Cantor Confusion
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From: William Hughes on 15 Jan 2007 12:03 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1168779834.276761.153130(a)a75g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > > Hardly. > > Which digit marks an infinite initial segment? Given an infinite (in your terms potentially infinite) sequence of digits (e.g Cantor's diagonal), there is one initial segment that is not "marked by a digit". This is the sequence (in your terms potentially infinite sequence) of all digits. Thus one cannot say Cantor's diagonal "consists only of finite initial segments". (Note, the fact that every element of the diagonal, is an element of a finite initial segment is not relevant). - William Hughes
From: Franziska Neugebauer on 15 Jan 2007 12:12 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> >>> An infinite union of finite sets is an infinite object. >> > >> >> But that object does *not* have an infinite _member_. > > The union of all finite trees > contains all terms of the sequence (a_n) > with n in N and a_n in {0, 1} representing t real number r. The finite set Sigma = { 0, 1 } contains (even as member) every term of the binary representation of every real in [0, 1]. Nonetheless no real of [0, 1] is member in Sigma. > The infinite path is the limit. > Are there more limits than sequences? Define limit. F. N. -- xyz
From: William Hughes on 15 Jan 2007 12:13 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >> Recall: Do get it out you must have put it in. > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > Either it does not exist - or there are infinite paths in the union of > > > finite trees. > > > > False dilemma. It "consists" of only finite initial segments (not > > finitely many therof, of course) _and_ it does exist _and_ there are > > only _finite_ paths _as_ _members_ in the union of all finite trees. > > > The union of all finite trees and the complete infinite tree are > identical with respect to nodes and edges, but note with respect to > paths. Could you please specify the asserted difference concerning the > paths p in terms of their definition? The paths are: > p = Sequence (a_n) with n in N and a_n in {0, 1} > while the real numbers are represented by: > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > > > > > > Again: To get it out you must have put it in. Since noone put infinitely > > deep trees into the union of trees it does not have such paths as > > members. > > But as limits. I cannot believe that countable many paths can have > uncountably many limits. Why not? A limit of finite paths is not defined by a single finite path but by a sequence of finite paths. There are certainly a lot more sequences of paths than there are paths. - William Hughes
From: Andy Smith on 15 Jan 2007 03:07 > > > I can define a series: > > > S = (a1-a0) + (a2-a1) + ... + (a(N-1) - a(N-2)) > > > which necessarily integrates to a(N-1)-a(0) = 1 > > Why does a(N-1)-a(0) = 1? Sorry, should have said, with a(0) = 0 and a(N-1) = 1. > > > If I could choose a (countably) infinite N, then I > can have an infinite series with a defined last term. > > Suppose a(k) = 1 - 1/2^k for each natural number k, > and suppose we set > a(w) = 1. Then what is your "series"? For each > finite k > 1 you have a > term a(k) - a(k-1), but you don't have a "last" term > because w has no > predecessor. > > > But, you will say, I cannot let N become > (countably) > > infinite, because there will be terms in the series > > that I cannot identify? > > Each point can be identified, but your > difference-of-successive-points > scheme breaks down. > > > i.e. the kth term in the sequence: > > > 0,1/2,3/4,..((2^k-1)/2^k) can be made arbitrarily > > close to 1, but cannot be 1. But even if I define > > a sequence > > > 0,1/2,3/4,..((2^k-1)/2^k),..,1 > > > by including the additional element > > I can't say that the corresponding series > > S = 1/2 + 1/4 + ... = 1 > > has a last term in it, presumably because I can't > > define it? > > There's no problem with the last member of the > sequence (1), but you > can't define a difference of successive terms at that > point. > Yes, OK. I am finally getting myself tuned in, I think. Just to clarify, you can't have a polygon with a countably infinite number of vertices (but you can have a circle with an uncountably infinite number of points) ? Also, just to confirm understanding again, mathematicians speak of an open and a closed interval e.g. [0,1] I think closed means including the points at 0 and 1 in the interval 0 to 1; maybe you use open to describe the inclusive set, whatever. Anyway, if, for example, you took the set of points from (0,1) excluding the point at 1, you would not then be able to recreate the ordered set inclusive of the point at 1 by re-appending it because you would not know where to put it? --- Andy
From: Franziska Neugebauer on 15 Jan 2007 13:21
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> > Obviuosly yoou intermingle "unary" and "binary". >> >> Indeed I mean "binary represenation". To resume: If your set of >> binary representations does not contain the representation of 1/3 >> your proof is meaningless already before starting to write it down. > > The representation of 1/3 is in the union of all finite trees. The path of 0.[01] is not a member of Sigma* with binary alphabet Sigma = { 0, 1 }. > It is the limit of a set of finite paths like N is the limit of all of > its finite initial segments. This should read: The binary representation of 1/3 is _not_ a member of the set of all finite paths _as_ N is _not_ a member of the set of all natural numbers. F. N. -- xyz |