Cantor Confusion
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From: David Marcus on 15 Jan 2007 19:17 Andy Smith wrote: > > I'm not using a Web interface. Try using a non-Web > > interface. > > > Such as e.g. ? (excuse the ignorance) Try Googling "usenet news reader". I'm using Gravity on Windows. I've used it since before it became free. To use a news client, you need to ask your ISP for the name of their news server. -- David Marcus
From: Andy Smith on 15 Jan 2007 09:09 Dave Seaman wrote: ... > > No, sin(pi/x) is undefined at x = 0. > > You can define a function f: R -> R by > > f(x) = sin(pi/x), if x != 0, > = 0, if x = 0. > > and in that case, it's true that f(x) = 0 (but > because the definition of > f says so, not because of antisymmetry. You can > deduce antisymmetry from > the definition, not the other way around. > If f(x) is such that at any eta, f(eta) = -f(-eta), this would imply that f(0) = 0? If f(0) was anything other than 0, the antisymmetry would be destroyed? How could something that is antisymmetric not be other than 0 at its mirror point - even if it is multivalued, it will retain antisymmetry in inverting x ? Alll this is a bit "angels on a pin", but is ultimately important I think from a philosophical perspective on the nature of infinity. ---- Andy
From: David Marcus on 15 Jan 2007 19:22 Andy Smith wrote: > Just to clarify, you can't have a polygon with > a countably infinite number of vertices I suppose that depends on how you define "polygon". Usually, a polygon has only a finite number of vertices. > (but > you can have a circle with an uncountably infinite > number of points) ? Any circle of positive radius consists of an uncountable number of points. You can prove this by bijecting an arc of the circle to an interval of the reals. > Also, just to confirm understanding again, mathematicians speak of an open > and a closed interval e.g. [0,1] > > I think closed means including the points at 0 and 1 in > the interval 0 to 1; Correct. > maybe you use open to describe > the inclusive set, whatever. Open is the interval without the end points. "Open" and "closed" have more general definitions that can apply to sets other than intervals. > 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? -- David Marcus
From: Dik T. Winter on 15 Jan 2007 19:59 In article <1168870204.087161.242050(a)m58g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168533543.556066.104940(a)p59g2000hsd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > > But you do not see this small difference. You > > > > > claim that the union of all finite numbers or of all finite initial > > > > > segments {1,2,3,...,n} is the infinite set {1,2,3,...}. > > > > > > > > Elaborate on the difference. > > > > > > Why do you see it different in case of the union of all finite trees or > > > paths? > > > > I am not talking about unions of paths. I am talking about unions of > > *sets* of paths. That is something completely different. > > But it is quite irrelevant, because every path of the set of infinit > paths is a union of finite segments of paths. There are only countably > many sequences. Whether there are only countably many sequence remains to be decided (I say: no), 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. > > But the sets of the paths in the finite trees do *not* contain infinite > > sequences, and so the union of those sets can not contain infinite > > sequences. > > The union of all finite trees and the complete infinite tree are > identical with respect to nodes and edges, but differ with respect to > paths? 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. > Could you please specify the asserted difference concerning the > paths p in terms of their definition? The paths are: > p = Sequence (a_n) with n in N and a_n in {0, 1} > while the real numbers in general are represented by: > r = SUM (a_n * 2^-n) with n in N and a_n in {0, 1}. > Identical representations imply identical numbers. 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. 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. (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. BTW, we get the same results if we start with the open intervals (1/n, 1-1/n). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 15 Jan 2007 20:09
Andy Smith wrote: > Dave Seaman wrote: > > No, sin(pi/x) is undefined at x = 0. > > > > You can define a function f: R -> R by > > > > f(x) = sin(pi/x), if x != 0, > > = 0, if x = 0. > > > > and in that case, it's true that f(x) = 0 (but > > because the definition of > > f says so, not because of antisymmetry. You can > > deduce antisymmetry from > > the definition, not the other way around. > > > If f(x) is such that at any eta, f(eta) = -f(-eta), > this would imply that f(0) = 0? 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. > If f(0) was anything other than 0, the antisymmetry would be destroyed? f doesn't have to be anything at zero. Functions have whatever domain you give them when you define them. > How could something that is antisymmetric not be other than 0 > at its mirror point - even if it is multivalued, it will retain > antisymmetry in inverting x ? 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. > Alll this is a bit "angels on a pin", but is ultimately important > I think from a philosophical perspective on the nature of infinity. You seem to have a funny idea about math. Mathematical objects are as we define them. You seem to want to argue in reverse. -- David Marcus |