Cantor Confusion
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From: Newberry on 21 Dec 2006 22:15 mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Newberry schrieb: > > > > > > > > > > > > > > > > >The edges have cardinal number aleph_0. > > > > > > > > > > > > Certainly. > > > > > > > > > > > > > A > > > > > > > subset of the paths can be shown to be in bijection with the real > > > > > > > numbers of the interval [0,1]. > > > > > > > > > > > > Which subset? Path as a set of intervals or path as a set of eges? How > > > > > > does the above follow from this? > > > > > > > > > > The paths are binary representations of all real numbers of the > > > > > interval [0,1]. Some rational reals have double representations, namely > > > > > such ending by 000... and such ending by 111... like 1.000... and > > > > > 0.111... . So there are not less real numbers in the interval [0, 1] > > > > > than paths in the binary tree. We have > > > > > > > > > > aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 > > > > > > > > > > while according to set theory > > > > > > > > > > aleph_0 < 2^aleph_0 > > > > > > > > > > Regards, WM > > > > > > > > Well, a Cantorist would say that this part does not hold > > > > > > > > |{edges}| >= |{paths}| > > > > > > > > because you considered only the finite paths but not the infinite paths. > > > > > > In the infinite tree there are no finite paths. > > > > Yes, but you arrived at |{edges}| >= |{paths}| by calculating > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > In this limit all the n are finite. So a Cantorist would claim that you > > examined only the finite sub-paths, > > > Every real number (like, e.g., pi) has digits r_1, r_2, r_3, ..., only > at finite digit positions: 1,2, 3, ...(because an infinite digit > position could not be identified and the corresponding digit could not > be determined).. > > > and not the infinite paths. I do > > not know if this Cantorist claim makes sense, > > It helps them to veil the contradictions appearing in set theory, > because they can easily jump from "every natural number is finite" to > "there are infinitely many natural numbers". > > > but the question is > > whether you can generate a direct contradiction of the type P & ~P in > > ZFC. > > I do not think that this is of any interest after it can easily be > shown that the results of ZFC are inacceptable. I just needed to understand what we were talking about: unacceptability or formal contradiction. But it is important. Showing P & ~P is more unacceptable. Compare for instance > the resuls I have recently prepared in a discussion with Virgil: > > You accept a diagonal of a matrix longer than any line. > You accept more separated paths than splitting positions (= origins of > separated paths) [in the binary tree]. > You accept that 1 + 1/2 + 1/4 + ... > 10 is possible. > And at last you accept the existence of an actually infinite finite > number. > > Enough? Well, no. I apoligize for having joined this discussion too late. There are 4000 entries and I cannot go through all of them. I do not know if Virgil accepted those absurd statements above. When I was thaught the diagonal argument in school my first reaction was to construct a binary tree and start numbering the nodes from top to bottom and from left to right. They laughed maliciously and said "you left out all the infinite paths." I said "I exhausted the whole tree." They said "no, you did not., there are finite subsets, co-finite subsets, and infinite subsets." And I saw how dumb I was. Again, I do not know if this argument makes any sense. But it is a verbalization of the fact that in ZFC the diagonal argument and numbering of the tree nodes can co-exist without generating a formal contradiction. It seems to me though that your "geometric" argument (mapping the paths on the edges) shows the Cantorist verbalization to be entirely implausible. It is clear that the entire infinite paths have been mapped and that all of the paths have been mapped. There is nothing left out. > > Regards, WM
From: Dik T. Winter on 21 Dec 2006 22:48 In article <1166755731.692357.302950(a)i12g2000cwa.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > Dik T. Winter wrote: .... > > > It is irrelevant for my proof. It is as irrelevant (and silly) as the > > > question: in how many shares is the unit divided in the last term of > > > the series 1/2^n. Nevertheless we can calculate the limit of this > > > series. > > > > It is very relevant for the proof. You assume you can split in shares and > > later on recombine those shares. When each edge is divided in finitely > > many shares that is allowed. It is not allowed if you divide in infinitely > > many shares, which you are doing. And the series 1/2^n is not relevant. > > We can calculate the limit of that series because for every eps there is > > an n0 such that for n > n0 the finite sum is on a distance not larger > > than eps from 2, there are no shares involved at all. > > The issue here is whether the paths can be mapped onto the edges. More > precisely can each path be mapped on a set of edge segments such that > all the sets are disjoint. It seems to me that it can. You are wrong. Wolfgang claimed a surjection from the edges to the paths. I.e. can we map the edges to the paths. There are surjections from the paths to the edges, but Wolfgang claimed the other way around. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Newberry on 21 Dec 2006 23:39 Dik T. Winter wrote: > In article <1166755731.692357.302950(a)i12g2000cwa.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > Dik T. Winter wrote: > ... > > > > It is irrelevant for my proof. It is as irrelevant (and silly) as the > > > > question: in how many shares is the unit divided in the last term of > > > > the series 1/2^n. Nevertheless we can calculate the limit of this > > > > series. > > > > > > It is very relevant for the proof. You assume you can split in shares and > > > later on recombine those shares. When each edge is divided in finitely > > > many shares that is allowed. It is not allowed if you divide in infinitely > > > many shares, which you are doing. And the series 1/2^n is not relevant. > > > We can calculate the limit of that series because for every eps there is > > > an n0 such that for n > n0 the finite sum is on a distance not larger > > > than eps from 2, there are no shares involved at all. > > > > The issue here is whether the paths can be mapped onto the edges. More > > precisely can each path be mapped on a set of edge segments such that > > all the sets are disjoint. It seems to me that it can. > > You are wrong. Wolfgang claimed a surjection from the edges to the paths. > I.e. can we map the edges to the paths. There are surjections from the > paths to the edges, but Wolfgang claimed the other way around. OK, and why exactly can't we map the edges onto the paths? > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Franziska Neugebauer on 22 Dec 2006 00:04 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: ,----[<4589b948$0$97268$892e7fe2(a)authen.yellow.readfreenews.net> ] | >> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] | >> >> | >> >> > If an unbounded diagonal exists, then obviously an | >> >> | >> >> > unbounded line must exist. [(*)] | [...] | >> >> That does not prove (*) since the whole diagonal is not bounded | >> >> by any such n. | >> > | >> > If you doubt my proof, | >> | >> Which proof? You simply stated: | >> | >> (1) forAll elem in diag Exists line (m is in line) | >> (2) forAll elem in diag (elem is finite natural number) | >> (3) forAll natural n (forAll elem in diag <= n | >> is in line n) | >> | >> So please show us step-by-step how (*), which states | >> | >> (*) notExists natural n (forAll m in diag (m <= n)) | >> -> Exists natural n' (notExists natural n(forAll m | >> in line n' (m <=n)), | >> | >> follows from (1), (2) and (3). | > | > See my recent reply to William Hughes. | | Please copy and paste. I am not able to guess what you have in mind. `---- [...] > We talked about the potentially infinite set N. >> Which means that for any n we can find a single line L(n) that works >> for this n. It does not mean that that we can find a single >> line L_D, that works for all n. > > That is impossible, beause there is no "all n". If you can't proof (*) you should write so. > If "all n" were somewhere, then we had no problem to put them in one > single line. We are not disputing "all n are somewhere" but the validity of (*). Until now (*) is not valid. > Due to the finiteness of every n, every position of the > line was finite. And due to the fact that a number with only finite > indexes is a natural number, the whole number was a natural number. I don't know hat "whole number" you are writing about but it seems to me that you compulsively try to apply your well-known unproved AND invalid convictions ("N cannot be infinite because A n e N (n is finite)"). > Recently we had the following claim (by Virgil) > > The number of equations > > 1+1+1+...+1 = n > > with finitely many 1 is infinite. Considering the fact that there are > exactly as many 1 as different numbers, it is impossible that the > number of numbers can be infinite while the sme number of 1 is finite. The number of sums (left to the "=") and the number of n's (right to the "=") is not bounded. What precisely is your problem? [...] > In short: If any natural number n does exist, then there exists a line > in which this and any smaller number does exist. > This holds for every number n. Every natural n. > You know that the logical symbol for every is forall. I don't see how this could support your "famous" quantifier reversal. Hence that does not prove the existence of a _single_ line (*). F. N. -- xyz
From: Virgil on 22 Dec 2006 00:06
In article <1166755731.692357.302950(a)i12g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Dik T. Winter wrote: > > In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1166617425.221440.15570(a)80g2000cwy.googlegroups.com> > > > > mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > > > We are talking about shares, but we do not need more than 3/2 per > > > > > path. > > > > > > > > 3/2 of what? But again you do not answer my question. In how many > > > > shares is the edge going left from the root divided, and how do you > > > > do that division? > > > > > > It is irrelevant for my proof. It is as irrelevant (and silly) as the > > > question: in how many shares is the unit divided in the last term of > > > the series 1/2^n. Nevertheless we can calculate the limit of this > > > series. > > > > It is very relevant for the proof. You assume you can split in shares and > > later on recombine those shares. When each edge is divided in finitely > > many shares that is allowed. It is not allowed if you divide in infinitely > > many shares, which you are doing. And the series 1/2^n is not relevant. > > We can calculate the limit of that series because for every eps there is > > an n0 such that for n > n0 the finite sum is on a distance not larger > > than eps from 2, there are no shares involved at all. > > The issue here is whether the paths can be mapped onto the edges. More > precisely can each path be mapped on a set of edge segments such that > all the sets are disjoint. It seems to me that it can. The issue here is whether for an infinite binary tree the set of edges (or, equivalently, the set of nodes) can be surjected to the set of paths, which it cannot. |