Cantor Confusion
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From: Andy Smith on 16 Jan 2007 02:15 Dave Seaman wrote: > 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). 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 ?) 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. > 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... --- Andy
From: Franziska Neugebauer on 16 Jan 2007 12:32 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > The trees are sets of nodes. Trees usually consist of a set of nodes _and_ a set of *edges*. The set of paths in a particular tree is defined only if _both_ of the sets are given. F. N. -- xyz
From: Franziska Neugebauer on 16 Jan 2007 12:36 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: [...] >> The set of nodes is the same for each tree. However, a tree is more >> than its nodes. > > Yes, some breath of God must be in there. There is. And His name is "set of edges". F. N. -- xyz
From: William Hughes on 16 Jan 2007 12:57 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > No. The (potentially) infinite sequence of all digits is not > > > > > > in the list, nor is it in the union of all finite binary trees. > > > > > > > > > > Simply draw the path of 1/3 with a slope of 45°: > > > > > > > > > > 0. > > > > > 0 > > > > > 1 > > > > > 0 > > > > > 1 > > > > > ... > > > > > > > > > > Then you see that it is in the tree which contains all levels > > > > > enumerated by all n in N, if and only if it is the diagonal of Cantors > > > > > list enumerated by all n in N. > > > > > > > > > > > > > > > > > > > > > Thus one cannot say > > > > > > > > Cantor's diagonal "consists only of finite initial segments". > > > > > > > > > > > > > > One must say so. It is similar to saying N consists of finite numbers > > > > > > > only. > > > > > > > > > > > > No. The statement > > > > > > > > > > > > All elements of N are finite numbers > > > > > > > > > > > > is true. The statement > > > > > > > > > > > > All initial segments of Cantor's diagonal are finite > > > > > > > > > > > > is false. There is one (potentially) infinite initial segment. > > > > > > > > > > in a list enumerated by all natural numbers? But there is no infinite > > > > > path in a tree the levels of which are enumerated by all natural > > > > > numbers? > > > > > > > > There are two such trees. > > > > > > > > T1: The union of all finite trees > > > > > > > > T2: The inifinite tree. > > > > > > > > Both T1 and T2 have levels which are enumeratred by all natural > > > > numbers. > > > > However, the trees are not the same. T1 does not contain an infinite > > > > path. > > > > T2 does contain an infinite path. > > > > > > The trees are sets of nodes. > > > > Plus a structure on the nodes. The set of nodes in T1 is the > > same as the set of nodes in T2. The structure placed on the nodes > > in T1 is not the same as the structure placed on the nodes in T2. > > Ah, well. Dik had just propsed some paths running horizontally. > > > > > Please give a node which is in one but not > > > in the other tree. > > > > No such node exists. > > Nice to hear. > > > The set of nodes is the same for > > each tree. However, a tree is more than its nodes. > > Yes, some breath of God must be in there. > > > > > Otherwise withdraw your assertion of T1 =/= T2 as > > > wishful believing. > > > > No. There can be other differences. > > I deeply deplore that you and others are allowed to call your religious > beliefs mathematics. But it is clear now why there is no contradiction > in ZFC and why there will never be such a contradiction. In fact, there > can always be other things in there which cannot be determined but > which invalidate an obvious contradiction. > > > > > Paths are sets of nodes too, they are subsets of > > > trees. > > > > Just because a set of nodes is in T1, does not mean > > that the path represented by that set of nodes is in T1. > > All paths run downwards. There is no other possibility than being > longer or shorter. But this property is easily translated in having > more nodes or less. And this leads to at least one node which is > different. > > > > > > > > > > For Cantor's diagonal we can say: > > > > > > > > The (potentially) infinite set of all finite initial segments of > > > > Cantor's > > > > diagonal does not contain an (potentially) infinite initial > > > > segment. > > > > > > > > Cantor's diagonal does contain an (potentially) infinite initial > > > > segment. > > > > > > For the tree we can say: The infinite set T1 of all finite initial > > > segments of paths does not contain an infinite path. T2 does contain > > > all possible infinite paths. > > > > Correct. > > If there is no difference in terms of nodes, why does T1 not contain > all possible paths? Because the paths define the nodes, not the other way round. > > What must be changed so that T1 gets T2? Concerning the hard ware (not > the breath of God). > > > > > If you cannot distinguish T1 and T2 by > > > finding a node which is in T2 but not in T1, then T2 = T1. > > > > Incorrect. Two different trees can have the same nodes. > > What must be changed in order to make T1 being T2? A tree is a set of paths. (there are other possible definitions, however, we need T1 to be a tree, so we can compare the trees T1 and T2) The tree contains paths, the paths contain nodes. Therefore a tree contains nodes. However, the fact that a set of nodes M is in the tree does not mean that path, P1, corresponding to the nodes M is in the tree. M may also be composed of nodes from more that one path. Call a tree 'node-complete' if Whenever the nodes M corresponding to a path P are in the tree, then path P is in the tree. Given any tree (set of paths) we can add paths to it to make it node-complete. T1 is not node-complete. T2 is node-complete. To make T1 into T2 we have to add paths to T1 to make it node complete. - William Hughes
From: David Marcus on 16 Jan 2007 13:14
Andy Smith wrote: > David Marcus wrote: > > I suppose that depends on how you define "polygon". > > Usually, a polygon > > has only a finite number of vertices. > > Sometimes pi is estimated by computing the circumference > of a regular convex polygon with N vertices, and letting > N->oo . But actually such a limit does not exist? The limit of the length of the circumferences certainly exists. Whether the limit of the polygons exists depends on how you define such a limit. However, the limit of a sequence needn't have the same properties as the elements of the sequence. Consider the sequence x_n = 1/n. The limit as n goes to infinity is zero. Each term of the sequence is positive, but the limit is not positive. So, assuming we define a suitable limit so the limit of polygons is a circle, this doesn't mean the limit is a polygon. > > > Anyway, if, for example, > > > you took the set of points from (0,1) excluding > > > the point at 1, you would not then be able to recreate the > > > ordered set inclusive of the point at 1 by re-appending it > > > because you would not know where to put it? > > > > I'm sorry, but I have no clue what you mean. > > Recreate? Re-append? > > Wouldn't know where to put it? > > All I meant was that if you have the ordered finite set > e.g. {0,1,2,3,..,N} you could partition it into exclusive subsets > {0,1,2,3,..,N-1} and {N}, We say "disjoint", not "exclusive". > and then, if you wished, > recreate the original set by appending {N} to > {0,1,2,3,..,N-1}. Not sure what you mean by "appending". Let "u" mean union. Then {0,1,2,3,..,N} = {0,1,2,3,..,N-1} u {N}. > But when you form the open subset > of [0,1) by e.g. excluding the point {1}, you could > not form the ordered closed set [0,1] from > [0,1) and {1} - because [0,1) does not have a last member, We say "largest", not "last". > so you would not know where to append {1}. [0,1] = [0,1) u {1}. So, I still don't know what you mean. What do "not form" and "append" mean? -- David Marcus |