Cantor Confusion
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From: David Marcus on 22 Jan 2007 15:56 Virgil wrote: > In article <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net>, > Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > > In a finite tree, there is a necessary bijection between paths and > terminal edges (or leaf nodes). So that the set of paths contains > exactly the same information about a finite tree as does the combination > of the set of nodes and set of edges. > > In an infinite tree, at least one in which no path ends, there are no > such things as terminal edges or leaf nodes. So the set of paths > contains more information than does the combination of the set of nodes > and set of edges, and different sets of paths lead to the same sets of > nodes and edges. > > Since it is sets of paths of a tree that WM has been going on about, it > seems more reasonable to consider those sets of paths from the start. Perhaps to a mathematician. However, WM's understanding of the words "set", "tree", and "path" is too primitive to make doing so productive. > > I would call that trees "path-confined" or so. A usually defined tree > > (set of nodes plus set of egdes) is by no means path-confined. Even > > finite trees can be path-confined in the way you propose: > > > > Let M = {0, 1, 2} and E = {(0, 1), (0, 2)}. This unconfined tree > > obviously has P = { (0, 1), (0, 2) }. You may _define_ the a path- > > confined tree by T' = ( M, E, P' ) for example by explicitly _setting_ > > P' = { }. Then T' contains no paths at all. Nonetheless this is not a > > property of the origial usually defined tree G = (M, E). > > On the other hand, your path-confined tree does not use all of its nodes > and edges in its paths, as mine are required to do. Ah, now I understand what you are doing. I think this is too subtle. However, if you wish to discuss this with WM, I suggest you come up with another name than "tree" for the object. -- David Marcus
From: Virgil on 22 Jan 2007 16:04 In article <1169489199.691793.259120(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > > Max n,m is not defined either, > > > > Aha > > > > ,----[ WM in <1169111380.377993.67320(a)l53g2000cwa.googlegroups.com> ] > > | The union of two finite trees T(m) and T(n) with m and n levels, > > | respectively, where m < n, is the tree with n levels. > > `---- > > > > So you mean m < n is not defined? Then it makes no sense at all to write > > about trees? > > > Sorry, this should read: Max (n,m) is not defined *other* (than for > finite m and n). The union of m and n is the maximum of both. > Nevertheless the union of all natural numbers exists as well as the > union of segments {1,...,n} and {1,..., m} and the infinite union of > all segments. The union of any finite number of finite segments is a finite segment, but the union of infinitely many finite segments is not a finite segment, so that WM's argument fails. An infinite union need not be what every finite union is. Otherwise such an infinite union would have to be finite.
From: Virgil on 22 Jan 2007 16:09 In article <1169489693.112797.92660(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > WM is not competent to pass judgement on mathematicians. > > That could be decided by mathematicians only. Right! And WM is not one of them. > > This is false for infinite trees > > This assertion alone is capable of showing the non-existence of > infinity. If it is false, it is equally capable of showing the existence of infinity.
From: Virgil on 22 Jan 2007 16:55 In article <MPG.201eef028339b7b0989bd8(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > In article <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net>, > > Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > > > > In a finite tree, there is a necessary bijection between paths and > > terminal edges (or leaf nodes). So that the set of paths contains > > exactly the same information about a finite tree as does the combination > > of the set of nodes and set of edges. > > > > In an infinite tree, at least one in which no path ends, there are no > > such things as terminal edges or leaf nodes. So the set of paths > > contains more information than does the combination of the set of nodes > > and set of edges, and different sets of paths lead to the same sets of > > nodes and edges. > > > > Since it is sets of paths of a tree that WM has been going on about, it > > seems more reasonable to consider those sets of paths from the start. > > Perhaps to a mathematician. However, WM's understanding of the words > "set", "tree", and "path" is too primitive to make doing so productive. > > > > I would call that trees "path-confined" or so. A usually defined tree > > > (set of nodes plus set of egdes) is by no means path-confined. Even > > > finite trees can be path-confined in the way you propose: > > > > > > Let M = {0, 1, 2} and E = {(0, 1), (0, 2)}. This unconfined tree > > > obviously has P = { (0, 1), (0, 2) }. You may _define_ the a path- > > > confined tree by T' = ( M, E, P' ) for example by explicitly _setting_ > > > P' = { }. Then T' contains no paths at all. Nonetheless this is not a > > > property of the origial usually defined tree G = (M, E). > > > > On the other hand, your path-confined tree does not use all of its nodes > > and edges in its paths, as mine are required to do. > > Ah, now I understand what you are doing. I think this is too subtle. So far I have seen little that isn't too subtle for WM to grasp. > However, if you wish to discuss this with WM, I suggest you come up with > another name than "tree" for the object. I'll work on it.
From: David Marcus on 22 Jan 2007 16:59
Virgil wrote: > In article <MPG.201eef028339b7b0989bd8(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > Ah, now I understand what you are doing. I think this is too subtle. > > So far I have seen little that isn't too subtle for WM to grasp. Very true. -- David Marcus |