Cantor Confusion
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From: Virgil on 15 Jan 2007 16:42 In article <1168870204.087161.242050(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1168533543.556066.104940(a)p59g2000hsd.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > > But you do not see this small difference. You > > > > > claim that the union of all finite numbers or of all finite initial > > > > > segments {1,2,3,...,n} is the infinite set {1,2,3,...}. > > > > > > > > Elaborate on the difference. > > > > > > Why do you see it different in case of the union of all finite trees or > > > paths? > > > > I am not talking about unions of paths. I am talking about unions of > > *sets* of paths. That is something completely different. > > But it is quite irrelevant, because every path of the set of infinit > paths is a union of finite segments of paths. There are only countably > many sequences. We may embed every finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the direction of its last branching. E.g., a 3 edge path that branches left from its root then right then left again, will be extended to a path that continues branching left endlessly. Then all finite binary trees become extended to restricted infinite binary trees each of which is a proper subtree of the complete infinite binary tree. The union of all such restricted infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the whole tree. Since every path in every one of the restricted infinite binary trees in the union is eventually constant, every path in the union of all those trees will also be eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then show that the set of paths that the union actually contains has will be countable, but the set of paths in the complete infinite binary tree that the union omits will be uncountable.
From: Virgil on 15 Jan 2007 16:43 In article <1168870832.926957.321570(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > >>> An infinite union of finite sets is an infinite object. > > > > > >> But that object does *not* have an infinite _member_. > > The union of all finite trees contains all terms of the sequence (a_n) > with n in N and a_n in {0, 1} representing t real number r. The > infinite path is the limit. Are there more limits than sequences? > > > > > > > So your question "Why should it?" is easily answered: It should not, but > > as Virgil has pointed out: "you invalidly assume an infinite [...] > > union of finite sets must contain an infinite object". > > The union of all finite trees and the complete infinite tree are > identical with respect to nodes and edges, but differ with respect to > paths? Could you please specify the asserted difference concerning the > paths p in terms of their definition? The paths are: > p = Sequence (a_n) with n in N and a_n in {0, 1} > while the real numbers in general are represented by: > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > Identical representations imply identical numbers. We may embed every finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the direction of its last branching. E.g., a 3 edge path that branches left from its root then right then left again, will be extended to a path that continues branching left endlessly. Then all finite binary trees become extended to restricted infinite binary trees each of which is a proper subtree of the complete infinite binary tree. The union of all such restricted infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the whole tree. Since every path in every one of the restricted infinite binary trees in the union is eventually constant, every path in the union of all those trees will also be eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then show that the set of paths that the union actually contains has will be countable, but the set of paths in the complete infinite binary tree that the union omits will be uncountable.
From: Virgil on 15 Jan 2007 16:47 In article <1168871141.646580.60890(a)s34g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1168779322.952374.232460(a)q2g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > Which ones of those specifically "rooted finite paths" does WM claim > > > > contain infinitely many nodes? > > > > > > Which segment {1,2,3,...,n} of N contains infinitely many elements? > > > Is N the union of all initial segments? > > > > If one makes a claim about finite initial segments of N, that claim need > > not be true for anything that is not a finite initial segment of N. > > > > And N is not a finite initial segment of N. > > And an infinite path is not a finite path. > > > > > > > > > > Each of the sets {1}, {1,2}, {1,2,3}, etc., is both a member and a > > > > subset of N, but N is not. > > > > > > And, in addition, it is not green. Yes. Relevance? > > > > So that WM claims every property common to all finite initial segments > > of N must also be a property of N? How about finiteness? > > Why then should the finity of all the paths in the finite trees exclude > the infinity of the paths of the union of all paths / trees? We may embed every finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the direction of its last branching. E.g., a 3 edge path that branches left from its root then right then left again, will be extended to an endless path that continues branching left endlessly. Then all finite binary trees become extended to restricted infinite binary trees each of which is a proper subtree of the complete infinite binary tree. The union of all such restricted infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the whole tree. I claim that it is a proper subtree in a way which destroys all of WM's arguments. Since every path in every one of the restricted infinite binary trees in the union is eventually constant, every path in the union of all those trees will also be eventually constant. So that, for example, the union can not contain any path which alternates between branching left and branching right as such a path cannot be in any of the trees being unioned. Further, it will not contain any path which has infinitely many branchings in both directions. One can then show that the set of paths that the union actually contains has will be countable, but the set of paths in the complete infinite binary tree that the union omits will be uncountable.
From: Virgil on 15 Jan 2007 16:48 In article <1168871541.528925.47880(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Virgil schrieb: > > > In article <1168779834.276761.153130(a)a75g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > > Hardly. > > Which digit marks an infinite initial segment? > > > If the diagonal construction did not produce an endless string > > of digits, it would not produce a number known not to be in the list, > > so finite initial segments don't cut it. > > If the union of finite trees (as defined) did not produce an infinite > tree, then a natural number n would exist, which would not label a > level in the union of trees. Such a number n does not exist. So the > union of the trees is an infinite tree. An infinite tree has no finite > path. Every infinite binary path is in this infinite binary tree. > > Arguments? > > Regards, WM We may embed every finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the direction of its last branching. E.g., a 3 edge path that branches left from its root then right then left again, will be extended to a path that continues branching left endlessly. Then all finite binary trees become extended to restricted infinite binary trees each of which is a proper subtree of the complete infinite binary tree. The union of all such restricted infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the whole tree. Since every path in every one of the restricted infinite binary trees in the union is eventually constant, every path in the union of all those trees will also be eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then show that the set of paths that the union actually contains has will be countable, but the set of paths in the complete infinite binary tree that the union omits will be uncountable.
From: Virgil on 15 Jan 2007 16:49
In article <1168872164.184807.218390(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Franziska Neugebauer schrieb: > > > > >> Recall: Do get it out you must have put it in. > > > > > > Recall Cantor's diagonal. It consists only of finite initial segments. > > > Either it does not exist - or there are infinite paths in the union of > > > finite trees. > > > > False dilemma. It "consists" of only finite initial segments (not > > finitely many therof, of course) _and_ it does exist _and_ there are > > only _finite_ paths _as_ _members_ in the union of all finite trees. > > > The union of all finite trees and the complete infinite tree are > identical with respect to nodes and edges, but note with respect to > paths. Could you please specify the asserted difference concerning the > paths p in terms of their definition? The paths are: > p = Sequence (a_n) with n in N and a_n in {0, 1} > while the real numbers are represented by: > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. We may embed every finite binary tree in a complete infinite binary tree by extending each finite path infinitely in the direction of its last branching. E.g., a 3 edge path that branches left from its root then right then left again, will be extended to a path that continues branching left endlessly. Then all finite binary trees become extended to restricted infinite binary trees each of which is a proper subtree of the complete infinite binary tree. The union of all such restricted infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the whole tree. Since every path in every one of the restricted infinite binary trees in the union is eventually constant, every path in the union of all those trees will also be eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then show that the set of paths that the union actually contains has will be countable, but the set of paths in the complete infinite binary tree that the union omits will be uncountable. |