Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Franziska Neugebauer on 18 Jan 2007 06:49 mueckenh(a)rz.fh-augsburg.de wrote: > My trees are defined to be a set of nodes and of edges. Lesson learned? Not quite. A tree is an ordered pair of a set of vertices and a _set_ of egdes. F. N. -- xyz
From: William Hughes on 18 Jan 2007 07:28 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > I will switch to your defintion of a tree. A tree consists > > of a set of nodes, and contains all paths that > > the nodes define. > > > > > Of course not. The complete tree T1 contains all paths. > > > > T1 is not a tree. > > Here are some examples and the definition. According to this > definition, T1 is a tree. > > This is a tree of type Weeping Willow with n = 1 level before getting > constant. > >>> > 0. > >>> > / \ > >>> > 0 1 > >>> > / \ / \ > >>> > 0 1 0 1 > >>> > | | | | > >>> > 0 1 0 1 > >>> > | | | | > >>> > 0 1 0 1 > >>> > .......................... > > This is a cut tree with one level > > >>> > 0. > >>> > / \ > >>> > 0 1 > > > 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. This definition > unites sets of nodes (and sets of edges, respectively) and it is valid > for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). Correct: Completely analgous to: the union of two bounded initial segments with largest elements m <n respectively is a finite initial seqment with largest element n. > > 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 ... And to say that is is a tree is false. Such a statment is completely analagous to: the union of all bounded initial segments with largest element n where n is a natural number is contained within a bounded initial segment L_D. UT is not a tree. > > 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 > their union as well as in WWT. > Only correct for trees. Since UT is not a tree, this does not apply to UT. The paths in UT are not determined by the nodes in UT. - William Hughes
From: Franziska Neugebauer on 18 Jan 2007 07:48 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> As I have mentioned in my last post: You have not given a definition >> of a union of all finite trees. > > 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. OK. 1. Let T(i) denote the tree of (finite) depth i e N. 2. Define depth(T(i)) := i. 3. Let u(E, F) denote the function which yields the deepest of two given trees E and F: / E if depth(E) >= depth(F) u(E, F) := | \ F else Introduce the notation E U F := u(E, F). So obviously Ai,j e N (T(i) U T (j) = T(max(i, j)). 4. Extend max to an arbitrary set S of integers: If E m e S A i e S (m >= i) then m is called the maximum of S, max (S). (note: this definition does not state that every S has a maximum). 5. Let V be some (finite) set of trees { T_1, ..., T_n } n e N. 6. Define the set of depths of a set of (finite) trees: D(V) := { depth(t) | t e V } 6. We now introduce the notion of the "union of a finite set of trees" (sloppyly "union of trees") as: U V := T_1 U ... U T_n (n e N) Proof as an exercise that forall sets of finite trees V having card(V) e N it holds that U V = T(max(D(V)). 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 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? > 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. F. N. -- xyz
From: Dik T. Winter on 18 Jan 2007 07:48 In article <1169117024.775595.256400(a)11g2000cwr.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Further with the finite trees as > > you currently define them (with all having infinitely many nodes, so I > > the only finiteness is the number of splitting), the union of sets of > > paths in two finite trees is the set of paths in the union of those > > two trees. *But* that does not hold when you unite the whole collection > > of finite trees. > > No? > > 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. This definition > unites sets of nodes (and sets of edges, respectively) and it is valid > for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). > > 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 ... > > 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 > their union as well as in WWT. In what way does that contradict what I wrote? I was talking about sets of paths and their union. What you write is completely unrelated to that. > > > But *all* paths are in the union of the > > > finite trees, because it is simply silly to suspect different sets of > > > paths in a tree with all nodes. > > > > Eh? There is a whole collection of different sets of paths in a tree with > > all nodes (assuming the definition of tree as I stated above, that is the > > pair [set of nodes, set of edges]). Of course it is silly to suspect > > different complete sets of paths. > > So it is. And there is not. > > > Therefore the only conclusion is that > > > the path 0.010101... does not exist. > > > > Wrong. > > If it exists, then it exists in every tree with all nodes and edges. Yes, and I never contradicted that. > > > So you say. If you were right, there should be some indication for the > > > difference. > > > What makes the path 0.010101... be in a tree with all nodes but not in > > > a tree with all nodes? > > > > Again, I do not state that. What I state is that that path is (with the > > definitions above) in the complete tree. > > It is in every tree which contains all nodes and edges (as defined by > me), because there is no end when we follow the path always swithing > betwen 0 and 1. Yes, so what? I never said that was not the case. > > And I also state that that path > > is not in the union of the sets of paths of the finite trees (also again > > according to the above definition). And I say so (and am allowed to say > > so) because that path is in none of the finite trees. > > That is the same as saying N is not in the union of all finite segments > {1,2,3,...,n}. Not so, there is no equivalence between the statements. But this statement about N is right. > N is this union. This one is also right. But N is not *in* that union. There is a difference between stating that N is the union and N is in the union. Or do you assert that N is an element of N? What is the justification? > The complete tree is the union of all > finite trees. The complete tree contains all nodes and edges. Therefore > it contains all paths. You apparently are not willing to read what I write. I am talking about sets of paths. And what path is contained in what set of paths. > > > What can we do to switch its existence on and off? > > > > I do not do that. Where do I state that it does not exist? I only state > > that it exists but that it is not in the union of the sets of diagonals of > > finite squares. > > Is it or part of it outside of this union? This is a silly question. Something is an element of a set or somthing is not an element of a set. You can not have something being partly an element of a set. > > > > Right. But the complete diagonal is not in the union of the sets of > > > > diagonals of the finite triangles. > > > > > > Where is it? Or is it nowhere? > > > > Why should it be in that union? What reasoning are you going to assume > > that it *must* be in that union? It is simply not in that union, and > > that is easily enough to prove. But it exists, but not in that union. > > If it is not in the union, then it must be outside of that union. At > least part of it. Which part is outside? Again some very basic misinterpretation. You can not have things that are only partly element of a set. > You say it exists. It exists not in the union. Where does it exist? Outside the union. Let's illustrate. Call the diagonal of the triangle of order n: d_n. The union of the sets of diagonals of the triangles is: {d_1, d_2, d_3, d_4, ...} the diagonal d of the infinite tree is not in it. > > > > It does contain that path (according to your definition above). > > > > > > Yes, that is obvious. But, for a moment, switch to the cut trees, > > > please. Does the union of all cut trees contain the paths 0.111...? You > > > would say no. > > > > And again, I would say yes. > > That is remarkable. Are you really sure? Yes. I did even show that in an earlier article I wrote. > > I even provided a definition of union of trees > > for that case and with that definition that was easily enough to prove. > > But then we have that the union of the sets of paths of two different > > finite trees is not the set of paths of the union of those two trees. > > What is lacking? Consider a tree of level 1 and a tree of level 2. The tree of level 1 contains only paths going through two nodes, the tree of level 2 only paths going through three nodes. So none of the paths in the level 1 tree is a path in the level 2 tree. So the union of the sets of paths contains both all paths of the level 1 tree and all paths of the level 2 tree. The set of paths of the level 2 tree contains only the paths of the level 2 tree. > > > If this path exists, then it must be constructible from all the nodes > > > present. There is a bit, 0 or 1, for every index n in N. That is not > > > sufficient? > > > > Not sufficient to be in the union of the sets of paths of the finite trees. > > All nodes and all edges are in the union. There can be no path being > utside. There are no nodes or edges in the union. The union is a union of sets of paths, so it only contains paths. How you manage to get nodes and edges as elements of a set of paths escapes me. > > The reason is that a path is in that union *if and only if* it is in one > > of the sets that are used to form the union. > > The reason is: If every subset of a path in the union, then the whole > path in that union. Wrong. That is not how set theory works. If every subset of a path is in the union, then the union contains (as elements) a whole lot of subsets of paths. But not the complete path. Consider the following set: {{1}, {2}, {3}, ...} this set does *not* contain (as element) {1, 2}. > Every subset is in the union. Or not? Depends. Every subset of the sets of paths is in the union. But every subset of paths is not in the union. > > > There are only countable many sequences. > > > > I disagree. There are uncountably many limits. You are trying to prove > > that there are countably many sequences and limits. If there are countably > > many sequences there are countably many limits, and the other way around. > > But until now you have failed to prove either the first or the other. > > Every initial segment of a sequence in the union of trees. All initial > segments are countable. Indeed, all (proper) initial segments are countable. But this does show nothing. > > Your proof of countably many sequences hinges on the assumption that the > > union of the sets of paths in the finite trees is the set of paths in the > > completed trees. But that one is false, as I did show above. > > No. If every subset of a path in the union, then the whole path in that > union. Wrong. A subset of a path is a subset of an element of the union. Such a subset is not necessarily an element of the union. What you state is like saying that the set {{1}, {2}} also contains {1, 2} as element. And that is fundamentally wrong. > > Now you are talking about unions of paths. The theorems of set theory are > > about unions of *sets* of paths. There is quite some difference. What > > is the union of all paths? How do you *define* the union of two paths? > > The union of all paths is the tree, because the paths are subsets of > the tree. But the union of paths does *not* tell us anything about unions of sets of paths. > > Again, I never stated that there are paths missing from that set of paths. > > I only stated that there are paths that are not in the union of the sets > > of paths of the finite trees. And it is simple to state why, those are > > the paths that are not present in the finite trees. > > Every subset of a path in the union of trees. ==> The path in the union > of trees. That does not contradict what I state. I state that if a path is not in any of the finite trees it can not be in the union of the sets of paths in the finite trees. That is the definition of union. Something is an element of the union if and only if it is element of at least one of the sets that make the union. > > > But paths which are not determined by the set of all nodes must stand > > > out by some indication, > > > > There are no such paths. > > -- > > Correct. All paths are in the union of all finite trees. And I did never contradict this. But not all paths are in the union of the sets of paths from the finite trees. There is quite some distinction. And in your proof of the countability of paths you assume the second. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Franziska Neugebauer on 18 Jan 2007 08:02
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> > The union of {1} and {1,2} is {1,2}. This holds for initial >> > segments of N as well as for finite trees, cut or as Trauerweide. >> >> { 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. >> > But there is no infinite path by definition? >> >> In a tree of finite hight: no. > > In the union of all finite trees: yes. Define union of all finite trees. > (If every subset of an infinite path (= set of nodes) is in the uninon > of finite trees, then the complete path in the union.) Definite union of all finite trees first. >> >> > 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. >> Define "union of all finite trees". We are currently talking about >> finite paths. > > I did. "Define" here means to formally define. >> > and all finite paths which could extend any of its paths. >> > >> > The union of all finite paths is an infinite path, > >> >> Don't see that even a union of two paths is a path. > > A path is an ordered set of nodes. The union of two paths is he union > of these nodes. 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. >> > 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. > there is a node at level n+1 which also belongs to the path. if you do > not agrre, then please say where a path ends in the union tree. Please order your thoughts before writing. F. N. -- xyz |