Cantor Confusion
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From: mueckenh on 1 Feb 2007 07:02 On 30 Jan., 17:43, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Jan 30, 8:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > You can try to utter again and agan this nonsense, but after a while > > aI will cease to reply. Every such thing including only natural > > numbers is covered by induction. All natural numbers are subject to > > induction. > > True, you can use induction to prove something about > any natural number. However, the question is: "Can > you use induction to prove something about a set > of natural numbers?" This question has a trivial answer: If the set of all natural numbers is nothing than all natural numbers, then yes. If the set of all natural numbers is more than all natural numbers, then no. > > There are two types of sets of natural numbers > > I: sets of natural numbers that are not (potentially) infinite > > II: sets of natural numbers that are (potentially) infinite > > Induction can only prove things about sets of type I. So you believe that type II sets have some esoteric supplement? We can prove that every number is a number while a set of several numbers is not a number. There is no reason to distinguish between finite and infinite sets. > > By running induction longer and longer you can get more and more > sets of type I. By running induction for an infinite time you can > get all > sets of type I. No. By two steps of induction, namely P(1) P(n) ==> P(n+1) you can prove P for all natural numbers and for all sets of natural number (except quantitative statements as I mentioned above). Your strong belief in the inaccesible infinite uttered above is your (and some other people's) personal opinion but has nothing to do with mathematics. > But you will never get a set of type II. (You may get > a collection of sets whose union contains the same elements > as a set of type II, but you will never get a single set of type > II). Small wonder. There is no set of type II. > > The set of all natural numbers, the union of all natural numbers, N, > is a set of type II. You cannot use induction to prove anything about > the union of all natural numbers. Small wonder. This set does not exist, but for every existing set of natural numbers, induction is sufficient. Regards, WM
From: mueckenh on 1 Feb 2007 07:07 On 30 Jan., 20:25, Virgil <vir...(a)comcast.net> wrote: > In article <1170156979.321128.184...(a)j27g2000cwj.googlegroups.com>, > > I can't tell whether the union of all lengths is finite or not. > > As the union of all lengths is obviously equivalent to the union of all > finite ordinals and the union of all finite ordinals is clearly not > finite, WM can't tell much. The union of all lengths is not the sum of all lengths, which, according to your standpoint might be infinite. But the union of all lengths is a length. You may say that it is infinite or not. If it is infinite, then it corresponds to an infinite natural number. If it is not infinite, then the set N does not exist other than as a potentially infinite (= finite but unbounded) set. Regards, WM
From: mueckenh on 1 Feb 2007 07:08 On 30 Jan., 20:40, Virgil <vir...(a)comcast.net> wrote: > In article <1170157336.490678.144...(a)v45g2000cwv.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 28 Jan., 20:44, Virgil <vir...(a)comcast.net> wrote:> > In fact? If > > there is a path of lengths n then there is a path of > > > > length n+1. And there is a path of length 1. What is the length of > > > > the union of all these paths (which contains only finite paths)? > > > > If that union is to be regarded as a path at all > > > Yes, please regard it as such! > > > > then it cannot be a > > > finite path as it would have to be be longer than every finite path > > > (for every path of length n, the union wold have to be of length at > > > least n+1 ). So its length is not finite. > > > On the other hand, it must not be longer than every natural number > > because it is simply the union of all natural numers. > > So that WM claims something that must be longer than any natural but can > not be longer than every natural? Good luck finding it! It has been known for centuries. Longer than any *existing* natural is the potentially infinite. Regards, WM
From: Franziska Neugebauer on 1 Feb 2007 07:12 mueckenh(a)rz.fh-augsburg.de wrote: > On 30 Jan., 17:18, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> >> 2. BECAUSE it is _undefined_ until the "sum of all natural >> >> numbers" is defined. >> >> > It is defined if the set of all natural numbers s defined, >> > > Here is the definition of the set N > > 1 > 2 > 3 > ... > > Here is the definition of the sum of all elements of N > > 1 > 23 > 456 > ... Abstract art? F. N. -- xyz
From: mueckenh on 1 Feb 2007 07:17
On 30 Jan., 21:50, Virgil <vir...(a)comcast.net> wrote: > In article <1170164725.794981.290...(a)a34g2000cwb.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > Franziska Neugebauer schrieb: > > > Continued > > > > 2. BECAUSE it is _undefined_ until the "sum of all natural numbers" is > > > defined. > > > This sum is defined by > > I > > II > > III > > ... > > WRONG! As usual. > > A listing of all natural numbers is not the same as the summing of all > natural numbers, and WM has, at most, produced a listing. The sum is a listing: II + III = IIIII = 1,2,3,4,5. The sum of all natural numbers can be obtained (if all natural nunbers exist) by the following "diaogonalization" 1 23 654 .... It is countable. Regards, WM |