Cantor Confusion
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From: stephen on 22 Dec 2006 15:49 Newberry <newberry(a)ureach.com> wrote: > 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))) Every path passes through an infinite number of edges. There are two edges incident on the root, and every path passes through one or the other of these two edges. So you are not going to find a bijection in such a naive fashion. Stephen
From: Virgil on 22 Dec 2006 16:00 In article <1166796951.212591.257990(a)73g2000cwn.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > 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. If we have a contradiction the at least one of the assumptions leading to that contradiction is false. The assumptions involved in this contradiction are (1) the axioms of ZFC and (2) there is a surjection from the set of edges (or nodes) of an infinite binary tree to the set of paths of that tree. As I have nowhere seen any alleged construction of such a surjection that was not terminally flawed, I prefer to believe that no such surjection can exist.
From: Lester Zick on 22 Dec 2006 16:52 On Thu, 21 Dec 2006 15:22:56 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Dik T. Winter schrieb: >> >> > Indeed. No axiom is necessarily true. Fortunately nothing you say is necessarily true. >> How then should the truth of AC be proved? > >One does not prove axioms, one either assumes them or does not assume >them. Unless unlike neomathematikers one is not too lazy or stupid to prove them true. >For a given axiom system, those statements in that system which have >been proved from those axioms are called theorems, at least within that >system. So instead of truly true truth we have modely modeled models. ~v~~
From: Lester Zick on 22 Dec 2006 16:54 On 21 Dec 2006 17:08:14 -0800, cbrown(a)cbrownsystems.com wrote: > >Virgil wrote: >> In article <1166737382.495151.299530(a)i12g2000cwa.googlegroups.com>, >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Newberry schrieb: >> > > >> > > 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. >> >> Except that the only demonstration of such "inacceptability" would be >> the discovery of some internal contradiction within ZFC or NBG. >> > >You are discounting the fact that for WM, an assertion is unacceptable >iff it "feels" unacceptable. Whereas for you an assertion is acceptable iff it is an assertion. ~v~~
From: Lester Zick on 22 Dec 2006 16:54
On Thu, 21 Dec 2006 18:37:09 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <1166749694.734273.263470(a)73g2000cwn.googlegroups.com>, > cbrown(a)cbrownsystems.com wrote: > >> Virgil wrote: >> > In article <1166737382.495151.299530(a)i12g2000cwa.googlegroups.com>, >> > mueckenh(a)rz.fh-augsburg.de wrote: >> > >> > > Newberry schrieb: >> > > >> >> > > > 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. >> > >> > Except that the only demonstration of such "inacceptability" would be >> > the discovery of some internal contradiction within ZFC or NBG. >> > >> >> You are discounting the fact that for WM, an assertion is unacceptable >> iff it "feels" unacceptable. >> >> Cheers - Chas > >Damn right! Such emotionally inspired blindness to the logic of his own >weird claims is itself unacceptable. How can an assertion be unacceptable? ~v~~ |