Re: [math-fun] Terminology
Franklin writes: << Ah, but it isn't that simple. Consider f, not as a function from X to Y, but just as a function. In set-theoretic terms, as a set of ordered pairs, such that nothing appears as the first element of any two distinct pairs. Now the domain is the set of elements appearing as the first element of any pair, and this is exactly the domain in any other sense. But the only reasonable definition for the range is the set of all elements appearing in the second element of any ordered pair; and this is the image in the other sense. Another way to look at it: if Z is the image of f: X -> Y, then f can be regarded as a function X -> Z (or X -> W for any W containing Y as a subset), and this is really the same function. So the image (often called the range) is a property of the function, while the "codomain" is a property of the function as viewed in a particular context, namely as a function X -> Y.
I don't know about this. (For me, "just a function" always includes a domain & range, in the traditional terminology -- but I know what you're saying.) If every function had its codomain by definition shrunken down to its image, then every function would be onto, and questions about whether certain functions are, or could be, onto would become meaningless. But above all: In order to define a specific function, one needs to begin with a set to which the second element of its ordered pairs belong; it is not until after the function is defined that one can even ask what its image is. At that point, you have the option of defining a new function whose ordered pairs have their second elements in the image of the first function. But you also have the option of retaining the original such set (codomain) if there is some reason to do so. E.g., this makes it much easier to ask questions like this: Given a polynomial function of degree d, in n variables, with integer coefficients P: Z^n -> Z, characterize the set I(n,d) of all possible images. (I don't know whether this problem has been solved or not. Anyone know if it has?) --Dan
Assuming Dan means the set of images as the polynomial and its degree take (dome?) all possible values, my distant and foggy memory claims it's the recursively enumerable sets, by Matasevic's theorem --- I think n = 26 suffices. Having recently been sat upon for my poor historical research, I have just attempted to verify this on Google --- but oddly can raise not a single hit. Any logicians out there? Fred Lunnon On 3/19/06, dasimov@earthlink.net <dasimov@earthlink.net> wrote:
Given a polynomial function of degree d, in n variables, with integer coefficients
P: Z^n -> Z,
characterize the set I(n,d) of all possible images.
(I don't know whether this problem has been solved or not. Anyone know if it has?)
--Dan
I'm not saying that viewing a function as X -> Y is meaningless, just that there is another way to view it. And, yes, questions of whether a function is onto make sense only in this sort of context. Certainly, when talking about all functions from X to Y, one does not mean only the ones that are onto. Regarding defining a specific function, I am thinking in terms of ZF set theory. There, a function is just a set of ordered pairs (of sets) with the requisite uniqueness property. Such a function does not bring with it any concept of its codomain, just the image - which is normally called the range. The only general range would be the set of all sets - which does not exist. Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net I don't know about this. (For me, "just a function" always includes a domain & range, in the traditional terminology -- but I know what you're saying.) If every function had its codomain by definition shrunken down to its image, then every function would be onto, and questions about whether certain functions are, or could be, onto would become meaningless. But above all: In order to define a specific function, one needs to begin with a set to which the second element of its ordered pairs belong; it is not until after the function is defined that one can even ask what its image is. At that point, you have the option of defining a new function whose ordered pairs have their second elements in the image of the first function. But you also have the option of retaining the original such set (codomain) if there is some reason to do so. ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com
Found it --- spelling Matyasevic or Matijasevic. E.g. there is a 2006 page at http://primes.utm.edu/glossary/page.php?sort=MatijasevicPoly giving the 26 variable generator I was thinking about --- which unhappily only generates primes, and only as its positive values. I haven't seen the 1971 paper referred to there, but I guess it shows that any r.e. set can be generated as the positive values of a polynomial. The page also mentions that there is no polynomial which generates the primes, so Dan's question is evidently still open. Fred Lunnon On 3/19/06, dasimov@earthlink.net <dasimov@earthlink.net> wrote:
... Given a polynomial function of degree d, in n variables, with integer coefficients
P: Z^n -> Z,
characterize the set I(n,d) of all possible images.
(I don't know whether this problem has been solved or not. Anyone know if it has?)
--Dan
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