Cantor Confusion
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From: Newberry on 26 Dec 2006 22:02 Virgil wrote: > In article <1167167820.156449.44540(a)42g2000cwt.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > David Marcus wrote: > > > Newberry wrote: > > > > stephen(a)nomail.com wrote: > > > > > Newberry <newberry(a)ureach.com> wrote: > > > > > > David Marcus wrote: > > > > > >> Newberry wrote: > > > > > >> > It certainly makes it highly couterintutive that there are more > > > > > >> > paths > > > > > >> > then edges although I do not know if it generates a flat > > > > > >> > contradiction. > > > > > >> > > > > > >> Yes, it is counterintuitive (depending on your intuition). No, there > > > > > >> is > > > > > >> no contradiction. > > > > > > > > > > > Just because a system avoids a contradiction of the type P & ~P does > > > > > > not mean that it is justified. For example an omega-inconsistent > > > > > > system > > > > > > may not produce any P & ~P and would be still unacceptable. Similarly > > > > > > a > > > > > > system in which we can prove that > > > > > > > > > > > #edges = 2 * #paths > > > > > > > > > > > and at the same time that the cardinality of paths is greater than > > > > > > the > > > > > > number of edges is unacceptable. > > > > > > > > > > In what system can you prove that #edges = 2 * #paths? > > > > > > > > In ZFC. Theorem: In an infinite binary tree there are twice as many > > > > edges as there are paths. > > > > Proof: At the level n the ratio of eges over paths is expressed by > > > > 2*2^n - 2)/2^n > > > > For the entire tree we calculated the limit for {n-->oo} > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2. QED > > > > > > That's fine up to where you write "QED". Why does the limit you > > > calculated tell us anything about the cardinality of the edges and paths > > > in the infinite tree? > > > > I was only proving that the ratio of edges to paths in an infinite > > binary tree is 2. > > But you have proved no such thing. > > > The number 2 has been claculated for the entire > > infinite tree. > > Your claim is based on the assumption that > > lim n -> oo (f(n)/g(n) = (lim n -> oo f(n))/(lim n -> oo g(n)) It is not. > > but that is only valid when both limits on the right hand side exist, > meaning that they are finite. > > If you want to assume it where the limits on the right side do not > exist, you will have to prove it in some formal system first.
From: Newberry on 26 Dec 2006 22:08 Virgil wrote: > In article <1167167417.484441.153260(a)i12g2000cwa.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > Virgil wrote: > > > > Newberry cannot deny that the logic of as many leaf nodes as paths is > > > the same as the logic of more nodes than paths, so if one holds they > > > both must, and if one fails, both fail. > > > > Let me get this straight. How many terminal nodes are there in an > > infinite binary tree? > > Depends on whose rules one is going by. > > By my rules, none. > > By your rules, one per path. I have explicitly stated that there are no terminal nodes in an infinite tree. The only thing that prevents us from reaching consensus on this one your insistence that I have said the opposite of what I have clearly written above (that there are no terminal nodes in an infinite tree.) Now let us assume that there are no terminal nodes in an infinite tree. Correct me if I am wrong but it seems to me then that the set of the terminal nodes in an infinite tree is empty and its cardinality is zero. Having said that, what does it have to do with the cardinality of the paths?
From: Newberry on 26 Dec 2006 22:14 David Marcus wrote: > Newberry wrote: > > David Marcus wrote: > > > Newberry wrote: > > > > stephen(a)nomail.com wrote: > > > > > In what system can you prove that #edges = 2 * #paths? > > > > > > > > In ZFC. Theorem: In an infinite binary tree there are twice as many > > > > edges as there are paths. > > > > Proof: At the level n the ratio of eges over paths is expressed by > > > > 2*2^n - 2)/2^n > > > > For the entire tree we calculated the limit for {n-->oo} > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2. QED > > > > > > That's fine up to where you write "QED". Why does the limit you > > > calculated tell us anything about the cardinality of the edges and paths > > > in the infinite tree? > > > > I was only proving that the ratio of edges to paths in an infinite > > binary tree is 2. The number 2 has been claculated for the entire > > infinite tree. The number of nodes in each infinite path is aleph0 same > > as the number summands in our infinite sequence. There is one-t--ne > > mapping between N and the levels in the tree. The limit uses all > > members of N. If you disagree please tell me which node in which path > > has been left out. > > No idea what you mean by "left out". However, you didn't answer my > question: Why does the limit you calculated tell us anything about the > cardinality of the edges and paths in the infinite tree? Because if #edges/#paths = 2 then #edges > #paths. Everybody on this board, including the Cantorists, insist that |edges| = aleph0. > > I really don't see how you can think that you've proved that "the ratio > of edges to paths in an infinite binary tree is 2". The number of edges > and paths in the infinite tree are both infinite, so why is the ratio > even defined? That is what limits are for. > > Are you trolling? > > -- > David Marcus
From: Dik T. Winter on 26 Dec 2006 23:01 In article <1167188902.725881.21050(a)79g2000cws.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > Virgil wrote: > > In article <1167167417.484441.153260(a)i12g2000cwa.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > Virgil wrote: > > > > > > Newberry cannot deny that the logic of as many leaf nodes as paths is > > > > the same as the logic of more nodes than paths, so if one holds they > > > > both must, and if one fails, both fail. > > > > > > Let me get this straight. How many terminal nodes are there in an > > > infinite binary tree? > > > > Depends on whose rules one is going by. > > > > By my rules, none. > > > > By your rules, one per path. > > I have explicitly stated that there are no terminal nodes in an > infinite tree. The only thing that prevents us from reaching consensus > on this one your insistence that I have said the opposite of what I > have clearly written above (that there are no terminal nodes in an > infinite tree.) > > Now let us assume that there are no terminal nodes in an infinite tree. > Correct me if I am wrong but it seems to me then that the set of the > terminal nodes in an infinite tree is empty and its cardinality is > zero. > > Having said that, what does it have to do with the cardinality of the > paths? In the finite case all paths end at a terminal node. So when you compare number of edges to number of paths at finite levels you are comparing edges to paths with terminal nodes. -- 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 26 Dec 2006 23:05
In article <1167189286.045893.92280(a)h40g2000cwb.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > David Marcus wrote: .... > Because if #edges/#paths = 2 then #edges > #paths. > Everybody on this board, including the Cantorists, insist that |edges| > = aleph0. Not that you conclude things about the infinite here based on the finite. > > I really don't see how you can think that you've proved that "the ratio > > of edges to paths in an infinite binary tree is 2". The number of edges > > and paths in the infinite tree are both infinite, so why is the ratio > > even defined? > > That is what limits are for. You state above: #edges/#paths = 2, assuming the infinite tree. But this is only true if lim{n -> oo} #edges@n/#paths@n = = (lim{n -> oo} #edges@n / lim(n -> oo) #paths(a)n. You have to prove that that is true. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |