Cantor Confusion
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From: Franziska Neugebauer on 23 Jan 2007 04:30 Virgil wrote: > In article <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net>, > Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> > wrote: > >> Virgil wrote: >> >> > Franziska Neugebauer wrote: >> >> Virgil wrote: >> [...] >> >> Do you agree that a tree (finite and infinite) is completely >> >> determined by the nodes and edges? >> > >> > Finite trees are. Infinite trees are not. >> >> Please correct me if I am wrong: >> >> 1. Let the set of nodes M = omega\{0}. >> >> 2. Introduce the nodes of /level/ n e omega: >> >> L(n) = {2^n, ..., 2^(n+1) -1} >> >> n=0 1 >> / \ >> n=1 2 3 >> / \ / \ >> n=2 4 5 6 7 >> ... >> >> 3. Let the set of edges of all nodes from level m-1 to level m, >> m e omega\{0}: >> >> E(m) = U {{(i, 2i), (i, 2i + 1)} | i e L(m - 1)} (finite union) >> >> 4. Let the set of all edges be >> >> E = U {E(m) | m e omega\{0}} (countable infinite union) >> >> According to the usual definitions of graph theory G = (M, E) is an >> infinite graph. It is even an infinite binary tree. >> >> So I would like to ask you what _in_ or _of_ G is not completely >> determined by its sets M and E? > > Whether the set of all possible pats is required to involve every node > and every edge. You don't agree that the tree is given and the set of paths is derived from that? My central question is: How exactly do you define tree? > For finite trees it is. For infinite trees it is not. Sure but all that does by no means support your objection to my statement "Do you agree that a tree (finite and infinite) is completely determined by the nodes and edges?". >> >> Do you agree that the set of paths if kind of a "derived" >> >> property? >> > >> > For finite trees, yes. For infinite trees no! > At least not if we are allowed to excude unneeded paths. >> To begin with let us agree on what kind of paths we are talking >> about: In the present context ("representation of real numbers") we >> are concerned with paths which originate in the root node of the tree >> (0). I would like to define an infinite path as an infinite sequence >> of nodes (s(i))_i e omega having the property >> >> a) s(0) = 0 >> b) (s(i-1), s(i)) e E for all i e omega\{0} >> >> A finite path is a finite sequence of nodes over some domain >> D(m) = {0} U { n | n < m } m e omega having >> >> a) s(0) = 0 >> b) (s(i-1), s(i)) e E for all i e D\{0} >> >> >> I asked since I have the apprehension that you take the structure >> >> (M, E, P) (M = set of nodes, E = set of edges, P = set of paths) >> >> for the tree. Whereas I take (M, E) for the tree. >> >> >> >> F. N. >> > >> > For finite trees (M,P) works nicely, but it does not for infinite >> > trees. >> >> (M, P)? You mean (M, E), do you? > > Sorry. Yes! What exactly does not /work/ with the definition G = (M, E)? >> > Consider the infinite binary tree limited to paths which are >> > "eventually constant", >> >> IMHO in the usual definition of tree, G = (M, E), there is no >> facility to restrict the tree to only those paths which share a >> certain property. So you obviously have a variant definition of tree >> in mind? > > In a finite tree, there is a necessary bijection between paths and > terminal edges (or leaf nodes). Since neither terminal edges nor paths are explictly put into the tree this is a "derived" property and it is true, sure. > 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. What exactly do you mean by information? I can only guess what you mean: For every finite tree (M, E) there is a _unique_ set of paths P with the properties: a) A m e M (E p e P -> m is node in p) and b) A b e E (E p e P -> b is edge in p). > In an infinite tree, at least one in which no path ends, there are no > such things as terminal edges or leaf nodes. Sure, but since in the usual approach no paths (but nodes and edges) are put into the tree that does not support your objection that G does not completely _define_ the tree in the infinite case. > So the set of paths contains more information than does the > combination of the set of nodes and set of edges, In the usual infinite binary tree G = (M, E) there is no room for different sets of paths (in the same set theoretical framework) since P is a function of (M, E). So you obviously use a different notion of tree. > 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. Then we need to use a revised definition of tree. F. N. -- xyz
From: Franziska Neugebauer on 23 Jan 2007 04:40 David Marcus wrote: > Virgil wrote: [...] >> > 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. Yes. F. N. -- xyz
From: Carsten Schultz on 23 Jan 2007 04:41 G. Frege schrieb: > On Sun, 21 Jan 2007 18:55:37 +0100, Carsten Schultz > <carsten(a)codimi.de> wrote: > >> It would save everybody a lot of time if you would be precise in >> your statements. However, you do not seem to be capable of this. >> >> Also, the errors in your arguments would be even more obvious, >> maybe even to you, if your arguments were stated precisely. >> > I doubt that WM is interested in any errors in his own argumentation. > He's only seeking for errors ("contradictions") in modern set theory. > Remember, if errors in his "arguments" are pointed out to him, he > usually prefers to ignore that. I agree. That is why precision is not an option for him. Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: mueckenh on 23 Jan 2007 04:46 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> >> By induction you only define (if at all) every _finite_ union > >> >> > >> >> T(1) U ... U T(n) > >> > > >> > That is completely sufficient as long as there is no upper bound, > >> > but > >> > n --> oo (potential infinity). > >> > >> Then I would like to have an explanation what > >> > >> T(1) U T(2) U ... > >> > >> shall mean in contrast to > >> > >> T(1) U T(2) U ... U T(n) n e N. > > > > There is no contrast but identity (for n --> oo). > > "Mückenheim-Limes"? There is a tree T which contains the root node at level 0 and if it contains the (finite tree T(n) down to) level n then it contains the (finite tree T(n+1) down to) level n+1. The existence follows from the existence of the real numbers and all their initial segments The union of all levels = the union of all finite trees = tree T. > > U { T(i) | i e N } > > is not defined. > > > V* is not in the union (as omega is not in the union over all natural > > numbers or over all finite intial segments of omega). > > I want you to define > > T(1) U T(2) U ... It is defined. It is the tree which contains the root node and if it contains the tree with n levels, then it contains the tree with n+1 levels. > > > The union of all natural numbers does not include omega. > > The union of all segments does not include it. > > The union of all finite trees does not include it. > > I am asking for a definition of "the union of all fintie trees". > You have not yet given one. Define the union of all segments of paths which always turn right. Define the union of all segments of paths which always turn left. Define the union of all segments of paths which always alternate. So you get 0.111... and 0.000... and 0.101010... Try to doubt the unions of all segments of any other infinite path. Regards, WM
From: mueckenh on 23 Jan 2007 04:48
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> > nevertheless the union of {1,...,n} and > >> > {1,..., m} and the infinite union of all segments are defined. > >> > >> We are writing about trees. > > > > The trees have levels. For them we have the same as for the initial > > segments given above. > > Again: Your notations > > T(1) U T(2) U ... > > and > > U {T(i) | i e N } > > are undefined. You are in error. The union of the trees T(n) and T(n+1) is defined. n is a natural number. Therefore the union of all finite trees is defined. If you try to construct the tree with n levels, do you fail at some number of levels? No. Therefore the union is defined for every n. More is not feasible. > > > >> >> | > >> >> | The questions is: How do you define U V_omega? > > > > V_omega is not in the union. > > What is V_omega? The union of all finite trees. > > > The union is only over omega elements, but that is not a problem > > because it is covered by the axiom of infinity. > > Unions are covered by the axiom of union. But only if real > set-theoretical unions are involved. > > It is an equivocation (fallacy) to claim that your tree-union (which > selects the deepest out of two trees as "union") is a union. It has > been explained by Virgil, WH and me that a set-theoretical union of > trees is hardly a tree. Hence you are writing on undefined notations. > It has been claimed, but falsely. If two trees, T(n) and T(m) are idential down to level n but T(m) contains some moere levels, then the union of both is T(m). Further, both Virgil and William understand that the union defined by me is identical to te complete tree T as far as nodes and edges and levels are concerned. They merely doubt the identity of path due to some inexplicable religious belief in a death religion. Regards, WM |