Cantor Confusion
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From: mueckenh on 30 Jan 2007 08:33 Fuckwit schrieb: > 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. :-) Why should it be a natural number ??? We know that the set of natural numbers already has not a finite cardinal number. Their sum cannot be less as can be shown by induction. > > > > > 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). Complete it (if you can). The number of units given there is countable. I is th sum of all natural numbers. > > 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. It is defined up to EVERY natural number. > (With other words, the sum of all finite numbers does not exist.) (With other words, the sum of all finite numbers, including EVERY finite number, does exist.) Regards, WM
From: mueckenh on 30 Jan 2007 08:36 Bob Kolker schrieb: > 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. So you would say that the sum of each n in N is the same as the sum of all n in N? I would say the first is finite for each n while the second is infinite. Regards, WM
From: mueckenh on 30 Jan 2007 08:42 Franziska Neugebauer schrieb: > 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. Why should we repeat such self-evident truth again and again? > > >> 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. 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. What you erroneously try to claim is that N is more than the assembly of all natural numbrs. But that is phantasy. > > 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, namely by I II III .... which is a countable set of symbols "I" which are to be distinguished by their in
From: mueckenh on 30 Jan 2007 08:45 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 .... i.e. by this set of symbols "I", each of which is to be identified by its index. Regards, WM
From: Andy Smith on 30 Jan 2007 08:45
Just following up the idea of countability. Am I right in thinking that any set of "finitely describable" objects is necessarily countable? - on the grounds that "finitely describable" implies that there exists some (multi-dimensional) address space which uniquely defines an object, and that that address is indexable by a finite set of natural numbers, and then that that multi-dimensional address (as with the "standard arrangement" for counting the rationals) can be ordered such that each address has a unique index in the natural numbers? On that basis I would surmise that e.g. the set of all roots of all polynomials with rational coefficients are countable - is that correct? -- Andy Smith |