Cantor Confusion
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From: mueckenh on 15 Dec 2006 02:21 cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > The ordinal 2^omega is a countable set. Even omega^omega is a countable > > set. > > You could learn this elementary knowledge from my book: > > http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 > > > > Unfortunately, following this link led me to a page saying "Das Buch > ist nicht in unserer Datenbank gespeichert", which I don't understand, > but guess means "That book is not in our database, sadly". Unfortunately this link was truncated. I try to post it again. http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 Regards, WM
From: Han de Bruijn on 15 Dec 2006 02:57 William Hughes wrote: > Han de Bruijn wrote: > >>William Hughes wrote: >> >>>Han de Bruijn wrote: >>> >>>>Let's repeat the question. Does there exist more than _one_ concept of >>>>infinity? Isn't unbounded the same as infinite = not finite = unlimited >>>>= without a limit? Please clarify to us what your "honest" thoughts are. >>> >>>You are *way* in deficit on clear answers. Try answering >>>the following question with yes or no. >>> >>> Is there a largest natural number? >> >>No. > > I there an unbounded set of natural numbers? Suppose you mean "Is". What does it mean that a set is unbounded? http://groups.google.nl/group/sci.math/msg/13795822737a77ca Han de Bruijn
From: Han de Bruijn on 15 Dec 2006 03:14 Virgil wrote [ his gospel, worth to be quoted as a whole ]: > Unless one is working in some system in which the axioms are set out in > advance, one is operating in a fog. > > Without some assumptions (other than the rules of logic) one cannot > deduce anything about anything. So one must start with SOME assumptions. > > It is sensible to require all those assumptions to be explicitly stated, > so that one can be sure of precisely what is being assumed. Such > collections of statements are called axiom systems. > > And it is plain that no sound mathematics can be developed unless based > on some axiom system as its solid foundation. > > The only issue then is what system(s) of axioms. Disagreed with almost anything above. Look up: http://web.maths.unsw.edu.au/~norman/views2.htm And go to the section "Does mathematics require axioms?". > For arithmetic and analysis, ZFC and NBG have been found useful. > > For geometry, Hilbert's refinement of the Euclidean axioms has been > found useful. > > It would be of great advantage to find a single axiom system sufficient > to cover all of mathematics, but that has not yet been shown to be > possible. > > However ZFC and NBG have each been shown to be sufficient for a great > deal of mathematics, which is why so much attention is paid to them. > > There are some other possibilities along the lines of Category Theory or > Topos theory that I understand are in the running as well, but are as > well developed. Current axiom systems are almost exclusively associated with Set-like theories. I've argued more than once that this starting point is very much socially biased. I call it the Warehouse or Supermarket Paradigm: http://hdebruijn.soo.dto.tudelft.nl/jaar2006/Supermarkt.jpg Within a society unlike Capitalism, such a bias for mathematics would only be half of the truth. The other half would be a Labour Paradigm, as is implemented, basically, in nowadays Constructivism. Han de Bruijn
From: Han de Bruijn on 15 Dec 2006 03:15 Virgil wrote: > In article <d1914$45811514$82a1e228$2825(a)news1.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > >>William Hughes wrote: >> >>>Therefore, as there are at least as many digit positions in 0.111... >>>as there are primes, there are an infinite number >>>of digit positions in 0.111... >> >>Binary or decimal? > > Irrelevant! So? Han de Bruijn
From: Virgil on 15 Dec 2006 03:18
In article <1166166146.392714.316100(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > There is a mapping. This is proven by the fact that there are more > > > edges than paths. > > > > But the latter is simply false. Moreover, you have stated that you > > had constructed a mapping. That statement was wrong? > > A bit sloppy. Being precise, I emphasized: I have construcuted a > surjective rational relation from a subset of the set of edges onto the > set of paths. This surjective rational relation proves the existence of > a surjection. Not to anyone except yourself. > > The notion "rational relation" is a further development of the notion > "relation". While a relation connects elements of a domain to elements > of a range, the rational relation connects shares of the elements of > the domain to (shares of the) elements of the range. A simple example: > > A) Consider the relation: > > 1 ---> a > 2 ---> b > > B) Now a rational relation could be: > > 1.1 ---> a > 1.2 ---> b > > 2.1 ---> a > 2.2 ---> b > > If 1.1 and 1.2 are the only shares 1, i.e., if 1.1 u 1.2 = 1 and if 2.1 > and 2.2 are the only shares of 2, i.e., if 2.1 u 2.2 = 2, then one full > element of the domain is available for every element of the range. > Therefore B proves that a surjective mapping like A is possible, even > if it cannot be found. > > This proof is similar to the proof of a well-ordering of the real > numbers. It cannot be done (and it can even be proved that it cannot be > done) but it has been proved to exist. > > The rational relation devised in the binary tree proves that a > surjection of edges onto paths exists. In so far my assertion "there is > a surjective mapping" was not quite wrong. > > > > But you have not proven that the tree does not contain more paths than > > edges. > > See above. We see it but do not believe that it does what you say it does. Your example assumes a bijection and constructs a "rational relation". But in the absence of any original bijection there is no assurance of any final bijection by such construction. And assuming what you are trying to prove is a no-no. |