Cantor Confusion
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From: Dave Seaman on 16 Jan 2007 19:20 On Tue, 16 Jan 2007 21:58:54 GMT, Andy Smith wrote: > David Marcus writes > Doubtless you lot have some legally watertight definition. As far as I > am concerned, a function is a formula that provides a value for a given > input (or inputs, in a multidimensional situation). It doesn't have to be a "formula". Certainly not a single formula. For example, here is a perfectly legitimate function f: [0,1] -> R: f(x) = 1, if x is irrational, = 0, if x is rational. This function is discontinuous everywhere. >>However, we can't simply define derivatives to be whatever we want. >>Derivatives are determined by the values of the function. The function k >>is not continuous at zero, so it is not differntiable at zero. >>> (incidentally >>> how does exp(-1/x^2) get off the ground anyway? OK, it's not >>> defined at x = 0, but you CAN define a function f by >>> e.g. Taylor expansion in x about some x0, such that f(x) = >>> exp(1/x^2) for all x!=0, and that IS defined at x =0, >>> and that has f and all its derivatives = 0 at x =0 ?) If a function has a Taylor expansion that converges to the function on some neighborhood of a point, the function is said to be "analytic" on that neighborhood. Not every (real) function that has a Taylor series is analytic. In other words, the Taylor series, even when it exists, is not obligated to converge to the original function. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dik T. Winter on 16 Jan 2007 19:14 In article <acd4a$45acdd34$82a1e228$24751(a)news2.tudelft.nl> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: .... > It is remarked that, on the right side of the tree, there are only the > natural numbers: 1/1, 2/1, 3/1, 4/1, 5/1 .. n/1 . Each row in the tree > contains 2^(n-1) fractions. When we have arrived at n/1, there are 2^n > fractions in total (also counting 0/1 = 0). It is known that this tree > exhausts all fractions. It exhausts all naturals as well. At each stage > (n-th row) in the S-B tree, there are n naturals and 2^n fractions. We > conclude that n naturals correspond with 2^n fractions. But, since the > cardinality of the naturals is Aleph_0, the cardinality of the rational > numbers (= "reals") must be 2^(Aleph_0): Continuum Hypothesis proved .. First, the S-B tree does not contain irrational numbers (as you well know). And you are assuming that lim{n -> oo} 2^n = 2^oo = 2^aleph_0 Care to give a proof for those equalities? The limit of a function is not necessarily the function at the limit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: MoeBlee on 16 Jan 2007 19:47 Andy Smith wrote: > > Doubtless you lot have some legally watertight definition. As far as I > am concerned, a function is a formula that provides a value for a given > input (or inputs, in a multidimensional situation). Dump that definition. It's picture language. Use mathematical definitions instead: {x y} = z <-> At(t in z <-> (t=x v t=y)) {x} = {x x} <x y> = {{x} {x y}} p is an ordered pair <-> Exy p = <x y> r is a relation <-> Ax(x in r -> r is an ordered pair) f is a function <-> (f is a relation & Axyz((<x y> in f & <x z> in f) -> y = z)) MoeBlee
From: Dik T. Winter on 16 Jan 2007 19:57 In article <1168959047.119560.220700(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168870204.087161.242050(a)m58g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > but, yes, a set of paths is a set of unions of segments. The union of > > a collection of sets of paths is also a set of unions of segments. But > > conside a, b and c three segments. Define p1 = union(a, b) and > > p2 = union(a, b, c). Further define s1 = {p1} and s2 = {p2}. In that > > case: > > union(s1, s2) = {p1, p2} = {union(a, b), union(a, b, c)}; > > no further simplification is possible. It is *not* equal to s2. > > If you unite levels of trees, then the union is not the greater level. > But if you unite all nodes of trees, then the union is the greater > tree. In your proof of the countability of the set of paths of the complete tree you are using unions of sets of paths. So that is what we are talking about. Not about something unrelated. You should really focus on the *sets* of paths, otherwise your proof holds no water. > > Depends on your definition. If in the union of the finite trees the set > > of paths is the union of the sets of paths, then they differ. If the > > set of paths in the union is not related to the sets of paths in the > > the finite trees, they are equal. > > Every path is a subset of the set of all nodes. The set of al nodes is > the union of all finite trees. That is enough. If you insist that only > finite paths are possible in the union of all finite trees, then you > say that there are no infinite paths at all. But I *do* not state that. I am talking (and I will repeat it again and again) about the union of *sets* of paths. > > But again, you are now using a limit, not a union. Those things are > > different. Consider the intervals in the real of rational numbers: > > [1/n, 1-1/n] > > all closed, indexed by natural numbers. We can look at those intervals > > as sets of numbers, so we can take their union. We find: > > union{n in N} [1/n, 1-1/n] = (0, 1) > > while > > lim{n -> oo} [1/n, 1-1/n] = [0, 1] > > by some reasonable definition of limit. > > That is a good example. > > Note two things. > > (1) difference in notation. In the union there is no order implied in > > the uniting, it is independed on the order and can even be done > > about collections that are not countable. With the limit you > > have to state an order. > > In the union of paths there is an order (by length of the paths (before > getting constant)). But in the uniting ordering playes *no* role at all. > > (2) 0 and 1 are in the limit, but not in the union. They are not in > > the union because they are in none of the intervals. > > 1 = 0.111... is in the union of all finite trees, also 0.000... Yes, when you define the union as follows: Consider a tree as a collection of nodes and edges with some particular structure (left undefined for now, but it should be clear, for instance, the edges are directed edges in the graph). So it consists of two sets: set n of nodes and set e of edges. The union of a collection of trees is the graph consisting of the union of the sets of nodes and the union of the sets of edges. When (as in your case by equivalencing nodes from different trees and edges from different trees) in the union you can always find that with trees t1 and t2, the set of nodes in t1 is a subset of the set of nodes in tree in t2, and the set of edges in t1 is a subset of the set of edges in t2, or everything just the other way around, the union of the trees is also a tree. In a tree we can identify a root node, the only node without any edge pointing to it). In a tree we can identify paths. We can do it in two ways. I prefer the first, you prefer the second. (1) A path is a sequence of edges that starts at the root and where two edges can follow each other when one edge points to a node from which the other edge emanates. (2) As (1), but a path does not terminate at a node from which other edges emanate. Now set up an ordering: for trees t1 and t2: t1 < t2 when for the sets of nodes n1 and n2 and the sets of edges e1 and e2 holds: n1 subset n2 and e1 subset e2. Next define the set of paths p for each tree. According to either (1) or (2). We easily find when (t1 and t2 finite and) t1 < t2 that union(t1, t2) = t2. When we define paths according to (1) we also find that union(p1, p2) = p2. But when we define it according to (2) we have that union(p1, p2) != p2. Because (according to (2)) all paths in t1 have the same number of edges, say k1. The same holds for t2, say k2. As there is no path in t2 that has k1 edges, there is no element in p2 that has k1 edges, and as all elements in p1 have k1 edges p1 is not a subset of p2. Actually their intersection is empty, so their union consists of all elements of p1 and p2 and is not equal to p2. Now let us get at T = union{n in N} (t_n). It is well defined according to the definitions above. It consists of the set of nodes: union(n_n) and the set of edges: union(e_n). Next we can look at the set of paths in this union. If we use definition (2) of paths there is no chance that the set of paths P in T is also union(p_n), because it does not even hold for two trees. So we have to shift to (1) to get even a fighting chance. Can we show that P is union(p_n) according to definition (1)? And the answer is: no. We can't because it is false. Every p_n contains only paths of finite length, so their union also contains only paths of finite length. But P contains paths of infinite length. In a way, P is a limit, not a union. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 Jan 2007 20:15
In article <1168961119.168661.39130(a)38g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > The diagonal is *not* the union of the sets of diagonals of the finite > > matrices, because that is a set with infinitely many elements. It is also > > not in that union, because all elements of that union have finite length. > > To use unions here you have to be careful about defining the sets you > > are uniting. Consider the ordered (multi)-sets consisting of the initial n > > digits of that diagonal for each n. Each set represents an initial > > segment of the diagonal. The union of them can properly be defined and > > represents the diagonal. (The definition of union in this case would be > > something along: > > The union of two ordered (multi)-sets is defined if for every > > n either both have an n-th element and they are equal, or at most > > one has an n-th member. > > You can extend this to an arbitrary collection.) > > Like the paths in a tree, for instance. No, pray try to do it for sets of paths. > > But when you try to prove that the set of paths is countable you are *using* > > the union of sets of paths, and that is not the set about which you want > > to prove something. > > No. What I am using is this: > 1) The union of all finite trees contains the union of all finite paths > and this set is countable. I think you mean: 1) The union of all finite trees contains the union of the sets of finite paths and this set is countable. otherwise it makes no sense. Indeed, the union of the sets of all finite paths is countable. Alas, it does not contain any infinite path. If you did not mean this, please define what you understand under "the union of all finite paths". > 2) The union of all finite trees contains all nodes. > 3) You cannot add another tree to the set of all finite trees without > adding at least one node. > (The paths are subsets of the set of nodes. Two different sets must be > distinguished by at least one element.) Yes, right. > 4) But there is no node to add. Indeed. > 5) Therefore the union of trees contains all possible paths. Right. > 6) We know that the set of all paths contained in his union is > countable. Wrong. The union of the sets of paths is countable, but there is no infinite path in that union. Again, and again, and again, the union of the set of paths is not the set of paths in the union. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |