Cantor Confusion
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From: Newberry on 26 Dec 2006 00:16 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. > > 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: Newberry on 26 Dec 2006 00:23 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? Is it true that at each level 2*2^n - 2)/2^n? Is it true that lim{n-->oo} (2*2^n - 2)/2^n = 2? Is is true that the cardinality of the index n is the same as the cardinality of the edges in an infinite path? > > So that to be consistent, you must claim that every infinite path has a > terminal node, i.e., what never has a last node always has a last node. > > Trees in ZFC have no such blatant inconsistencies.
From: Virgil on 26 Dec 2006 00:58 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 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: Virgil on 26 Dec 2006 01:03 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. > > > > > 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: stephen on 26 Dec 2006 02:46
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. You cannot just assume that "the limit" is meaningful. Limits tell you what happens has n becomes large, but remains finite. Stephen |