As sets, this is false, as both sets contain only a single element. No, they both contain the left bar and the right bar. III in addition contains a middle bar. > Only in Ancient Rome does II + I = III. There was not yet the "+" sign. In fact it is not necessary for unary numbers. I II is III > The name is not the thing named. If there really is a *thing*, this may be true. Regards, WM
From: William Hughes on 27 Jan 2007 11:38
On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > Induction can possibly > > prove that all the members of V* have some property, but can prove > > nothing about V* itself.We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. > Therefore all the path-*lengths* in the union are natural numbers. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, then at least > one of the paths in the union must be infinite. > > Is this so? No. Think of the EIT. The diagonal is the union of the lines. None of the lines is an (potentially) infinite set. The diagonal is an (potentially) infinite set. A union of finite sets can be (potentially) infinite. The fact that p is a union of finite paths does not tell us whether p is finite or (potentially) infinite. -William Hughes |