Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: stephen on 26 Dec 2006 02:51 Newberry <newberry(a)ureach.com> wrote: > Virgil wrote: >> In article <1167094506.703211.116860(a)73g2000cwn.googlegroups.com>, >> "Newberry" <newberry(a)ureach.com> wrote: >> >> > David Marcus wrote: >> > > Newberry wrote: >> > > > paths edges >> > > > level 1: 2 = 2^1 2 >> > > > level 2: 4 = 2^2 6 >> > > > level 3: 8 = 2^3 14 >> > > > >> > > > level n: 2^n (not sure what the formula is) >> > > > >> > > > Does the ratio edges/paths converge to 2 for n --> infinity? >> > > >> > > Yes, as WM is fond of repeating ad nauseum. >> > > >> > > > 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. >> >> If you wish to argue that because the ratio of paths to edges (or nodes) >> in finite trees is bounded that it must by some sort of limit argument >> remain bounded for infinite trees, then you must allow the same argument >> for the ratio of paths to terminal nodes. > What exactly is wrong with the limit I (or rather WM, got to give him > credit) have calculated? It has nothing to do with the infinite case. > Is it true that at each level 2*2^n - 2)/2^n? There is no finite level n for an infinite tree, and you cannot just toss transfinite cardinals into an equation willy nilly and expect it to make sense. > Is it true that lim{n-->oo} (2*2^n - 2)/2^n = 2? Yes, but what does that have to do with anything? Explain why you think that limit is relevant? > Is is true that the cardinality of the index n is the same as the > cardinality of the edges in an infinite path? What index n? There is no aleph_0 level. You are apparently just trolling at this point, as all of this has been explained to you and you have simply ignored the explanations. Do you think that an infinite tree has zero leaf nodes? Do you agree that the limit of the ratio of nodes to leaf nodes is 2 as the tree gets large? I guess you do think that 2*0 = oo afterall. Stephen
From: Newberry on 26 Dec 2006 11:33 stephen(a)nomail.com wrote: > Newberry <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > >> Newberry <newberry(a)ureach.com> wrote: > >> > David Marcus wrote: > >> >> Newberry wrote: > >> >> > paths edges > >> >> > level 1: 2 = 2^1 2 > >> >> > level 2: 4 = 2^2 6 > >> >> > level 3: 8 = 2^3 14 > >> >> > > >> >> > level n: 2^n (not sure what the formula is) > >> >> > > >> >> > Does the ratio edges/paths converge to 2 for n --> infinity? > >> >> > >> >> Yes, as WM is fond of repeating ad nauseum. > >> >> > >> >> > 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 is not a proof. I can just as easily "prove" that an infinite > tree has an infinite number of leaf nodes, despite the obvious > fact that it has zero leaf nodes. If it has 0 leaf nodes then it seems to me that their cardinality is 0, not aleph1. You cannot just assume that "the limit" > is meaningful. Limits tell you what happens has n becomes large, > but remains finite. > > Stephen
From: Newberry on 26 Dec 2006 11:40 Virgil wrote: > In article <1167109569.272807.321820(a)h40g2000cwb.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > > > Newberry <newberry(a)ureach.com> wrote: > > > > David Marcus wrote: > > > >> Newberry wrote: > > > >> > paths edges > > > >> > level 1: 2 = 2^1 2 > > > >> > level 2: 4 = 2^2 6 > > > >> > level 3: 8 = 2^3 14 > > > >> > > > > >> > level n: 2^n (not sure what the formula is) > > > >> > > > > >> > Does the ratio edges/paths converge to 2 for n --> infinity? > > > >> > > > >> Yes, as WM is fond of repeating ad nauseum. > > > >> > > > >> > 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 > > In ZFC an equally valid theorem:In an infinite binary tree every endless > path has a terminal node. > > Proof: A the level n, the ratio of paths to terminal nodes is n/n, as > every path in such a tree has a terminal node. > For the entire tree we calculate the limit for {n --> oo} > lim{n-->oo} n/n = 1. QED By calculating the number of path and edges at the level n of an infinite tree we are not making any assumption that the paths terminate at that level. > > Now it should be obvious that if either "proof" is valid, they both are, > and if either is flawed, they both are. > > So it is only in such infinite trees in which every infinite path has a > terminal node that there can be as many nodes (or edges) as paths.
From: Newberry on 26 Dec 2006 11:44 Virgil wrote: > In article <1167110178.666662.149790(a)73g2000cwn.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > Virgil wrote: > > > In article <1167094162.439384.295810(a)79g2000cws.googlegroups.com>, > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > David Marcus wrote: > > > > > Newberry wrote: > > > > > > Virgil wrote: > > > > > > > In article <1166895046.650593.195620(a)a3g2000cwd.googlegroups.com>, > > > > > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > > Virgil wrote: > > > > > > > > > In article > > > > > > > > > <1166854303.474151.267360(a)h40g2000cwb.googlegroups.com>, > > > > > > > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > > > > Is it true that the ratio of edges over paths converges to > > > > > > > > > > two as > > > > > > > > > > we > > > > > > > > > > approach infinity? > > > > > > > > > > > > > > > > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > > > > > > > > > > > > > > > It is true that the ratio of terminal nodes to paths converges > > > > > > > > > to 1 > > > > > > > > > as > > > > > > > > > the path lengths increase towards infinity. > > > > > > > > > > > > > > > > What about the ratio of all the edges to all paths? Does it > > > > > > > > converge > > > > > > > > to > > > > > > > > 2? > > > > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > > > > > > > > > > > It does not matter. > > > > > > > > > > > > Why does it not matter? > > > > > > The cardinality of the inexes in the limit is aleph0, and the > > > > > > cardinality of the nodes in any infinite path is aleph0. It means > > > > > > that > > > > > > in calulating the limit > > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > > > > > we transversed all the infinite paths. > > > > > > > > > > What does "traversed" mean? And, how is it relevant to determining the > > > > > cardinality of the set of paths? > > > > > > > > It means that we have taken into account the entire tree and we > > > > determined that the number of edges in said entire tree is twice as > > > > higher as the number of paths. > > > > > > Then it means that you are producing nonsense. > > > > It is not me who is producing it. > > You are making a claim which has as a consequence the same sort of limit > argument that every path in the infinite tree has a terminal node, as > well as a root node. You seem to be very troubled by the terminal nodes. If you realized that there is no actual infinity you would understand that there are no terminal nodes in an infinite tree. > > > > > > > > The same " limit" argument will conclude the in a tree in which no path > > > has a terminal node, every path has a terminal node. > > > > That is the contradiction we were talking about. > > > > > > > > For every finite tree, the number of terminal nodes (leaf nosed) is > > > exaclty equal to the number of paths, so that if the limit argument is > > > valid for the edge to path ratio, it must equally be valid for the > > > terminal node to path ratio.
From: Virgil on 26 Dec 2006 13:16
In article <1167150798.246311.268000(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: > > Newberry <newberry(a)ureach.com> wrote: > > > > > stephen(a)nomail.com wrote: > > >> Newberry <newberry(a)ureach.com> wrote: > > >> > David Marcus wrote: > > >> >> Newberry wrote: > > >> >> > paths edges > > >> >> > level 1: 2 = 2^1 2 > > >> >> > level 2: 4 = 2^2 6 > > >> >> > level 3: 8 = 2^3 14 > > >> >> > > > >> >> > level n: 2^n (not sure what the formula is) > > >> >> > > > >> >> > Does the ratio edges/paths converge to 2 for n --> infinity? > > >> >> > > >> >> Yes, as WM is fond of repeating ad nauseum. > > >> >> > > >> >> > 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 is not a proof. I can just as easily "prove" that an infinite > > tree has an infinite number of leaf nodes, despite the obvious > > fact that it has zero leaf nodes. > > If it has 0 leaf nodes then it seems to me that their cardinality is 0, > not aleph1. So is Newberry claiming aleph_1 = 0? Newberry cannot deny that the logic of as many leaf nodes as paths is that same as the logic of more nodes than paths, so if one holds they both must, and if one fails, both fail. > > You cannot just assume that "the limit" > > is meaningful. Limits tell you what happens has n becomes large, > > but remains finite. > > > > Stephen |