Cantor Confusion
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From: Virgil on 22 Dec 2006 00:54 In article <1166765461.976879.105200(a)i12g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1166755731.692357.302950(a)i12g2000cwa.googlegroups.com>, > > 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? Given any listing of such paths, i.e., a mapping, f, whose arguments are natural numbers and whose values are infinite sequences of L/R symbols (the nth symbol representing branching left or right at the nth node). We construct a sequence g according to the rule that the nth branching of g is to be the opposite of the nth branching of the nth listed path. This rule of construction applied to any clearly defines a path that differs from each path in that list in at least one branching. Thus every listing omits at least one path. In fact, with some considerable modifications, this rule can create as many paths not in any list as there are paths.
From: Virgil on 22 Dec 2006 00:54 In article <1166765631.154421.111470(a)i12g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > 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. Exactly.
From: Virgil on 22 Dec 2006 00:58 In article <1166765789.824953.204330(a)79g2000cws.googlegroups.com>, "Gc" <Gcut667(a)hotmail.com> wrote: > 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. You never get to 2^aleph_0, as the number at every level is still finite. And it is a theorem in ZFC or NBG that a countable family of finite sets has a countable union.
From: Gc on 22 Dec 2006 01:02 Virgil kirjoitti: > In article <1166765789.824953.204330(a)79g2000cws.googlegroups.com>, > "Gc" <Gcut667(a)hotmail.com> wrote: > > > 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. > > You never get to 2^aleph_0, as the number at every level is still finite. You are correct. I noticed my mistake right away after posting. > And it is a theorem in ZFC or NBG that a countable family of finite sets > has a countable union. Yes I noticed that too. Every level the set of nodes is finite.
From: Gc on 22 Dec 2006 01:14
Virgil kirjoitti: > In article <1166765789.824953.204330(a)79g2000cws.googlegroups.com>, > "Gc" <Gcut667(a)hotmail.com> wrote: > > > 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. > > You never get to 2^aleph_0, as the number at every level is still finite. > > And it is a theorem in ZFC or NBG that a countable family of finite sets > has a countable union. OK. When n grows, finite E(p)/P(n) simply get`s closer to 2. The limit doensn`t concern the infinite case. Is this what the proof above really says? |