Cantor Confusion
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From: Dik T. Winter on 25 Dec 2006 19:58 In article <1167082920.650511.242870(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > It was Dik who insisted that a well ordering of the reals can be done > > > (not only be proved), if AC is true. Here is the discssion: > > > > Indeed, it can be done *when AC is true*. But perhaps we have a different > > view on the meaning of the sentence "can be done"? > > "can be done" is not "does exist". We had cleared this point long ago. You missed my message where I already did correct this? Under ZFC+V=L a well-ordering does exist. > > So again my question. To how many paths is the first left branch from the > > rood assigned? To how many paths is the second left branch assigned? > > What is an infinite path? Why do you not answer my question? (Note: rood -> root.) The point is that you think that also in the infinite tree you can assign shares of edges as 1 + 1/2 + 1/4 + ... to a path. But that is false. There is *no* edge that is shared by finitely many paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: stephen on 25 Dec 2006 20:10 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? Stephen
From: Virgil on 25 Dec 2006 22:57 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. The same " limit" argument will conclude the in a tree in which no path has a terminal node, every path has a terminal node. 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 25 Dec 2006 23:03 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. 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: Newberry on 26 Dec 2006 00:06
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 > > Stephen |