Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: stephen on 22 Dec 2006 20:58 Newberry <newberry(a)ureach.com> wrote: > 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| ? No. As I said, every path passes through one of the two edges incident to the root. There are only two edges incident to the root, but there are an infinite number of paths. So it does not follow that if every path passes through at least one edge in some set of edges, that the set of edges is not larger than the set of paths. Stephen
From: Dik T. Winter on 22 Dec 2006 21:15 In article <1166837581.381928.281240(a)i12g2000cwa.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > stephen(a)nomail.com wrote: .... > > > 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| ? No. Why should that follow? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 22 Dec 2006 21:28 In article <1pkoo2dd7u1tcqeoidr3u9jdl0iio3gr31(a)4ax.com> Lester Zick <dontbother(a)nowhere.net> writes: > 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. Of course not. Neither what you state. > >> 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. Did you prove the first postulate of Euclid? Or are you too lazy or stupid to do so? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Newberry on 22 Dec 2006 22:27 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.
From: stephen on 22 Dec 2006 22:32
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 |