Cantor Confusion
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From: Dik T. Winter on 15 Jan 2007 20:02 In article <45abc73c$0$97265$892e7fe2(a)authen.yellow.readfreenews.net> Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> writes: > Dik T. Winter wrote: > > mueckenh(a)rz.fh-augsburg.de writes: > > > 0.111... and its initial segments are *representations* (according > > > to you, they cannot be numbers). They can represent natural numbers > > > as well as certain real numbers. Obviously both sets of numbers are > > > isomorphic. > > > > What is the isomorphism? And 0.111... does *not* represent a natural > > number in unary notation. Do you claim there is an isomorphism > > between the set of natural numbers and the set of real numbers: > > {1/2, 3/4, 7/8, ..., 1/9} > > That is false. There is a bijection, but that bijection does *not* > > work through the representation. > > There _is_ a bijection? Between which sets? WM says that they can represent both sets of (natural intended here) numbers and certain real numbers. There is a bijection between the natural numbers and the set I quote above {(2^n-1)/(2^n), 1/9}. But you do not get it when you go from the rational number to its binary representation and use that as a unary representation for the natural numbers. 1/9 is left out. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Jan 2007 21:17 In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: .... > If f(x) is such that at any eta, f(eta) = -f(-eta), > this would imply that f(0) = 0? No (assuming any eta > 0). It would only imply that lim{x->0} f(x) = 0. > If f(0) was anything other than 0, the antisymmetry would be destroyed? It could be anything. And if f(0) is defined as anything different from 0 there is no antisymmetry. > How could something that is antisymmetric not be other than 0 > at its mirror point - even if it is multivalued, it will retain > antisymmetry in inverting x ? Wrong. It can also be undefined. Consider the function 1/x. It is definitely antisymmetric on its range: R \ {0}. But 1/0 is undefined, and certainly not equal to 0, so the range does not include 0. > Alll this is a bit "angels on a pin", but is ultimately important > I think from a philosophical perspective on the nature of infinity. I think not. The whole idea stems from the basic idea that a true function should be continuous. You may think that only continuous functions are worth considering, but in that case you have to omit sin(1/x), because that one is not continuous for x = 0. There is even no way to make it continuous at x = 0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 15 Jan 2007 21:45 Dik T. Winter wrote: > In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: > ... > > If f(x) is such that at any eta, f(eta) = -f(-eta), > > this would imply that f(0) = 0? > > No (assuming any eta > 0). It would only imply that lim{x->0} f(x) = 0. ?? -- David Marcus
From: David Marcus on 15 Jan 2007 21:50 mueckenh(a)rz.fh-augsburg.de wrote: > Here is the formal proof: > > Theorem. The set of real numbers in [0, 1] is countable. > > Lemma. > Each digit a_n of a real number r of the real interval [0, 1] in binary > representation has a finite index n. > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > > Proof. > A natural number n can be represented in a special unary notation: n = > 0.111...1 with n digits 1 (the leading 0. playing no role). Example: 1 > = 0.1, 2 = 0.11, 3 = 0.111, ... > In this notation the definition of the set of natural numbers, (1, 2, > 3, ...} = N, reads > > {0.1, 0.11, 0.111, ...} = 0.111.... (*) > > Note that also the union of all finite initial segments of N, {1, 2, 3, > ..., n}, is N = {1, 2, 3, ...}. Therefore (*) can also be interpreted > as union of initial seqments of the real number 0.111.... > > A real number r of the real interval [0, 1] can be represented as one > (ore two) path in the infinite binary tree. The set of all real numbers > r of the real interval [0, 1] is then given by the infinite binary > tree: > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > ................... > > A finite binary tree is the infinite binary tree, cut off below a level > n with n in N. > Here is a tree with two levels: > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > namely level 1 and level 2. (The root at level 0 is conveniently not > counted, because 0 is not a natural number.) > The union of binary trees is defined as the union of levels. > The union of two or finitely many different finite binary trees simply > is the largest on. > Taking the uninion of all finite binary trees, we get the complete > infinite binary tree with all levels. All infinite paths representing > real numbers r of the real interval [0, 1] are in this union. We can > see this by the path always turning right, 0.111..., which is present > in the tree, according to (*). > > Conclusion: Every finite binary tree contains a finite set of path. The > countable union of finite sets is countable. The set of paths is > countable. The set of real numbers in [0, 1] is countable. QED. Using the same axioms and principles that you used in this proof, can you prove the following? There are only finitely many natural numbers -- David Marcus
From: Dik T. Winter on 15 Jan 2007 22:26
In article <MPG.20160647a0a74784989b3d(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > Dik T. Winter wrote: > > In article <18893272.1168906211385.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: > > ... > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > > > this would imply that f(0) = 0? > > > > No (assuming any eta > 0). It would only imply that lim{x->0} f(x) = 0. > > ?? It appears to be clear to me. When the eta are != 0, you can say nothing about f(0). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |