Cantor Confusion
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From: Franziska Neugebauer on 16 Jan 2007 11:29 mueckenh(a)rz.fh-augsburg.de wrote: > 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. 0.[01] (the alternating path) is a member of the infinite binary tree. > The union of all finite binary trees contans all nodes --- infinitely > many. This is obviously not the union in the sense of set theory. Set theoretically a tree is a directed graph which is defined as an ordered pair (V, E). The union of two trees is in general not a tree at all. So obviously you mean a different tree representation in terms of sets and/or a different definiens in the definition of "union". > Now apply your logic. To which definitons? >> > 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. Aha. > 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). Since no infinite string is a member of the set of all finite sequences of strings this is trivially true. So what? F. N. -- xyz
From: Andy Smith on 16 Jan 2007 01:19 David Marcus wrote: > I suppose that depends on how you define "polygon". > Usually, a polygon > has only a finite number of vertices. > Sometimes pi is estimated by computing the circumference of a regular convex polygon with N vertices, and letting N->oo . But actually such a limit does not exist? > > > 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? > > I'm sorry, but I have no clue what you mean. > Recreate? Re-append? > Wouldn't know where to put it? > All I meant was that if you have the ordered finite set e.g. {0,1,2,3,..,N} you could partition it into exclusive subsets {0,1,2,3,..,N-1} and {N}, and then, if you wished, recreate the original set by appending {N} to {0,1,2,3,..,N-1}. But when you form the open subset of [0,1) by e.g. excluding the point {1}, you could not form the ordered closed set [0,1] from [0,1) and {1} - because [0,1) does not have a last member, so you would not know where to append {1}. --- Andy
From: Andy Smith on 16 Jan 2007 01:37 David Marcus wrote: > Try Googling "usenet news reader". I'm using Gravity > on Windows. I've > used it since before it became free. To use a news > client, you need to > ask your ISP for the name of their news server. > > Ty. I shall investigate --- Andy
From: Franziska Neugebauer on 16 Jan 2007 12:01 mueckenh(a)rz.fh-augsburg.de wrote: > 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". True. With Sigma = { 1 } (unary) and Sigma = { 0, 1 } (binary) the cardinality of Sigma* is card(omega) which is equivalent to "Sigma* is countable" and "a bijection between Sigma* and omega exists". The issue is that 0.[01] is not in Sigma*. F. N. -- xyz
From: Franziska Neugebauer on 16 Jan 2007 12:07
William Hughes wrote: > mueckenh(a)rz.fh-augsburg.de wrote: >> William Hughes schrieb: >> >> > 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. >> >> Given the sequence of initial segments of binary 1/3: >> 0.0 >> 0.01 >> 0.010 >> 0.0101 >> ... > > No, this is not a list of all the initial segments of binary 1/3. > You are missing one. > The (potentially) infinite sequence of all digits is not in this > list. (cough) Is this any longer compatible with the usual definition of initial segment? http://mathworld.wolfram.com/InitialSegment.html F. N. -- xyz |