Cantor Confusion
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From: Virgil on 27 Jan 2007 14:27 In article <1169915856.541369.8170(a)p10g2000cwp.googlegroups.com>, 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: > > > 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. Which, collectively, form the set N. > > > > > 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. WM is again showing his ignorance of set theory. WM is not referring to mere sets but to sequences, or multisets, or at least something other than a set, in which the same object can be counted more than once. In a SET an object either is or is not a member, but cannot be a member more than once. So that, as sets with I as a member, I and II and III are all identically sets with one element. > > The name is not the thing named. > > If there really is a *thing*, this may be true. Even if there isn't, it is true. Because a name which does exist cannot be a thing which does not exist.
From: Virgil on 27 Jan 2007 14:46 In article <1169915968.970612.249360(a)a34g2000cwb.googlegroups.com>, 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". Depends on how it is stated. If one's induction is of the form: There is a set S such that (1) The first natural is a member of S, and (2) The successor of every member of S is a member of S Then one's conclusion should be that N is a subset of S. > >> > > > 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. There are also prepositions and adverbs, etc. 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. WM again argues that there must be an infinite natural in order that the number of naturals not be a finite natural. At this late date, I don't suppose WM will ever get things straightened out in his head. > > Is this so? It is *not so* that a union of sets must contain a not finite object if it is to be itself not finite. At least by any definition of finiteness for sets presently available. The two most common such definitions are, or are equivalent to: (1) A set is finite if it bijects with the set of all naturals less than some fixed natural. Each such natural, when regarded as a set, is therefore finite. or (2) A set is finite if there does not exist any injection from it into any of its proper subsets. Note that neither definition defines finiteness except for sets. Does WM have some alternate definition of finiteness? If so he should present it to us, as by either of the above definitions, the set of all naturals is a not finite set with all members finite.
From: Virgil on 27 Jan 2007 14:55 In article <1169917120.839454.172270(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. WM cannot keep his numbers straight. The union of all natural numbers, when regarding them as ordinal numbers, which are sets, is a set, and a number, but not a natural number. It is an ordinal number, and under some definitions of cardinality, a cardinal number, but it is not a natural number. Note that 2 is a natural number but that its reciprocal, while a number, is not a natural number. Certain types of operations on certain numbers, or sets of numbers, may result in numbers of a different type. > So a union of different > natural numbers must contain an infinite number should it ex�st? Not as a member, but the union itself may be a non-natural number as noted above. > > > > 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? Fact is that p, like N as the first limit ordinal, can be a set not satisfying any definition of finiteness. Which fact, WM has yet to face.
From: Virgil on 27 Jan 2007 14:57 In article <45bb8a1b$0$97236$892e7fe2(a)authen.yellow.readfreenews.net>, Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > 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. WM not only pursues it assiduously, he too often catches it, and then breeds it indiscriminately.
From: mueckenh on 27 Jan 2007 16:43
On 27 Jan., 17:41, "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: > > > 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. I do not want to prove something about a maximum of the set N. I am interested in the fact that every set of natural numbers has a finite maximum. Regards, WM |