Cantor Confusion
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From: Dik T. Winter on 20 Nov 2006 23:16 In article <1163705179.604901.314550(a)m7g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1163605613.609012.99490(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1163510733.868272.250410(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > No *finite* representation. But "Cantor's list" does allow *infinite* > > representations. More specific, his diagonal proof was about lists > > where each element consisted of an *infinite* sequence of symbols. > > Why do you think that my tree does not allow infinite sequences of > symbols (nodes)? It has. But each node is a finite distance away from the root. > > > i.e., Cantor's original list contains only the set of finite sequences > > > of m and w. I agree. > > > > No "i.e." and wrong. The list was a list of *infinite* sequences. > > Why do you think that my tree does not allow to represent such > sequences with 0 and 1 instead of m a nd w? But there is *no* node that represents an infinite sequence. Each and every node represents a finite sequence. > > > Above you said "The tree contains only the set of finite binary > > > representations." > > > > Yup. There is *no* node 1/3. > > There is no symbol 1/3 in Cantors list. Again confusion. But, 0.333... is a proper symbol in Cantors list. > > But you had said that it were the nodes that represented the numbers > > in the tree, so the paths are irrelevant. > > I never said so. The nodes represent the bits 0 or 1. So a node is either 0 or 1. > > There are infinite paths > > in your tree, but they do not contain a node that represents (for > > instance) 1/3. So, if the nodes represent numbers (as you have said), > > Do you have a reference? Not needed. Just above you state that the nodes represent the bits 0 or 1. I have shown how you could concatenate the representation of a node with the representations of its parent nodes to get a number. > > 1/3 is not in your tree. You are not clear about what the numbers in > > your tree are. Are they the nodes? Are they the paths? Sometimes > > you say one thing other times you say something different. So to get > > proper understanding. What are the things that represent numbers? > > Infinite paths. I do not understand. You stated the nodes represent bits. -- 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 20 Nov 2006 23:41 Dik T. Winter wrote: > In article <MPG.1fc5e10844be8cf2989911(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > > Dik T. Winter wrote: > > > In article <1163469888.057589.117040(a)k70g2000cwa.googlegroups.com> "William Hughes" <wpihughes(a)hotmail.com> writes: > > > > So lets count the set of all natural numbers {1,2,3,...} > > > > There are no natural number left. So we stop > > > > using natural numbers and use ordinals > > > > (and to nobody's surprise a few things change). > > > > > > This is wrong. There is no ordinal needed to count the elements of the > > > set of all natural numbers. You can count until you weigh an ounce (;-)) > > > but you will never finish. Neither the elements you wish to count will > > > be exhausted nor the numbers with which you count. I think this is > > > potential infinity. On the other hand, when you ask "how many" elements > > > there are in N, you need an infinity (and this is, I think, actual > > > infinity). But all this hinges quite closely on the semantics of the > > > word "count". If seen as a process, you do not need an infinity; when > > > seen as the result of a process, you do need an infinity. In many > > > languages (German and Dutch amongst others) there are different words > > > for the two meanings, but the meanings are conflated in English. > > > > Are you discussing languages or math? What would a mathematical meaning > > for "count the set of all natural numbers" be? > > I am discussing both language and maths. In German (as in Dutch) > there is a clear distinction between "Abz?hlen" and "Z?hlen". And as > you are in discussion with somebody from German origin, it is important > to keep this in mind. The first word means the process of counting, the > second means counting to get a result. In English for both the verb > "to count" is used (and is given as translation in dictionaries). I > think "to enumerate" is a better translation of "Abz?hlen". > > So let met rephrase the position of the opponents: > To enumerate the natural numbers you do not need ordinal numbers, > this is potential infinity. To count the natural numbers you do > need ordinal numbers. This is actual infinity. I think you are saying that enumeration is the process that we use to determine a count. So, a potential infinity is an infinite process while an actual infinity is an infinite count. I'm not sure whether there is such a distinction in mathematics. Maybe a potential infinity could be defined as a function with an infinite domain while an actual infinity could be defined as an infinite ordinal. With this definition, we don't need infinite ordinal numbers to define a potential infinity, e.g., we could define f:N -> N, f(x) := x + 1. Although, the domain of such a function would still be an infinite set. Usually, when mathematics drops distinctions that were important in the past, the decision is correct. If someone tries to reinvent all of mathematics for themselves, it isn't too surprising if they don't make it up to the present, but instead get stuck in the past. -- David Marcus
From: imaginatorium on 20 Nov 2006 23:56 mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > > > I know that at most 10^100 digits of sqrt(2) can be determined, in > > > principle. > > > > In principle, if a is the sqrt(2) to 10^100 digits, then > > 0.5*(a + 2/a) is the sqrt(2) to 2*10^100 digits. > > > > What "in principle" prevents me from calculating 2/a, > > or adding it to 1, or taking 0.5*(a + 2/a)? > > Lack of bits to represent 2/a if a already requires all bits available. > > If you don't believe me, simply try it. By about 330 calculations you > should be able to produce the first 10^100 digits. If you have a slow > computer you will need one second per calculation. So let it run for 5 > minutes and you will know what prevents you from calculating 2/a. I think the bottom line here is that Mueckenheim doesn't have a clue what "in principle" means. (Amongst other things) Brian Chandler http://imaginatorium.org
From: imaginatorium on 21 Nov 2006 00:00 Lester Zick wrote: > On 20 Nov 2006 12:00:42 -0800, imaginatorium(a)despammed.com wrote: > >Lester Zick wrote: > >> On Mon, 20 Nov 2006 01:39:59 -0500, David Marcus > >> <DavidMarcus(a)alumdotmit.edu> wrote: > >> >imaginatorium(a)despammed.com wrote: > >> >> Lester Zick wrote: > >> >> > On Sat, 18 Nov 2006 18:31:05 +0000 (UTC), stephen(a)nomail.com wrote: > >> >> > >David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > >> >> > >> Lester Zick wrote: > >> >> > >>> On Thu, 16 Nov 2006 02:02:49 -0500, David Marcus > >> >> > >>> <DavidMarcus(a)alumdotmit.edu> wrote: <yawn> > >> ~v~~ > > > >Arf arf! > > Bark on, Brian. You know, the "Mystery of the Dog who Didn't Bark in > the Night"? I've read the novel. Is that what you mean? Brian Chandler http://imaginatorium.org
From: Randy Poe on 21 Nov 2006 00:30
mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > > > I know that at most 10^100 digits of sqrt(2) can be determined, in > > > principle. > > > > In principle, if a is the sqrt(2) to 10^100 digits, then > > 0.5*(a + 2/a) is the sqrt(2) to 2*10^100 digits. > > > > What "in principle" prevents me from calculating 2/a, > > or adding it to 1, or taking 0.5*(a + 2/a)? > > Lack of bits to represent 2/a if a already requires all bits available. That means it's hard in practice. But what prevents 2/a from existing in principle? > If you don't believe me, simply try it. By about 330 calculations you > should be able to produce the first 10^100 digits. If you have a slow > computer you will need one second per calculation. So let it run for 5 > minutes and you will know what prevents you from calculating 2/a. What difference does the speed of an actual computer make? We aren't talking about implementation, we're talking about "in principle". How can a represent "all bits available" IN PRINCIPLE? In principle, there is no limit on the available bits. - Randy |