Cantor Confusion
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From: David Marcus on 16 Jan 2007 18:33 Andy Smith wrote: > > David Marcus wrote: > > > Andy Smith wrote: > 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. It is certainly a practical impossibility. I'm not sure what you mean by "logical impossibility" in this context. What would a bicycle chain with an infinite number of links be like? > >> 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? As all the elements of the sets we are discussing are real numbers, we can use the standard ordering of the real numbers for them. The number 1 doesn't suddenly become less than 1/2 just because we consider the set [0,1). If we let S = [0,1), then it is true that for all x in S, x < 1. Not sure if that answers your question because I'm still not sure what "put {1} back where you got it from" means. > I am just trying to clarify that you think > [0,1) does not have a last member, when viewed as an ordered set. If by "last" you mean "largest", then [0,1) does not have a last member. More formally, for all x in [0,1), there exists y in [0,1) such that x < y. A set T has a larget member if there exists x in T such that for all y in T, y <= x. -- David Marcus
From: David Marcus on 16 Jan 2007 18:44 Andy Smith wrote: > Dave Seaman wrote: > >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. Definition: A function f is continuous at a if lim_{x->a} f(x) = f(a). For the function f(x) = sin(1/x), lim_{x->0} f(x) does not exist. So, regardless of how we define f(0), f won't be continuous. That proves that the discontinuity is essential. > >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? We can define functions with whatever value of f(0) we want. > If you > define exp(-1/x^2) to be other than 0 at x = 0, then the derivatives > will be discontinous (and infinite), If the function isn't zero at zero, then it won't be continous, so the derivatives won't exist at zero. (Infinity is not a number.) > and such a defined function will > have no valid Taylor expansion, but so what? All you are saying is the function probably isn't useful. So, what? > You are happy to introduce > a discontinuity and an arbitrary value at x=0 on sin(pi/x) ? Don't follow. -- David Marcus
From: David Marcus on 16 Jan 2007 19:03 Andy Smith wrote: > David Marcus writes > >> 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, There is a single-valued version of asin. In fact, all functions are single valued (although in advanced math there are sometimes "multivalued functions", but this is an abuse of language). > 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. You are saying that different ways of analyzing a situation may have different attributes. And, some ways may work better than others. Very true. > >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. The question is whether your "antisymmetric" requires the domain to be all of R. How can h be antisymmetric? Doesn't the value at zero break the anti- symmetry? > 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. What does "correct" mean? > >> 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). Ah, you are at least a hundred years behind the times. No, a function is most definitely not a formula. A function is a rule which assigns, to each of certain real numbers, some other real number. For example, the rule that assigns to each number a the number 0 if a is irrational and the number 1 if a is rational is a function, but you will have a hard time coming up with a formula (nor is a formula required). > >> (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. Yes. The Taylor expansion can't get off the ground. However, the function can. > >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? An interesting question. It turns out that the behavior of the function for complex values of x prevents the Tayor series from working for real values. > >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... Actually, I take back what I said about the word "define". I think your main problem is that you aren't familiar with the modern concept of function. Functions are no longer formulas. They were in mathematics a couple of hundred years ago, and engineers and scientists still often think of them that way. But, things have changed. The modern concept clarifies things a lot. You might enjoy looking at the book Calculus by Michael Spivak. I'm sure that it explains Calculus in a way that is very different from how you learned it. -- David Marcus
From: Dave Seaman on 16 Jan 2007 19:08 On Tue, 16 Jan 2007 21:30:11 GMT, Andy Smith wrote: >>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. In order for a function to be continuous at a point x=a, three things have to happen: (1) f(a) must exist. (2) lim_{x->a} f(x) must exist. (3) The values in (1) and (2) must agree. When (2) fails, as is the case for sin(pi/x) at x=0, it means that no assignment of f(a) can possibly make the function continuous there. That's called an essential discontinuity. By contrast, if (2) holds but (1) or (3) fails, then the function can be made continuous by a suitable assignment of f(a). That's called a removable discontinuity. >>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? There is no distinction between the two cases. Each of those functions can be defined arbitrarily at 0, or left undefined there. > 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) ? If a function is not defined and continuous at x=0, then the derivatives do not exist at all at that point. There is certainly nothing wrong with that. It makes no sense to talk about derivatives being discontinuous or infinite if they are simply undefined. >>> 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? Discovery and definition are not mutually exclusive. Definitions provide us with a convenient way to state our discoveries. With a different set of definitions, the same discovery might be worded quite differently. I don't mean in the sense that English and French are different; English-speaking and French-speaking mathematicians tend to use mostly the same definitions in their respective languages. A mathematician from Tau Ceti might very well choose different fundamental concepts to assign names to, and therefore the statement of our familiar theorems might come out seeming quite different. -- 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 19:09
In article <1168955868.733166.287460(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > No initial segment of Cantor's diagonal number is infinite. Neverthelss > > > there is an infinite diagonal number? > > > > The diagonal number is not infinite. > > > Not wih respect to the hight of its digits and not with respect to the > value of the number, but it is infinite with respect to its number of > digits. Indeed. So you should better specify what you are meaning. > > > The limit of a sequence is not an element of the sequence. > > > > The limit of a sequence is *not* the union of a sequence. Union only works > > for sets. > > Yes, for the sets of nodes for instance. And for the sets of paths if > each path is appended by 000... and 111... as I described in my last > constribution. No. The first level tree has only paths that go through two nodes, the second level tree only paths that go through three nodes. They have no path in common (regardless the nomenclature), so the union contains both paths that go through two nodes and paths that go through three nodes. > > > Then also an infinite path with omega nodes can be created by a set of > > > finite segments of paths. > > > > As a limit, yes. But that does still tell us nothing about unions of > > sets of paths. When you create a union of two sets you take all elements > > from every set and remove the duplicates. > > We an do that with nodes and with paths (as originally defined). There are no duplicates to remove. A path that goes through two nodes is inherently different from a path that goes through three nodes. In a much earlier article, moreover, you stated that a path got its digits from the digits assigned to the nodes you pass through, so appending arbitrary digits to it is, eh, indeed arbitrary. > > > The relevance is: If the union of all nodes of a path is not an > > > infinite path, then the union of all digits of Cantor's diagonal is not > > > a real number. > > > > But I am not stating that. I am stating that "the union of the sets of > > paths in the finite trees is not the set of paths in the infinite tree". > > Please use my original definition yielding infinite paths in finite > trees. Your original definition stated that the digits are obtained by going through nodes. > > There is no consideration needed here about nodes, edges and whatever. > > Paths cannot exist where no nodes are. Therefore paths cannot exist > (partially) outside of the tree containing all nodes. Ah, so paths cannot exist (partially) outside the finite trees. So how can you even *think* about extending the paths in the finite trees to infinite paths? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |