Cantor Confusion
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From: Virgil on 22 Dec 2006 01:47 In article <1166768074.679477.239530(a)73g2000cwn.googlegroups.com>, "Gc" <Gcut667(a)hotmail.com> wrote: > 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? Yes. WM tries to argue that it always must be the case that Lim_{n -> oo} f(n)/g(n) =(Lim_{n -> oo} f(n)/(Lim_{n -> oo} g(n)) But if that were so, then endless paths wold have to have a last node.
From: Han de Bruijn on 22 Dec 2006 04:25 William Hughes wrote: > "the Naturals Construction Set" is not a definition, it is a property > that a set of natural numbers might or might not have. It also sheds > no light on the question of whether a potentially infinite set > of natural numbers exists as it is trivially obvious that both > bounded and unbounded sets can have "the Naturals Construction Set" > property. Yes. Nice, huh .. What else would be needed? > The question is not longer "are you trying to avoid the question" > but "why are you avoiding the question". Why should I put effort in trying to answer ill-defined questions? > However, the answer > to this lies in the psychology of the crank and is not > of present interest. Ah, your argumentation ends _there_. Now I wonder why you didn't simply START with "you are a crank". As has become the argument par excellence of the mainstream mathematics community for quite some time. Excuse me if I'm not much impressed by such "arguments". Look: you and your kind are rapidly becoming a minority. Han de Bruijn
From: Newberry on 22 Dec 2006 09:15 Virgil wrote: > 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. If we have a contradiction then ZFC is inconsistent.
From: Jesse F. Hughes on 22 Dec 2006 09:25 "Newberry" <newberry(a)ureach.com> writes: > Virgil wrote: >> 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: >> > > >> > > > 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. > > If we have a contradiction then ZFC is inconsistent. > Yes, if we can map edges onto paths, then we have a contradiction and if we have a contradiction, ZFC is inconsistent. But we can't map edges onto paths. -- Jesse F. Hughes "This Trojan appears to utilize a function of the Windows Media DRM designed to enable license delivery scenarios as part of a social engineering attack." -- MS candidly explains the role of DRM licenses
From: Newberry on 22 Dec 2006 15:44
Jesse F. Hughes wrote: > "Newberry" <newberry(a)ureach.com> writes: > > > Virgil wrote: > >> 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: > >> > > > >> > > > 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. > > > > If we have a contradiction then ZFC is inconsistent. > > > > Yes, if we can map edges onto paths, then we have a contradiction and > if we have a contradiction, ZFC is inconsistent. > > But we can't map edges onto paths. Does each path pass through at least one edge? Ax(Path(x) -> Ey(Through(x,y))) > > -- > Jesse F. Hughes > "This Trojan appears to utilize a function of the Windows Media DRM > designed to enable license delivery scenarios as part of a social > engineering attack." -- MS candidly explains the role of DRM licenses |