Cantor Confusion
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From: Virgil on 21 Nov 2006 19:20 In article <1164143247.933300.280980(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > Do you want to posit, that there is some line identical to the diagonal? > > That is not (yet) proven. > > > The number of elements of the diagonal is larger than all n: > E omega A n : omega > n. > > The lines have only finitely many elements. > > In a linear order of finitely many elements n exactly the following is > implied: > If for every n there exists an line L(n) such that L(n) contains all > elements m <=n > then there exists a line L such that for every n, L contains all > elements m <=n . > > This is valid for the finitely many elements of every line. (How many > lines there are is irrelevant.) As long as that "number" is finite. > > All elements of the diagonal are elements of the lines UL(n). > Therefore all elements of the diagonal are elements of a finite line. Consider "For all m in N there is an n in N such that D(m) in L_n" and "There is an n in N such that for all m in N D(m) in L_n." The former is trivial, the latter is false. This can more easily be seen in comparing "For all m in N there is n in N such that n > m" which is true with "There is an n in N such that for all m in N, n > m" which is false. Those who claim the former establishes the latter, or who fail to distinguish between the two, exhibit quantifier dyslexia. Some do it deliberately, knowing it to be wrong, others just do not know any better. In which category does WM belong?
From: Virgil on 21 Nov 2006 19:25 In article <1164143353.398396.215510(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > Perhaps D. Marcus' understanding in > > <MPG.1fcbb0f0d873ccdd989977(a)news.rcn.com> > > may help you to cope with my sentence under discussion. > > > You wrote: > 1) "The cardinality of omega is |omega| not omega." > And after learning from me that this is wrong, > 2) "The cardinality of omega, also written as |omega|, is omega". > > Now I am interested whether or not this is a contradiction in your > eyes. WM is guilty of much more serious trangressions of both logic and common sense that is involved with whether omega and |omega| are the same. Among others, WM's trangressions include repetitive instances of quantifier dyslexia.
From: Virgil on 21 Nov 2006 19:29 In article <1164143519.034139.154960(a)h54g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > > What about all contructible numbers or all words? They cannot be mapped > > > on N though they are a countable set. > > > > The set of all finite words, the set of polynomials, or any > > countable set can be put in bijection with N. Where do you > > get your view that it "can't be mapped to N"? > > The list of all finite words cannot be constructed. It can be constructed inductively, which means that in ZF or NBG, it can be constructed. > The list of all > constructible numbers cannot be constructed. The list of all naturals can be constructed inductively, which means that in ZF or NBG, it can be constructed. > That means, these > bijections cannot be constructed. In ZF they can! > > Regards, WM
From: Virgil on 21 Nov 2006 19:31 In article <1164143629.027905.46720(a)m7g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > The problem is the difference between saying > > > > The set of all natural numbers exists > > and > > We can generate an arbitrarily large set > > of natural numbers > > > > (i.e. the difference between actual and potential infinity) is > > mostly one of terminology. > > No. The first can be disproved, the second cannot be disproved. In what axiom system can the first be disproved?
From: David Marcus on 21 Nov 2006 19:35
Virgil wrote: > In article <1164143109.673986.202100(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > Note however that in order to attach a definite cardinal number to > > > > every set, well-ordering must be possible for every set. That is why > > > > Cantor insisted on well-ordering of all sets. > > > > > > Piffle. Well-ordering has nothing to do with the existence or not of > > > at least one bijection. > > > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > > definite cardinal number. > > The existence of a cardinal which bijects to a set S is not a necessary > condition for the identity function on S to exist. Every set, whether > well orderable or not, has an identity function, at least in such > relatively sane systems as ZF and NBG. > > If WM wishes to propose a system in which sets can lack identity > functions, he will have to explicitely present the axiom system for such > a system before anyone will accept its existence, much less its > consistency. And, much less its sanity. -- David Marcus |