Cantor Confusion
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From: Franziska Neugebauer on 30 Jan 2007 11:19 mueckenh(a)rz.fh-augsburg.de wrote: > Which proof are you talking about? What am I talking about??? F. N. -- xyz
From: William Hughes on 30 Jan 2007 11:43 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?" 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. 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. 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). 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. - William Hughes
From: Fuckwit on 30 Jan 2007 11:41 On Tue, 30 Jan 2007 17:18:59 +0100, Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: >> >> This sum is defined by >> >> I >> II >> III >> ... >> >> i.e. by this set of symbols "I", each of which is to be identified by >> its index. >> > I cannot see any sum here. > No one (other than WM) can. :-) F.
From: Virgil on 30 Jan 2007 14:25 In article <1170156979.321128.184840(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Jan., 20:18, Virgil <vir...(a)comcast.net> wrote: > > > > Please do not mix up numbers of elements and natural numbers > > > concerning path lengths. > > > Translated to our problem they say: There is an infinite number of > > > finite path length. > > > Since there are an infinite number of natural numbers to serve as path > > lengths, and path lengths for all those naturals, why not? > > > > > Now I am in doubt whether these path length when > > > put together (such that every path starts at 0 and ends at n) yield a > > > finite length or not. > > > So now WM can't tell whether a path staring at 0 end ending at n is > > finite or not? > > > 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. > > > > > > > > > > > > So a union of different > > > > > natural numbers must contain an infinite number should it ex�st? > > > > > > > > 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?So p does not have length n for > > > > > any > > > > > finite natural > > > > number n. So p does not have a finite path length. > > > > > But the length of p *is* a natural number. > > > If p is the union of all paths with finite lengths it must have a length > > at least as great as any natural, so which natural does WM claim is as > > great as every natural? > > There is no natural number greater than every natural. The path > lengths however must be such a number, if all natural paths are taken > together. If the finite natural numbers all taken together form a number which is not a natural nor finite but is still a number, why can't the finite paths when all taken all together form a path which is not a finite path but is still a path? What is WM's definition of "path". it would appear that his definition requires that it be finite, which means that in a maximal infinite binary tree there can be no paths at all. >The consequence is that you cannot take all natural numbers > together. Lots of people can, and do, in the form of the set N of all (finite) naturals or omega as the set of all finite ordinals. > There is not "every natural number". Maybe not in such backward areas as WM-ville, but everywhere else there is a set of all naturals. > > > > > > Either we say that p does not have a path length, > > > > or we say that the path length of p is an infinite number. > > > > > That's it! Either there are not all natural numbers ( = the union p > > > does not ave a length) > > > or there is an infinite number ( = there is an ininiutr finite > > > number). > > > There IS an infinite ordinal number, but it is not a natural number. > > Maybe. But the path-lengths are all finite numbers. Only if WM defines paths so that the maximal infinite binary tree doesn't have any. > > > There IS an infinite path but it is not a path of finite length. > > So we can say: If there is an infinite ordinal or cardinal, then there > must be an infinite path length? If an maximal infinite binary tree is to have any paths at all, they must be infinite. Does WM claim to have such a tree with only finite paths or with no paths at all? He has claimed weirder things.
From: Virgil on 30 Jan 2007 14:40
In article <1170157336.490678.144860(a)v45g2000cwv.googlegroups.com>, mueckenh(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! > > > Those of us who do not have > > coniption fits over the use of the word call that /infinite/. > > I have been knowing for long that the complete union of all finite > numbers would contain an infinite number, if this union existed. It ain't what you don't know that hurts you, its what you "know" that ain't so. And WM is hurting, however insensitive to that pain his may be. > > > > > > > > > > A set is finite if there does not exist any injection from it into > > > > any of its proper subsets. > > > > > Now apply this to paths. > > > The union of all finite paths allows the mapping of each finite path of > > length n to a path of length n+1 while giving the same union, so that > > that union is infinite (in the sense of being not-finite). > > This sense means potential infinity. It is a set and it is not a finite set. > > > > > The length of p is the union of the lengths of all finite paths. > > > Which is the first limit ordinal. > > No, lengths measured in terms of natural numbers are natural numbers. Since all natural numbers are also ordinal numbers, and arbitrary unions of ordinal numbers are ordinal numbers, the length of p is an ordinal. > > > > > > > > Correct. Think about it. > > > Thinking about it leads to the inescapable conclusion that infinite > > binary trees must have infinitely long (endless) paths. > > But some have more while others have less? All such paths are countably infinite in length (number of odes or number of edges). > > > > > The identity function on N, idn: N --> N : x |--> x, is such a > > surjection, Q.E.D. > > That is not a set "listed in its entirety". The complete definition of a function on N "lists in its entirety" all its values. That is what such listing are. > > Here is a set S of cardinality 2^2^aleph_15 listed and by this listing > well-ordered in its entirety > > N --> S : x |--> s Not until you can prove the function is surjective, it isn't. And the function is not surjective, nor even well defined. > > Regards, WM |