Cantor Confusion
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From: Newberry on 22 Dec 2006 20:25 Virgil wrote: > 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. Does each path pass through at least one node? Ax(Path(x) -> Ey(Through(x,y)) & Node(y))
From: Newberry on 22 Dec 2006 20:33 stephen(a)nomail.com wrote: > 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. So the answer is yes. If every path passes through at least one edge does it follow that the number of paths is less or equal than the number of edges? P = {x|Path(x)]; E = {y|Edge(y)} |P| <= |E| ? > 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 20:43 In article <1166837125.355950.83470(a)f1g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > 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. > > Does each path pass through at least one node? > Ax(Path(x) -> Ey(Through(x,y)) & Node(y)) Each infinite path starts at the root node (or initial node) and from there passes through an infinite alternating sequence of edges and nodes. At each node of a binary tree, except terminal nodes if any, one path follows an edge which branches 'left' and another follows an edge branching 'right'. In a finite tree, every path ends in a terminal node, and the maximum number of edges in any path of such a tree is called the order of the tree. In an /infinite/ binary tree there are no terminal nodes at all, and no path comes to an end..
From: Jesse F. Hughes on 22 Dec 2006 20:47 "Newberry" <newberry(a)ureach.com> writes: > So the answer is yes. If every path passes through at least one edge > does it follow that the number of paths is less or equal than the > number of edges? > > P = {x|Path(x)]; E = {y|Edge(y)} > |P| <= |E| ? Why would that follow? -- Jesse F. Hughes "Sigh. Back to figuring out how to solve the factoring problem and ending the world as we know it." -- James S. Harris
From: Virgil on 22 Dec 2006 20:58
In article <1166837581.381928.281240(a)i12g2000cwa.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > If every path passes through at least one edge > does it follow that the number of paths is less or equal than the > number of edges? Not when uncountably many paths pass through each edge. |