Cantor Confusion
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From: mueckenh on 27 Jan 2007 11:39 On 26 Jan., 18:51, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> context: > >> >> ,----[ <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> > >> >> ] > >> >> | >> Again: Your notations > >> >> | >> > >> >> | >> T(1) U T(2) U ... > >> >> | >> > >> >> | >> and > >> >> | >> > >> >> | >> U {T(i) | i e N } > >> >> | >> > >> >> | >> are undefined. > >> >> | > > >> >> | > You are in error. The union of the trees T(n) and T(n+1) is > >> >> | > defined. n is a natural number. Therefore the union of all > >> >> | > finite trees is defined. > >> >> | > >> >> | You have misunderstood the induction principle. It is not made > >> >> | for "counting over to the infinite". > >> >> `---- > > > Look: Above, there is written "i in N", not "i = N". The latter would > > be missed by induction. > ??? Induction covers (is valid for) all natural numbers, but not "the set of all natural numbers". >> > > You only try to huddle around, avoiding any concrete discussion. >I have a different understanding of what concrete means. Do you mean the > concrete in your head? You seem to know only one aspect of many things which have more. There are three meanings in English: noun, verb and adjective. The adjective has the same meaning as in German. > > > > Look at my tree. > Perhaps I need the Third Eye for that. Meanwhile I simplified the problem (for you) to the following simple question, considering only one path, for instance the path p on the outmost right hand side of the tree. This path p (in terms of nodes) is the union of all paths 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? Regards, WM
From: William Hughes on 27 Jan 2007 11:41 On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > Then you best conclusion is that it holds for all natural numbers but > > not that it holds for N. >Precisely that is what I want and what I do! No proving that something holds for every natural number is not the same as showing that it holds for N. Consider Every set L_n = {0,1,2,3,....,n} has the property that the maximum of L_n can never refer to more that one natural number. The maximum of the set N can refer to more than one natural number. - William Hughes
From: mueckenh on 27 Jan 2007 11:58 On 27 Jan., 17:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > 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. Fine. (Please use "numbers". The paths-lengths are numbers (of a unit). ) But set theorists deny this. They say: A union of finite numbers cannot yield an infinite number. So a union of different natural numbers must contain an infinite number should it exíst? > > The fact that p is a union of finite paths does not tell > us whether p is finite or (potentially) infinite. Fact is that p has no upper bound. Ok? Fact is that p is the union of only finite natural numbers. Ok? Regards, WM
From: Franziska Neugebauer on 27 Jan 2007 12:21 mueckenh(a)rz.fh-augsburg.de wrote: > 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: [...] >> > 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. Perhaps Virgil meant pursuit of nonsense? F. N. -- xyz
From: Franziska Neugebauer on 27 Jan 2007 12:29
mueckenh(a)rz.fh-augsburg.de wrote: > On 26 Jan., 18:51, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> >> context: >> >> >> ,----[ >> >> >> <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> ] >> >> >> | >> Again: Your notations >> >> >> | >> >> >> >> | >> T(1) U T(2) U ... >> >> >> | >> >> >> >> | >> and >> >> >> | >> >> >> >> | >> U {T(i) | i e N } >> >> >> | >> >> >> >> | >> are undefined. >> >> >> | > >> >> >> | > You are in error. The union of the trees T(n) and T(n+1) is >> >> >> | > defined. n is a natural number. Therefore the union of all >> >> >> | > finite trees is defined. >> >> >> | >> >> >> | You have misunderstood the induction principle. It is not >> >> >> | made for "counting over to the infinite". >> >> >> `---- >> >> > Look: Above, there is written "i in N", not "i = N". The latter >> > would be missed by induction. > >> ??? > > Induction covers (is valid for) all natural numbers, but not "the set > of all natural numbers". We know this for quite some time. The point is that you claim induction allows any assertion of U { T(i) | i e N }. So you eventually agree that is does not. >> > You only try to huddle around, avoiding any concrete discussion. > >>I have a different understanding of what concrete means. Do you mean >>the >> concrete in your head? > > You seem to know only one aspect of many things which have more. > There are three meanings in English: noun, verb and adjective. The > adjective has the same meaning as in German. The noun precisely represents the modus operandi you present in this newsgroup. >> > Look at my tree. > >> Perhaps I need the Third Eye for that. > > Meanwhile I simplified the problem (for you) to the following simple > question, considering only one path, for instance the path p on the > outmost right hand side of the tree. This path p (in terms of nodes) > is > the union of all paths 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? An in-depth discussion of your question you will find in <45bb7541$0$97231$892e7fe2(a)authen.yellow.readfreenews.net> F. N. -- xyz |