Cantor Confusion
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From: mueckenh on 16 Jan 2007 09:10 William Hughes schrieb: > 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. So we need even *more* (than one) objects of the tree to define *one* limit? This makes *more* limits than objects exist? > There are certainly a lot more sequences of paths than there are paths. Every sequence *is* a path. Therefore there cannot be more sequences than paths. Regards, WM
From: Han de Bruijn on 16 Jan 2007 09:12 mueckenh(a)rz.fh-augsburg.de wrote: > Cantor's diagonal can be represented as a (diagonal) path in an > infinite matrix. This matrix is the union of finite matrices. It is the > same as with the trees. Excuse me for jumping in somewhere, but I've discovered another thing with trees. Especially with the following one, the Stern-Brocot tree: http://www.cut-the-knot.org/blue/Stern.shtml It is remarked that, on the right side of the tree, there are only the natural numbers: 1/1, 2/1, 3/1, 4/1, 5/1 .. n/1 . Each row in the tree contains 2^(n-1) fractions. When we have arrived at n/1, there are 2^n fractions in total (also counting 0/1 = 0). It is known that this tree exhausts all fractions. It exhausts all naturals as well. At each stage (n-th row) in the S-B tree, there are n naturals and 2^n fractions. We conclude that n naturals correspond with 2^n fractions. But, since the cardinality of the naturals is Aleph_0, the cardinality of the rational numbers (= "reals") must be 2^(Aleph_0): Continuum Hypothesis proved .. Han de Bruijn
From: mueckenh on 16 Jan 2007 09:18 Franziska Neugebauer schrieb: > 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 }. But the path of 1/3 = 0.[01] is a member of a binary tree which contains all possible nodes --- infinitely many. The union of all finite binary trees contans all nodes --- infinitely many. Now apply your logic. > > > 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. Correct. One could add: Cantor's diagonal number is not a member of the set of all finite initial segments of Cantor's diagonal number. (And only such initial segments occur in Cantor's diagonal proof). Regards, WM
From: mueckenh on 16 Jan 2007 09:23 Franziska Neugebauer schrieb: > Dik! > > Dik T. Winter wrote: > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > [...] > > > 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? There is a bijection between N and the set of initial segments of 0.111..., which is an isomorphism: We have f(1) = 0.1 f(2) = 0.11 .... f(n) = 0.111...1 with n numerals "1". Regards, WM
From: mueckenh on 16 Jan 2007 09:30
Virgil schrieb: > > A union cannot have more elements than are united. > > No matter how hard WM tries to make it happen. I do not try to make it happen. > > > A set of sequencs cannot have more limits than there are sequences. > > Which leaves out uncountably many of them when unoining 'finite' trees. The union of all finite trees includes all nodes. (Every node is an element of the union of all finite trees.) Every path is a subset of the set of all nodes. Every path is in the union of all finite trees. Regards, WM |