Cantor Confusion
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From: Franziska Neugebauer on 19 Jan 2007 08:22 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: [...] >> 7. Now let V* denote the set of all finite trees { T(i) | i e N }. >> >> Since U V* is only defined for V having card(V) e N. Since V* does ^- should read "U V" (no star) >> not meet this requirement we (you?) have to define what >> >> U V* >> >> shall mean. The obvious definition >> >> "U V* = T(max(D(V*))" >> >> fails due to the reason that max(D(V*)) = max(omega) is not defined. >> >> The questions is: How do you define U V_omega? Answer? >> > This definition unites sets of nodes (and sets of edges, >> > respectively) >> >> Let T1 = (V1, E1) and T2 = (V2, E2). What you define here is >> >> T1 U T2 := (V1 U V2, E1 U E2) >> >> which is perfectly legal. But set theoretically a union of trees >> (ordered pairs) would read >> >> T1 U T2 = { V1, { E1 } } U { V2, { E2 } } >> = { V1, V2, { E1 }, { E2 } } >> >> which is hardly a tree. >> >> > and it is valid for Cut Trees (CT) as well as for trees >> > of type Weeping Willow (WWT). >> >> You should take care that union-operation defined so far _is_ _not_ >> identical with the set-theoretical union. >> >> > The union of all finite trees is the union of all trees with n >> > levels where n is a natural number: >> > UT = T(1) U T(2) U T(3) U ... >> >> This is undefined since your expression has card(omega) many terms. >> Please supply a finite substitute of that expression. Hint: >> >> When we (informally) write >> >> - {1, 2, ...} we mean { i | i e N } >> - { 1 } U { 2 } U ... we mean U {{1}, {2}, ...} >> = U {{i} | i e N } >> = N. >> >> But when we (informally) write >> >> - T(1) U T(2) U T(3) U ... we have not yet any definition at all >> _and_ we have no proof (or axiom) that this >> "infinite union" does exist. > > It depends simply on the question whether the uninon of all natural > numbers does exist. Whether a _meaning_ of T(1) U T(2) U T(3) U ... (inf-u) exists does _not_ depend on such question. It depends on a definition you have to provide. So please answer the question, what (inf-u) means and prove that it exists. > This is connected Whether that is "connected" is irrelevant to a possible definition of (inf-u). F. N. -- xyz
From: mueckenh on 19 Jan 2007 08:45 Franziska Neugebauer schrieb: > >> { 1 } U { 1, 2 } = { 1, 2 } is simply true, but does not "hold". > > > > No? > > There is no free variable in "{ 1 } U { 1, 2 } = { 1, 2 }" > for which it could hold for. It is not a statement form. Not an empty set allowed for in this case? Above we see a statement in every-day language. This statement is true, i.e., it holds in every-day language (although it will seldom be expressed in every-day language). > > >> >> > If there is an infinite diagonal, however, then there is no > >> >> > reason for the path 0.111... of the union of finite trees to be > >> >> > finite. > >> >> > >> >> The question is whether 0.[01] is in the "union of all finite > >> >> paths". > >> > > >> > If it exists, > >> > >> It exists and it is 0.[01]. > > > > Then it exists in the union of all finite paths. (Every segment of > > 0.010101... exists there, so the whole number does.) > > Define union of all finite paths first. The union of all paths turning always right gives the path: 0.111... This union does exist without any futher definition. But the union of all alternating paths does not exist? > >> Define "union of all finite trees". We are currently talking about > >> finite paths. > > > > I did. > > "Define" here means to formally define. The union of all paths turning always right gives the path 0.111... This and some other paths do exist in the tree. Where is the paths due to 1/pi? If it exists, formally defined, then it exists as defined by the initial segments of the series of digits or bits (Cantor's Fundamentalreihen). Then it exists in the tree too, because the tree is nothing but a representation of real numbers. If 1/pi does not exist, as I suspect, then it does neither exist in the tree. Then, even the infinite tree does not exist. > > You shall once and for all commit yourself to whether your want to > argument with > > 1. Union of trees. > 2. Union of nodes. > 3. Union of edged. > 4. Union of paths. > No. Each one will be applied where approriate. Otherwise you cannot see the contradiction. > >> > because for every end node n of a path, there is a path crossing n. > >> > >> Please elaborate. > > > > For every node on level n of a path, > > You switch between trees and paths. and levels and edges and sets and subsets. If you are not familiar with these things, please ask for further hints. Paths are subsets of the set of nodes. The tree, as I defined it, is an oredered set of nodes. Edges are not required, because the order is given by the original definition. But we can use edges to distinguish paths of the tree from those not in the tree. > > > there is a node at level n+1 which also belongs to the path. if you do > > not agree, then please say where a path ends in the union tree. No idea? Regards, WM
From: mueckenh on 19 Jan 2007 08:53 Virgil schrieb: > In article <1169111252.322725.263330(a)a75g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > T1, the union of all finite trees, contains only finite paths. > > > > It would be nice if you could show the existence of at least one finite > > path in this union. Can you? Where does a path end? > > With its last (terminal or leaf) node, which in a finite tree every path > has. But which node is that if we have the infinite union? Every node of a path is a terminal node in some tree. But the paths in the union simply refuse to stop on Virgil's command! In eternity! > > > > > If you define a tree to be a set of paths, then > > > two different trees can have the same nodes. > > > > My trees are defined to be a set of nodes and of edges. > > > That is an improvement. Previously, WM's trees only had nodes, Trees of that special kind which I defined need no edges, because I exclude all trees with different paths. It is only when some correspondents come up with other paths that the explicit mention of the possible connecion of nodes is preferable. > Non sequitur. But your stopping node is a sequitur? > > > > > > But I do not define a tree to be a set of paths ! > > But you do not define it the way mathematicians do, either. I said what I do. Of course it is something new. > So WM's "union" still does not contain any infinite paths, and certainly > not all infinite paths as he wrongly claims. Then you claim that the union of initial sements {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n}... U ... is not the same as the union of the ends of the initial segments {1} U { 2} U {3} U... U {n}... U .. which is N, which is certainly an infinite initial segment? >> The paths in a tree are completely defined by the sequences of nodes > >(or edges) which can be followed to an end in CT or without an end in >> the union of all CTs as well as in the WWT and the union of all WWTs. > But only by the full sequence, which is, in essence the path itself. > So that paths define paths. Sequences define limits. Regards, WM
From: mueckenh on 19 Jan 2007 08:57 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > A path is a subset of the set of nodes. > > Since every subset of the set of nodes of a tree is a set of nodes you > eventually say: "A path is a set of nodes". This is not the whole truth. > > A path _has_ a set of nodes (besides the information about the > connectivity of these nodes). > > > That does not mean that every subset of the set of nodes is a path. > > Actually no set of nodes _is_ a path. In a well defined tree, as my CT and my WWT are, every path is an ordered set of nodes. It is a subset of the complete ordered set of nodes which is the infinite tree. The sets are even well ordered. Regards, WM
From: William Hughes on 19 Jan 2007 09:04
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > UT is not a tree. > > > > Call it as you like. Simply call it T1. (If I speak of "tree" blow, > simply read "eert".) It may be what you like. In any case it is the > same by nodes, edges Yes > and paths as T2. No. The paths in T1 are not determined by the nodes in T1. The paths in T2 are determined by the nodes in T2. The paths in T1 are not the same as the paths in T2. > Only the "number of paths" is > assertede to be different. > > > 0 0. > / \ > 1 0 1 > / \ / \ > 2 0 1 0 1 > ... ..... > > > The union T1 of all finite binary trees covers all levels enumerated by > natural numbers. With respect to nodes and edges it is identical with > the infinite binary tree > T1 = T2 > According to set theory (including the axiom of choice) a countable > union of countable sets is a countable set. The set of paths in the > union tree T1is merely a countable union of finite sets, and, > therefore, T1 contains only a countable set of paths. But does T1 > contain only infinite paths? T1 contains only finite paths. T1 does not contain a single infinite path. The fact that there is an infinite path that can be contructed using the nodes from T1 is not relevant. The paths in T1 are not determined by the nodes in T1. > An index denotes the level to which a node belongs. The union of all > indexes of nodes of finite paths is the union of all initial segments > of natural numbers > {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n}... U ... = {1, 2, > 3, ...}. > This is also the set of all last elements of the finite segments, i.e., > it is the set of all natural numbers N. This is the set of all indexes > - there is no one left out. With respect to this observation we > examine, for instance, all finite paths of the tree T1 which always > turn right: 0.1, 0.11, 0.111, ... If considered as sets of nodes, their > union is the infinite path representing the real number 0.111... = 1. No. Their union is a set of finite paths. 0.111... is an (potentially) infinite path. 0.111... can be seen as a limit of a sequence of finite paths. It is not the union of a sequence of finite paths. Union and limit are not the same thing. (The union of {1/2},{3/4},{4/5} .... is not {1}). > >From this we can conclude that also every other infinite path belongs > to the union T1 of all finite trees. Only if you confuse union and limit. > We see, the trees T1 and T2 are identical with respect to all nodes, > all edges, and all paths (which would already have been implied by the > identity of nodes and edges). But the set of all paths is countable in > the tree T1 and uncountable in the same tree T2. > No the set of paths in T1 is countable. The set of sequences of paths in T1 is uncountable. There is no contradiction. The set of paths and the set of sequences of paths are two different things. - William Hughes |