Cantor Confusion
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From: William Hughes on 21 Jan 2007 17:45 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > The directions of the paths are the same in T1 and T2. If you insist on > > > a difference, then it can only result from the length of the paths. > > > > Yes. T1 contains only finite paths. > > Which path is finite? Where does a path end? Every path in T1 ends at a node. The fact that you can continue a path does not mean that you must continue a path. > > > T2 contains only > > (potentially) infinite paths. > > > > > > > The paths in T1 are not determined by the nodes in T1. > > > > > > By its nodes and edges every path is determined. > > > > Yes, but the fact that these nodes and edges are in T1 > > does not mean the path determined by these nodes > > and edges in in T1. > > Even if the paths are in T1 this does not mean that the paths are in > T1, I assume? You assume wrong. The statement was that even if the nodes and edges of path P1 are in T1, this does not mean that path P1 is in T1. More formally. T1 is the union of all finite paths. Let M be the set of nodes in T1, (that is if a node m is in M, then there is a path p(m) in T1, such that the node m is in p(m)). Let p1 be a (potentially) infinite path. Then Every node in p1 is contained in some path in T1. However There is no path in T1 that contains every node in p1 Proof: Let a path in T1 that contains every node in p1 be called L_D L_D is in T1 so L_D is finite. L_D contains every node in p1, so L_D is (potentially) infinite. Contradiction. Therefore L_D does not exist. - William Hughes
From: David Marcus on 21 Jan 2007 17:50 Virgil wrote: > In article <1169396771.216800.71250(a)38g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Couldn't you get to a consensus? It would spare me a lot of work. > > Why should sparing you work be any sort of goal for us as you do nothing > by way of serious work or logical thought to spare us work. That's a good point! > In fact, you seem to go out of your way to misunderstand and > misrepresent what we do and say. Indeed. It is remarkable how little WM's postings have changed. -- David Marcus
From: Virgil on 21 Jan 2007 17:52 In article <1169413190.884269.174180(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > The directions of the paths are the same in T1 and T2. If you insist on > > > a difference, then it can only result from the length of the paths. > > > > Yes. T1 contains only finite paths. > > Which path is finite? Where does a path end? At its leaf node, of course. > > Even if the paths are in T1 this does not mean that the paths are in > T1, I assume? You assume so many contrary to fact things, you might as well. > > We do have "to create or make paths", in particular because you cannot > define what we would have to do in order to make paths without changing > nodes and edges. If there *are nodes* (connected by edges or my general > pescription), then there *are the paths* corresponding to them. It may be false that every possible path trough a given node or edge exists, At least in an infinite tree. > > T1 is not a tree. > > The paths in T1 do not have to "go to the bottom". > > No? What about influence of gravity? How much mass do nodes and edges have? > Every path in T1 is, by definition, the union of an infinite set of > finite sets of nodes > {K(1)} U [K(1), (K(2)} U ... But not conversely. > > > > In mathematics a sequence is proven finite if and only if there is a > > > natural number n_0 which not belonging to its domain. Otherwise a > > > sequence is infinite. > > > > > > > All irrelevant. A sequence is not its domain. > > If you have a set of sequences S, you can take the union > > of the domains, D, then construct T to be the set of sequences > > whose domain is contained in D. However, you cannot conclude > > that any sequence contained in T is contained in S. > > You can, if, by definition, all terms (which are defined) are 1. Which in a binary tree that had better not be. > > > > Correct, but a domain is not a sequence. > > Here, we are interested in domains only (length of paths). Some of us are interested in more. > > No. A union of intervals is not an interval. > > The interval is a set of points. The union of two sets contains all the > elemets which are in at least one of the sets. This can be one > interval. But need not be.
From: Virgil on 21 Jan 2007 18:01 In article <1169414214.157040.226900(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > But if you, personally, think it is necessary, then simply use the > > > axiom of infinity. There is a set of all natural numbers, a set of all > > > levels, a set of all finite trees. > > > > But nothing that makes a union of finite trees into an infinite tree. > > Therefore the set of paths of the union tree T1 is countable. If T1 is the union of infinitely many finite trees, in WM's sense of being a tree with the union of node sets as node set and the union of edge sets as edge set, and no more, then there is no way of knowing which of infinitely many possible trees it is. > > > > While a union of sets is necessarily a set, trees are not merely sets, > > they have additional structure required, and the unioning of sets does > > not automate any appropriate amalgamation of the additional structures. > > > > WM is relying on his false assumption that an infinite tree is totally > > determined by its set of nodes and set of edges, > > That is the case. There is nothing else determining the tree and the > paths in it. False! I have given specific examples of two different infinite trees having precisely the same sets of nodes and precisely the same sets of edges, but having quite different sets of paths. In one the set of paths is countable and in the other uncountable. WM has given us no way of determining which of the infinitely many such different infinite trees his union should be. > > It shows that R is not an uncountable set. Of course, R is uncountable! Who ever said otherwise?
From: David Marcus on 21 Jan 2007 18:05
Virgil wrote: > In article <MPG.201d8bd1f8c7b013989bc1(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > It is interesting: > > > William knows that T1 and T2 exist, but T1 is different. > > > Virgil is also accepting T1 but, like William, sees only finite trees > > > therein. > > > Dik knows that T1 contains infinite trees, but doubts that its paths > > > are the union of the paths of the finite trees. > > > You doubt the existence of T1 (seeing that otherwise set theory is > > > contradiced?). > > > > > > Couldn't you get to a consensus? It would spare me a lot of work. > > > > Perhaps it would help if you look at the following and see if it is what > > you are saying. If it isn't, please note any corrections. > > > > An edge is an ordered pair of nodes. A tree T is an ordered pair (M,E) > > where M is a set of nodes, E is a set of edges. Assume that any tree > > we mention is a complete binary tree. > > > > Define a path P in T to be a function P: X -> M where X is either > > {0,...,n} or N, P(0) is the root node, (P(i),P(i+1)) is in E for all i > > such that P(i) and P(i+1) are both defined, and if the domain is > > {0,...,n}, then P(n) is a leaf node. Denote by Q(T) the set of paths > > in T. > > > > Define a nested sequence of trees to be a sequence {T_i} where T_i = > > (M_i,E_i), M_i is a subset of M_{i+1}, and E_i is a subset of E_{i+1}. > > Define the union of these trees to be T* = (M*,E*) where M* = U{M_i} > > and E* = U{E_i}. > > > > Claim: |Q(T*)| <= |U{Q(T_i)}|. > > The difficulty with that definition is that for "infinite" trees, the > set of nodes and the set of edges is not enough to determine the set of > paths, as it is in finite trees. I meant for Q(T) to be defined to be all paths that satisfy the definition above of being a path in the tree T. My definition above defines a tree just by specifying its nodes and edges. The paths are then determined. > The minimal set of paths consistent with a tree being infinite is > countable but the maximal one is not. I believe my Q(T*) is the same as what you call the maximal set. > As the distinction between the set of paths being countable and > uncountable is what WM is interested in, those sets of paths must be > taken into consideration. True. But, I don't think WM understands that we can tack on the set of paths as an explicit part of our definition of a tree. This is a rather advanced mathematical concept. It is analogous to the problem that Tony has not understanding that a set doesn't have to have an order. I'm pretty sure that WM believes that my Q(T*) is a subset of U{Q(T_i)}. In fact, I'll hazard a guess that he thinks this is obviously true. WM: Care to comment? Am I right about what you are saying? -- David Marcus |