Cantor Confusion
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From: Newberry on 26 Dec 2006 16:17 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. 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. > > -- > David Marcus
From: Virgil on 26 Dec 2006 17:17 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.
From: Virgil on 26 Dec 2006 17:25 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)) 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: David Marcus on 26 Dec 2006 18:09 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? 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? Are you trolling? -- David Marcus
From: David Marcus on 26 Dec 2006 18:14
Newberry wrote: > Virgil wrote: > > The Cantor construction rule produces for any countably infinite list > > a number which necessarily differs from every number in the list. > > Well, but two different strings can represent the same number e.g. > 0.9999999999999999999999999... > 1.0000000000000000000000000... > Not sure what the implicationsa are for Cantor's argument. This is easily handled. The number constructed has digits that are all four and five. If a real number has two decimal representations, then one representation ends in zeros and the other in nines. So, the number constructed can't be equal to either of them. -- David Marcus |