Cantor Confusion
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From: Virgil on 16 Jan 2007 15:10 In article <1168964868.658277.244510(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I deeply deplore that you and others are allowed to call your religious > beliefs mathematics. We deeply deplore both WM's religious mania and his delusion of infallability, particularly in the face of his many logical goofs. > But it is clear now why there is no contradiction > in ZFC and why there will never be such a contradiction. It is to be hoped that there will never be a contradiction within ZFC, but WM seems to have a good deal more faith in that than mathematicians do, and for quite the wrong reasons.
From: Virgil on 16 Jan 2007 15:17 In article <MPG.2016e6f7746f9f07989b49(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > 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? It doesn't, but due to the hitch in WM's giddyup, he is unable to discern the difference between what he wants to deduce from something and what is logically deducible from that something.
From: Andy Smith on 16 Jan 2007 15:20 >>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. > All that I was trying to confirm was, that you think a bicycle chain with a countably infinite number of links is a logical impossibility. > >We say "disjoint", not "exclusive". > Ta. >> 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? > When you say "union" does that include ordering? Of course, all the points in [0,1) plus {1} are all the points in [0,1], but you couldn't put {1} back where you got it from, because [0,1) doesn't have a last member? I am just trying to clarify that you think [0,1) does not have a last member, when viewed as an ordered set. -- Andy Smith
From: Andy Smith on 16 Jan 2007 16:30 >Actually it was David Marcus. > Sorry, sincere apologies, too many Davids. > > >That function cannot be made continuous at 0 at all. It is an essential >discontinuity. > Doubtless you are right, it is very badly behaved, but I would like to know the basis for the assertion. > >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. Yes, I agree, you can't define the derivatives, only the function value at the point, my mistake. But I don't see the distinction between this situation and sin(pi/x) - in one case, the function can apparently be defined arbitrarily at x = 0, and in the other it has to be 0? If you define exp(-1/x^2) to be other than 0 at x = 0, then the derivatives will be discontinous (and infinite), and such a defined function will have no valid Taylor expansion, but so what? You are happy to introduce a discontinuity and an arbitrary value at x=0 on sin(pi/x) ? > > >> 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. > > Well that I totally agree with - it is the underlying truths that are important, not the language or formalism in which they are expressed. So, you agree that you discover (like a scientist), rather than define, mathematical objects of universal interest? -- Andy Smith
From: Andy Smith on 16 Jan 2007 16:58
David Marcus writes >> Of course, I agree formally sin(pi/x) is undefined at x=0. > >What does "formally" mean? Because pi/0 is not defined, but thought that were good reasons for assigning sin(pi/x) a value of 0 at 0, as opposed to 0.3 , -.25 or any other possible number. (Actually, having read the intelligent and helpful post on the value of 0^0 I am suddenly not so sure about this, but anyway, that was what I meant by the "formally") > >> 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 think that you miss my point, or discount it. A line parrallel to the y axis has an infinite gradient; such a line viewed in (m,c) coordinates it has an "undefined" value (inf, inf). But the line is perfectly sensible at e.g. x = k. So, the same thing, I think (maybe), applies to functions. asin(y) is multivalued, difficult to comprehend, but a shift of perspective gives y = sin(x), which is readily comprehensible. Ditto, possibly, for situations involving infinities large or small - a change in variables potentially makes things more comprehensible. > >> 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. > I don't know about Wikipedia - I used the term "antisymmetric". f,g,h are all antisymmetric. g is "obviously" correct - because if you invert the universe left to right, there is no reason to suppose that sin(1/x) changes sign at x =0. Possibly not a killer argument. >> 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"? 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). > >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). Well maybe that would explain how it "gets off the ground". Excuse the loose talk, but I thought that would be clear - if a function and all its derivatives are zero at a point, then if you move an infinitessimal distance delta, to order (delta)**2 the function and all its derivatives are still all zero. And in a Taylor expansion to a finite offset, the function is still zero, because all its derivatives are zero. >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. Well that is undoubtedly the answer to the question. Why doesn't the Taylor expansion work might be a better way of phrasing the question? > >> 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. Yes. > >> > 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? > I am definitely not conversant with your technical definitions of words. Me, as far as Maths courses go, if you want to talk about anything dirty, like applied math, Fourier series, signal processing, vectors and differential equations, I am pretty clued up. But as far as set theory, number theory, pure maths in general, de nada. So from your perspective I am a Klutz. But I am not trying to engage/compete ... just curious, no ego here, happy to admit ignorance. Yes, I should read a book ( I am, actually) but that would just result in more questions... -- Andy Smith |