Cantor Confusion
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From: Franziska Neugebauer on 29 Jan 2007 11:09 mueckenh(a)rz.fh-augsburg.de wrote: > The proof has been given. In which peer-reviewed journal? F. N. -- xyz
From: MoeBlee on 29 Jan 2007 13:06 On Jan 26, 3:55 pm, "Gerard Schildberger" <Gerar...(a)rrt.net> wrote: >> No, My use of the elipses was to indicate missing words, > in this case: (World War I, World War II). I thought > was obvious, but one gets fooled all the time. I was only joking. MoeBlee
From: Virgil on 29 Jan 2007 13:10 In article <1170085231.817591.177450(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Jan., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Jan 28, 7:21 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > You claim that if something is true for every set L_n, > > > > then it is true for N > > > I do not talk about N. This symbol has become the aim of heaviest > > > abuse.You talk about the union of all natural numbers. N is > > the union of all natural numbers. > > Something that is valid for each n in N need not be valid for N. I am > interested in properties which are valid for every n. For instance, > every n is finite. Every segment {1,2,3,...,n} is finite. Every path > being he union of paths (segments) is finite. In the pseudo-union of all finite trees, as WM defines that pseudo-union, either there are no paths at all or those paths are not finite. Unless WM allows things to be "paths" which are shorter than other "paths", as every finite "path" is shorter than another finite "path" in that pseudo-union. > > > > > > > > > > We know something about the maximum that is true > > > > for every set L_n. > > > > > > So you do want to prove something about the maximum > > > > of the set N. > > > I want to see whether the union of all finite numbers can be an > > > infinite number. > > > A union of numbers is a set of numbers. It is > > also a number. The union of all finite numbers is an > > infinite number. > > No. Yes! Not a natural number but an ordinal number. > > > > > This question was raised in the framework of the > > > infinite tree. Set theorists asserted that a union of finite paths > > > cannot be / contain any infinite path. > > > A union of finite paths is a set of finite paths. A set > > of finite paths is not a path. > > A union of two sets of nodes is a set of nodes. > A union of two paths need not be a path, but can be a path. Depends on how one defines a path, and even of the union of two paths were a path that does not make the union of infinitely many paths a path, at least unless a path can be infinite. > > > A set of finite paths > > does not contain an infinite path. > > That is the problem. > The union of all paths having only nodes with value 0, is a path. Is > it infinite? Potentially infinite, yes. Actually infinite, no. As a set is either finite or infinite in mathematics, there is no distinction in mathematics between WM's potential versus actual. > > All finite path-lengths are natural numbers. > This is your last sentence: The union of all natural numbers contains > an infinite natural number. Not at all. That may be WM's last sentence, but as it is a false sentence, it is none of ours. > As there are no actually infinite finite numbers, there is only the > potential infinity. A set which is according to WM "potentially infinite" is not finite , which is the only alternative to finite that mathematics recognizes.
From: Virgil on 29 Jan 2007 13:15 In article <1170085646.542472.199010(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Jan., 15:00, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 27 Jan., 18:29, Franziska Neugebauer <Franziska- > > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > > > >> > 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. > > > > > Cant't you read? Please look closer. There is clearly spelled out that > > > every number *in* N is concerned, i.e., every finite numbver, not the > > > infinite number N. > > >First you cut the context and then you ask whether I can read it. > > My comment: No I can't read it anymore. > > > > The context was: > > > U { T(i) | i e N } and I did not cut it. > > > > > > > > > > >> ,----[ <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". > > >> `----Your claim is > > > > "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." > > > > Non sequitur. > > > > Your argument is of the type "counting over to infinity". > > There is no counting over. We count natural numbers, as far as > possible. > > > You may take > > notice of the similar claim: > > > > "The sum of the numbers n and n + 1. n is a natural number. > > Therefore the sum of all finite numbers is defined." > > If all natural numbers existed, why shouldn't their sum exist? Why shouldn't their square roots exist equally well? In fact the square roots off naturals have better pedigrees that the sum of all of them has. > > Of course we can show in unary representation (IET) that the sum of > all natural numbers is countable You can't even show that it exists, WM. > > Do you disagree? Show us that sum, and then we can discuss "countable". > > Regards, WM
From: Virgil on 29 Jan 2007 13:22
In article <1170085904.957438.52140(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > On 27 Jan., 22:52, Franziska Neugebauer <Franziska- > > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> mueck...(a)rz.fh-augsburg.de wrote: > > >> > I am interested in the fact that every set of natural numbers has a > > >> > finite maximum. > > [indendation corrected] > > >> Then you should perhaps not talk to contemporary set > > > theorists who are > > >> accustomed to the > > > > > > fixed idea being far from being a > > > > > >> fact that there *are* sets of natural numbers which do > > >> not have maxima at all. > > > > In the framework of ZFC it *is* fact that the set of all natural numbers > > does not have a maximum. Without a proof of a contradiction _within_ the > > ZFC framework there is no reason for your disfavour. > > The proof has been given. What WM presents as a proof is not accepted as proof by anyone else here. > Look at the infinite binary tree which > contains and not contains uncounatbly many paths. There is no such tree for any given definition of path. Depending on one's definition of path, and maximal infinite rooted binary tree either does or does not contain such paths, but never both. > Look at the big vase > which at noon is empty (did you read this delicious proof?). There is > no further reason to maintain ZFC. That WM is willfully blind in these matters does not mean that others are unable to enjoy the views. |