Cantor Confusion
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From: Virgil on 22 Dec 2006 00:19 In article <1166757335.711810.253790(a)42g2000cwt.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > I do not think that this is of any interest after it can easily be > > shown that the results of ZFC are inacceptable. Unacceptable to whom? And by what standard? So far it is only unacceptable to those who insist on taking as an axiom the negation of some axiom or theorem of ZFC. > 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. In ZFC, where one has existence of infinite binary trees, every path in such a tree models an example of such a "matrix". The path itself is the "diagonal". The "lines" are the finite subpaths from the root node of the tree to any node of the given path. Every line is a proper subpath of that diagonal, so is shorter than the diagonal. > > 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. Not in the way that WM misrepresents them at any rate. > > 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. Wm either sneaks in his requirement that the paths be only countably infinite or glissades over the details of how he can divide up his countably many edges to be recombined into uncountably many paths.
From: Virgil on 22 Dec 2006 00:28 In article <1166762377.524661.268400(a)f1g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > 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? There aren't enough edges ( or nodes). Each node can be represented uniquely by a finite string of left/right branchings which carries you from the root node to the node in question, with the empty string being the root node itself, and each edge by a finite non-empty sequence terminating at it terminal node. It has been shown many times that there are only countably many such strings. In the same manner of representation, it is clear that every different infinite sequence of such left/right branchings represents a different infinite path in the tree. It has been shown many times and in many ways that the set of such strings is not countable, in the sense that there is no way of surjecting the natural numbers onto that set.
From: Newberry on 22 Dec 2006 00:31 Virgil wrote: > 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. Why not?
From: Newberry on 22 Dec 2006 00:33 Virgil wrote: > In article <1166762377.524661.268400(a)f1g2000cwa.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > 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? > There aren't enough edges ( or nodes). > > Each node can be represented uniquely by a finite string of left/right > branchings which carries you from the root node to the node in question, > with the empty string being the root node itself, and each edge by a > finite non-empty sequence terminating at it terminal node. > > It has been shown many times that there are only countably many such > strings. > > In the same manner of representation, it is clear that every different > infinite sequence of such left/right branchings represents a different > infinite path in the tree. > It has been shown many times and in many ways that the set of such > strings is not countable, in the sense that there is no way of > surjecting the natural numbers onto that set. Right. So if he edges can be mapped onto the paths we have a contradiction.
From: Gc on 22 Dec 2006 00:36
mueckenh(a)rz.fh-augsburg.de kirjoitti: > Gc schrieb: > > > Every two > > paths separete on some finite level edge, but only when you got the > > whole countably infinite path (the union of it`s all finite subpaths > > starting from the beginning) all infinite long pathes separate from > > each other. > > Of course. And therefore all the paths can be counted by the number of > split positions. No. There is no point (except union of all the points) where a path separates from all the other pathes and becomes a unique path. The number of split positions isn`t therefore sufficent, you need infinite unions of them and your argument fails. > But we can look at the problem also from another point of view: > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > different pathes and E(n) = 2*2^n - 2 different edges. So we find > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > E(n)/P(n) >= 1 in any case. Hence, the assertion "E(n)/P(n) < 1 in an > actual infinite tree" has been disproved by simplest application of > mathematics. Why the set of edges needs to be countable? In the first level you have 2 edges, in the second 4 and so on and finally you got 2^aleph edges. > Regards, WM |