Cantor Confusion
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From: Virgil on 30 Jan 2007 15:15 In article <1170163983.376060.99370(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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 ??? Because the definition of addition for natural numbers requires that the set of naturals be closed under addition. Thus any operation whose result is not a natural cannot be addition. > > 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.) The set of naturals for which the sum up to that natural is defined includes all naturals, but the set of naturals, N, is NOT a member of itself so that the sum for N is not defined. WM is again conflating subset properties with membership properties. For n as /member/ of N, the sum over n = {0.1.,,,.n-1} is defined but not for every S as /subset/ of N.
From: MoeBlee on 30 Jan 2007 15:19 On Jan 30, 5:45 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > Am I right in thinking that any set of "finitely describable" objects is > necessarily countable? If the set of symbols is countable, then there are only countably many objects that can be defined ("described as UNIQUE") by finite strings of those symbols. > 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? The set of algebraic numbers is countable. MoeBlee
From: Virgil on 30 Jan 2007 15:28 In article <1170164184.870466.217970(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. For once, I agree with WM's analysis. I do not know how it works in other languages, but in English, "for each n" and "for all n" are not totally interchangeable, as his example demonstrates. One must be careful to specify unambiguously what the quantified variable represents to avoid such situations. In the above examples, the n is in one case the index set (for each n in N, sum the members of n) and in the other the term being summed ( sum of n's as members of N). Though WM has been known to conflate the two senses deliberately.
From: Andy Smith on 30 Jan 2007 15:37 Andy Smith <Andy(a)phoenixsystems.co.uk> writes > >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? I just googled that, and got the answer that Cantor showed that the finite union of countable sets is also countable, and hence that e.g. the set of all roots of all polynomials with rational coefficients are countable. "finite description" is possibly a more general term - it could be. e.g. a piece of (legalistic) prose, although one might possibly argue that the English language requires an infinite and self-referential basis as a starting point). -- Andy Smith
From: Virgil on 30 Jan 2007 15:42
In article <1170164555.633180.4440(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > Why should we repeat such self-evident truth again and again? Possibly because what seems so self-evident to WM seems so false to so many others. > > > > >> 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. WHile induction can show that the sum of any two naturals is a natural, or indeed, the sum of any finite number of naturals, by induction, there is nothing in any statement of any induction that requires the sum of an infinite number of naturals to be a natural. If WM claims that such a proof exists let him present if formally. If not, then WM should stop claiming what he cannot prove. > Every such thing including only natural > numbers is covered by induction. Only for finite sets of naturals can induction by WM, or anyone else for that matter, prove the sum of members to be a natural. If WM claims to be able to establish that the sum of members of an infinite subset of N is a member of N, let him prove his claims. > 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. WM misreads things, as usual. What everyone else is saying is that while one may sum the members of finite subsets of N to get a member of N, that one cannot sum the members of infinite subsets of N at all. > But that is phantasy. Then lets see WM find a member of N which is the sum of all members of N. |