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From: Newberry on 22 Dec 2006 22:51 stephen(a)nomail.com wrote: > 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. > > > What is wrong with WM's mapping? You divide each edge into two halves > > and pass one half to each branch. Then you divide the passed half again > > and again. > > So which edge is mapped to the path that always goes left? > Remember, you need to uniquely map each path to an edge. No, I don't. What I am showing here is that each path will accumulate the weight of two edges as it approaches infinity. > So which edge gets mapped to the path that always goes left? > > Stephen
From: stephen on 22 Dec 2006 22:55 Newberry <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: >> 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. >> >> > What is wrong with WM's mapping? You divide each edge into two halves >> > and pass one half to each branch. Then you divide the passed half again >> > and again. >> >> So which edge is mapped to the path that always goes left? >> Remember, you need to uniquely map each path to an edge. > No, I don't. What I am showing here is that each path will accumulate > the weight of two edges as it approaches infinity. That is not a mapping. You need to construct a surjection. Showing that each path will accumalate the weight of two edges has nothing to do with constructing a mapping. Stephen
From: Newberry on 22 Dec 2006 23:52 stephen(a)nomail.com wrote: > Newberry <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > >> 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. > >> > >> > What is wrong with WM's mapping? You divide each edge into two halves > >> > and pass one half to each branch. Then you divide the passed half again > >> > and again. > >> > >> So which edge is mapped to the path that always goes left? > >> Remember, you need to uniquely map each path to an edge. > > > No, I don't. What I am showing here is that each path will accumulate > > the weight of two edges as it approaches infinity. > > That is not a mapping. Correct. You need to construct a surjection. No, I don't. > Showing that each path will accumalate the weight of two edges > has nothing to do with constructing a mapping. But it shows that in an infinite tree there are two edges per path. > > Stephen
From: Virgil on 23 Dec 2006 00:28 In article <1166844479.124289.59150(a)48g2000cwx.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. > > What is wrong with WM's mapping? You divide each edge into two halves > and pass one half to each branch. Then you divide the passed half again > and again. But to do that, you must divide each edge into as many pieces as there are paths using it, and that is the same as the total number of paths, which number is not countable.
From: Virgil on 23 Dec 2006 00:30
In article <emi80g$e09$1(a)news.msu.edu>, stephen(a)nomail.com wrote: > 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. > > > What is wrong with WM's mapping? You divide each edge into two halves > > and pass one half to each branch. Then you divide the passed half again > > and again. > > So which edge is mapped to the path that always goes left? > Remember, you need to uniquely map each path to an edge. > So which edge gets mapped to the path that always goes left? > > Stephen And which edge to the path that alternates left and right? |