Cantor Confusion
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From: Bob Kolker on 29 Jan 2007 10:42 mueckenh(a)rz.fh-augsburg.de wrote: > > 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. Each n in N and all n in N mean exactly the same thing. They both express the universal quantifier. Bob Kolker
From: mueckenh on 29 Jan 2007 10:47 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? Of course we can show in unary representation (IET) that the sum of all natural numbers is countable: 1 11 111 .... Do you disagree? Regards, WM
From: mueckenh on 29 Jan 2007 10:51 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. Look at the infinite binary tree which contains and not contains uncounatbly many paths. Look at the big vase which at noon is empty (did you read this delicious proof?). There is no further reason to maintain ZFC. Regards, WM
From: Fuckwit on 29 Jan 2007 10:54 On 29 Jan 2007 07:47:26 -0800, mueckenh(a)rz.fh-augsburg.de wrote: >> >> 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? > Because it wouldn't be a natural number. :-) > > Of course we can show in unary representation that the sum of > all natural numbers is countable: > > 1 > 11 > 111 > ... > > Do you disagree? > Which sum? You have presented (the beginning of) a list of natural numbers (in unary representation). The sum of _all_ natural numbers might be formulated the following way: I + II + III + ... , though this expression is not defined for _infinitely_ many terms. (With other words, the sum of all finite numbers does not exist.) Fuckwit
From: Franziska Neugebauer on 29 Jan 2007 11:02
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. See below what you have cut. >> >> ,----[ <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 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." had been cut by your. >> 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? 1. Your reasoning if flawed. The latter does not follow from the former. It is *not* "covered" by induction. 2. BECAUSE it is _undefined_ until the "sum of all natural numbers" is defined. F. N. -- xyz |