Cantor Confusion
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From: Dave Seaman on 16 Jan 2007 13:21 On Tue, 16 Jan 2007 12:15:27 EST, Andy Smith wrote: > Dave Seaman wrote: Actually it was David Marcus. >> If f is defined at zero and you know that >> f(-x)=-f(x), then setting x to >> zero gives that f(0)=-f(0), which implies f(0)=0. >> But, if you have a >> function that is not defined at zero, then it isn't >> true that >> f(x)=-f(-x) for all x. It may be true for the x in >> the domain of f, but >> this by itself doesn't force you to increase the >> domain. In other words, >> if f isn't defined at zero, then writing f(0) is >> meaningless. >> f doesn't have to be anything at zero. Functions have >> whatever domain >> you give them when you define them. >> Functions are not multivalued. >> According to Wikipedia, an odd function has to have >> domain all of R. So, >> according to this definition, sin(pi/x) is not odd, >> since it isn't >> defined at zero. The function f that Dave Seaman >> defined above is odd. > OK, thank you David (and Dave Seaman, and Virgil et al). > Of course, I agree formally sin(pi/x) is undefined at x=0. > But, if something is undefined it is usually because > the coordinate system describing the underlying thing > is inappropriate at that point. e.g. y = mx + c is > not a good basis for describing all lines in a plane > (a better set of coordinates are the minimum distance from > the origin and a unit vector). > I don't know about sin(pi/x) since it is performing an > infinite number of oscillations around x = 0. But a value > of anything other than 0 at 0 would make the function > discontinuous at 0 because it is antisymmetric (or odd, if > you prefer). That function cannot be made continuous at 0 at all. It is an essential discontinuity. > exp(-1/(x^2)) is also "not defined" at x = 0, but > it would be a mistake to define the function and all > its derivatives as e.g. 42.0 at x=0, and then say > that the function and all its derivatives can be made less > than any value d by choosing some 0<x<eta. (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 ?) We can't define "the function and all its derivatives" at x = 0. We can define the function to be whatever is desired at x=0, but the derivatives (if they exist at all) are completely determined by the definition of the function. In particular, if the function is defined to be anything other than 0 at x=0, then the derivatives do not exist. This is not a mistake; this is mathematics. > Re example of 1/x, the discontinuity of 1/x at x = 0 suggests strongly > that in some sense + and -inf are the same location, > but I don't expect that suggestion to be viewed very sympathetically here. Why not? It's just the one-point compactification of the reals. >> You seem to have a funny idea about math. >> Mathematical objects are as we >> define them. You seem to want to argue in reverse. > I think that you lot discover meaningful mathematical > objects; I don't think you define them. If they are > universal truths, you could have an intelligent discussion > with the aliens when they land... The underlying truths will be the same, but the definitions will almost certainly differ. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: David Marcus on 16 Jan 2007 13:34 Andy Smith wrote: > Of course, I agree formally sin(pi/x) is undefined at x=0. What does "formally" mean? > But, if something is undefined it is usually because > the coordinate system describing the underlying thing > is inappropriate at that point. e.g. y = mx + c is > not a good basis for describing all lines in a plane > (a better set of coordinates are the minimum distance from > the origin and a unit vector). That is not what "undefined" means. > I don't know about sin(pi/x) since it is performing an > infinite number of oscillations around x = 0. But a value > of anything other than 0 at 0 would make the function > discontinuous at 0 because it is antisymmetric (or odd, if > you prefer). We say what function we are talking about. Let f(x) = sin(1/x), if x <> 0. g(x) = sin(1/x), if x <> 0, 0, if x = 0. h(x) = sin(1/x), if x <> 0, 1, if x = 0. f is not defined at zero. g and h are defined at zero. Using Wikipedia's definition of "odd", f and h are not odd, g is odd. It is meaningful to ask whether f can be extended to all of R so that its extension is odd. The answer to this question is yes. > exp(-1/(x^2)) is also "not defined" at x = 0, but > it would be a mistake to define the function and all > its derivatives as e.g. 42.0 at x=0, and then say > that the function and all its derivatives can be made less > than any value d by choosing some 0<x<eta. It is not a "mistake" to define a function as follows. k(x) = e^(-1/x^2), x <> 0, 42, x = 0. It is true that k is not continuous, but it is a perfectly valid function. Do you know the definition of "function"? 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 ?) What does "get off the ground" mean? I don't understand why you want to use a Taylor expansion to define the function (and I'm not sure if the Taylor expansion you want to use is zero at zero; I think it isn't). Why not just do it directly? I.e., let F(x) = e^(-1/x^2), x <> 0, 0, x = 0. This is a perfectly good function. It is infinitely differentiable, but not analytic. I.e., all of its derivatives exist, but its Taylor series at zero does not equal the function in any neighborhood of zero. > Re example of 1/x, the discontinuity of 1/x at x = 0 suggests strongly > that in some sense + and -inf are the same location, > but I don't expect that suggestion to be viewed very sympathetically here. The function 1/x isn't defined at zero. We can't extend it to be continuous because the limits as we approach zero from the left and the right don't exist. Infinity is not a number. Although, in real analysis, we do consider the extended real numbers where we include +inf and -inf as numbers. However, not all of the usual arithmetic properties extend to +/-inf, so you have to be careful. There are other contexts where a point at infinity is sometimes used, for example in complex analysis or geometry. > > You seem to have a funny idea about math. > > Mathematical objects are as we > > define them. You seem to want to argue in reverse. > > I think that you lot discover meaningful mathematical > objects; I don't think you define them. If they are > universal truths, you could have an intelligent discussion > with the aliens when they land... I think you don't understand how the word "define" is used in mathematics. What math courses have you had? -- David Marcus
From: David Marcus on 16 Jan 2007 13:43 mueckenh(a)rz.fh-augsburg.de wrote: > 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. > 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.) > 4) But there is no node to add. > 5) Therefore the union of trees contains all possible paths. > 6) We know that the set of all paths contained in his union is > countable. How does 6 follow from 1-5? -- David Marcus
From: David Marcus on 16 Jan 2007 13:58 mueckenh(a)rz.fh-augsburg.de wrote: > The union of all finite trees includes all nodes. > (Every node is an element of the union of all finite trees.) > Every path is a subset of the set of all nodes. > Every path is in the union of all finite trees. Let T_i = (N_i,E_i) be a sequence of trees where N_i is the set of nodes and E_i the set of edges. Assume N_i is a subset of N_{i+1} and E_i is a subset of E_{i+1}. Are you defining the union of these trees to be T = (N,E) where N = U_i N_i and E = U_i E_i? Are you also asserting that any path in T is also in one of the T_i? -- David Marcus
From: Virgil on 16 Jan 2007 14:28
In article <1168957823.678909.127030(a)38g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > A union cannot have more elements than are united. > > > > No matter how hard WM tries to make it happen. > > I do not try to make it happen. > > > > > A set of sequencs cannot have more limits than there are sequences. > > > > Which leaves out uncountably many of them when unoining 'finite' trees. > > The union of all finite trees includes all nodes. > (Every node is an element of the union of all finite trees.) > Every path is a subset of the set of all nodes. > Every path is in the union of all finite trees. > WRONG! We start by assuming that all trees and all paths issue from a common root node. We may embed every such finite binary tree in a complete infinite binary tree with the same root node by extending each finite path infinitely in the direction of its last branching, so each such infinite path is "eventually constant", in that past some point all its branchings are in the same direction. Then each finite binary tree extends to an /incomplete/ infinite binary tree which is a proper subtree of the /complete/ infinite binary tree . (Where in an incomplete tree not every node has the maximal allowable number of child nodes while in a complete tree every node does have the maximum number of child nodes). The every path in every finite subtree corresponds to a unique infinite path in the complete tree by this extension. The 'union' of all such incomplete infinite binary trees will be a subtree of the compete infinite binary tree. So far, I hope WM and I agree. The issue between us is whether that union will be a proper subtree of the complete infinite binary tree or will be the entire tree. Since every path in every one of the incomplete infinite binary trees in the union is 'eventually constant', every path in the union of all those trees will also be eventually constant. But only countably many of the uncountable paths in the complete binary tree are eventually constant. So that, for example, the union will not contain any path which alternates between branching left and branching right. Further, it will not contain any path which has infinitely many branchings in both directions. One can then easily show that the set of paths in the union will be countable, but the set of paths in the complete infinite binary tree will be uncountable. |