Cantor Confusion
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From: David Marcus on 19 Jan 2007 13:29 Andy Smith wrote: > I don't know about the cos(m! pi x) term, but (bearing in mind I don't > have a pilot's licence) I would say that: > > Lim n->oo {|x|^n} is definitely not 0 for |x|<1, 1 for |x|=1. > > At a first glance one is tempted to say, well, at any n, let x >= > 1-1/(2n). Then for any n, x^n >1/2 (terms of (1-y)^n are monotonic > decreasing for y<1 and alternating in sign), so > it cannot be true that the lim n->oo of |x|^n = 0 for all |x| <=x<1 You have demonstrated what people on sci.math refer to as dyslexia. You have switched the order of the operations. As far as the limit is concerned, x is a fixed number. Let's try x = 1/2. The question is what is the value of lim n->oo (1/2)^n ? To be more precise, saying lim n->oo |x|^n = 0, for |x| < 1 is different from saying lim n->oo sup_{|x|<1} |x|^n. "sup" is like max, but is used when the maximum isn't obtained. -- David Marcus
From: Virgil on 19 Jan 2007 15:05 In article <1169209815.493587.58970(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Carsten Schultz schrieb: > > > mueckenh(a)rz.fh-augsburg.de schrieb: > > > Carsten Schultz schrieb: > > > > > >>> You assume that the union P_i of paths contains more paths than can be > > >>> constructed from finite initial segments? > > >> I do not assume anything. I just note that being a path in the union of > > >> the T_i and being an element of the union of the P_i are a priori > > >> different things and that you would have to prove their equivalence in > > >> your setting should you claim this equivalence. > > > > > > > > > The union of all finite trees is an infinite tree. > > > The countable union of all finite paths is in the union of all finite > > > trees. > > > The "complete" tree containing all paths is identical to the union of > > > al finite trees, with respect to nodes and edges. > > > Identical trees cannot contain different sets of paths. > > > Therefore, both trees contain the same set of paths. > > > > > > > > >>>> the countability of the union of the P_i that there are only countably > > >>>> many paths in the union of the T_i. > > >>> Is Cantor's diagonal longer than the union of all of its initial > > >>> segments? > > >> Can you stay on the subject? > > > > > > That is the subject. Try to get it. > > >>> Is N more than the union of all of its initial segments {1, 2, 3, ..., > > >>> n}? > > >> Are there subsets of N that are not subsets of a proper initial segment? > > > > > > Of course. The set of primes, for instance. > > > > Wow. Even though the set N is identical to the union of its proper > > initial segments and identical sets cannot have different sets of subsets? > > Again you are mistaking one notion for another. > Trees which are identical with respect to nodes and edges cannot have > different *paths*. While that may be true for finite trees, it is demonstrably false for infinite trees. For example: Consider the infinite binary tree which is limited to paths which are eventually constant (for each path there is a last leftward branch, or a last rightward branch, but each path is endless). This limited infinite tree contains every node and every edge that the complete binary tree contains, but it is clearly not the same tree. \So WM is trivially and totally wrong in his claim above. It is WM's thoughtless assumption that every property of finite trees automatically also holds for infinite trees that leads him to make a fool of himself in this way.
From: Virgil on 19 Jan 2007 15:12 In article <1169210154.031676.79720(a)51g2000cwl.googlegroups.com>, 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 and paths as 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 But for infinite trees, nodes and edges are not enough to determine a tree uniquely, as there are different infinite trees which have the same set of nodes and the same set of edges, but different sets of paths. For example: The infinite binary tree which has only eventually constant paths ( from some node onward all branches are in the same direction) has the same sets of nodes and edges but not the same set of paths as the complete infinite binary tree. So that any deduction that WM makes based on this false claim that nodes and edges are enough to uniquely determine an infinite tree fails.
From: Virgil on 19 Jan 2007 15:14 In article <1169210289.200846.317840(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > 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. > > It depends simply on the question whether the uninon of all natural > numbers does exist. This is connected to the unuion of levels and > initial segments and finite trees. See below: > > 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 But for infinite trees, nodes and edges are not enough to determine a tree uniquely, as there are different infinite trees which have the same set of nodes and the same set of edges, but different sets of paths. For example: The infinite binary tree which has only eventually constant paths ( from some node onward all branches are in the same direction) has the same sets of nodes and edges but not the same set of paths as the complete infinite binary tree. So WM's arguments, being based on the false premise that infinite trees are uniquely determined by only nodes and edeges, fail.
From: Virgil on 19 Jan 2007 15:18
In article <1169211532.179686.171640(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > You say the set of paths in the union tree is ot the same as the set of > paths in the complete tree. If the sets of nodes and edges are > identical, then also the sets of paths are identical. But for infinite trees, nodes and edges are not enough to determine a tree uniquely, as there are different infinite trees which have the same set of nodes and the same set of edges, but different sets of paths. For example: The infinite binary tree of eventually_constant_paths ( from some node onward all branches of the path are in the same direction) has the same sets of nodes and edges but not the same set of paths as the complete infinite binary tree. So WM is basing his arguments on a false premise. |