mm-3719 === Subject: Re: RV convergence - Correction >> If we know that {X_n} and {Y_n} converge to X and Y respectively in >> distribution, can I say that the joint sequence {X_n,Y_n} also >> converges to (X,Y) in distribution? > Yes. A sequence (X_n) of random quantities in a Borel space > coverges to X in distribution iff (Ef(X_n)) converges ro Ef(X) > for every continuous, bounded, real-valued f. Oops. That only shows that if a sequence of random vectors converges in distribution, then so do the sequences of components. The answer to the original quesiton is no. Let Z and W be i.i.d. (e.g., independent and each Bernoulli(1/2)). Let Z_n = Z and W_n = W for all n. Then (Z_n) and (W_n) each converge to Z in distribution, but ((Z_n, W_n)) does not converge to (Z, Z) in distribution. If (X_n) and (Y_n) converge in distribution respectively to X and Y, X and Y are independent, and X_n and Y_n are independent for each n, then ((X_n, Y_n)) converges in distribution to (X,Y). === Subject: Re: RV convergence - Correction Also, if (X_n) and (Y_n) converge in distribution respectively to X and c, where c is a deterministic constant, then ((X_n, Y_n)) converges to (X,c). Of course, all of this presupposes that, for each n, X_n and Y_n are defined on the same probability space. -- === Subject: Re: Still problems when using change of base matrix >>I am trying to understand the use of the change of base matrix. In my >>previous post Representing f: R3 -> R3 by a matrix I think I >>confused >>myself so I am trying to reformulate my problem. >>f: R3 -> R3 is represented by the follwoing matrix, A, in the natural >>base: >>1 -2 -1 >>2 0 2 >>0 1 1 >>I then have this new base represented by the matrix S(-1) consisting of >>the >>vectors (q1,q2,q3): >>0 3 1 >>4 2 1 >>1 -1 -1 >>I would now like to find f(q1) and this can be done by: >>A*q1 = f(q1) >>but what is this and where does f(q1) belong? > If you write q1 as a column, and you left multiply it by A, you get > the result of adding 0 times the first column of A, plus 4 times the > second column, plus one times the third column. This corresponds to > the fact that the first column of A is f(1,0,0), the second column of > A is f(0,1,0), and the third column of A is f(0,0,1); and that > q1 = 0(1,0,0) + 4(0,1,0) + 1(0,0,1). > Thus, what you get is nothing more than > 0*f(1,0,0) + 4*f(0,1,0) + 1*f(0,0,1) = f(0,4,1) = f(q1). It is a > vector (written in column form) of R3. >>expressed in the old base right? That gives me A * q1 = (-9,2,5). >>(-9,2,5) is now the vector, f(q1), expressed in the old base. > Yes. >>Another way to get the vector f(q1) in the old base would involve finding >>q1 >>expressed in the old base and left multiply that with A. >>To get q1 in the old base I do: >>S(-1)*q1 = (11,-27,-16) > No. That's your mistake. > q1 is ALREADY expressed in the old base, the old base being the > standard basis. > The matrix S{-1}, which has the vectors q1, q2, and q3 as columns, is > the basis that translates from a vector expressed in terms of > Q={q1,q2,q3} to a vector expressed in terms of the old base. > To express q1 in terms of Q={q1,q2,q3}, you need to multiply S*q1, > which will give you (1,0,0). This is clear: the coordinate vector of > q1 with respect to Q={q1,q2,q3} is (1,0,0), because > q1 = 1*q1 + 0*q2 + 0*q3. >>I can then do: >>A*q1 (expressed in the old base) = (54,-10,-43) > No. The old base is the standard base. q1 is ALREADY expressed in > the standard base. You don't have to change it before applying A. god damn this is driving me nuts :-) I think I see your point. q1 are as you said expressed in the old base. What I think I am missing (and constantly seems to think q1 is) is a vector that is expressed in the base Q ={q1,q2,q3}. If I had a vector blop expressed in Q I would need to left multiply it with S(-1) to get it in the old base and then I could left multiply blop with A to get f(blop) in the old base.....hope I am finally getting this right. === Subject: Re: Still problems when using change of base matrix days. My association with the Department is that of an alumnus. >Arturo Magidin skrev i en meddelelse [.snip.] >> No. The old base is the standard base. q1 is ALREADY expressed in >> the standard base. You don't have to change it before applying A. >god damn this is driving me nuts :-) I think I see your point. q1 are as you >said expressed in the old base. What I think I am missing (and constantly >seems to think q1 is) is a vector that is expressed in the base Q >={q1,q2,q3}. >If I had a vector blop expressed in Q I would need to left multiply it >with S(-1) to get it in the old base and then I could left multiply blop >with A to get f(blop) in the old base.....hope I am finally getting this >right. Yes! I think your confusion stems from the fact that q1 is not expressed in terms of Q, it is expressed in terms of the old base. The expression of q1 in terms of Q is just (1,0,0) (add one times the first vector of Q, plus zero times the second vector of Q, plus zero times the third vector of Q, with Q= {q1, q2, q3}). Perhaps it might help you to think of it this way: you have two languages, sindarin (the old/standard base), and quenya (the new base Q). Matrix A only understands sindarin, and resplies in sindarin. Matrix S translates from sindarin to quenya; matrix S{-1} translates from quenya to sindarin. The reason why S{-1} has the vectors q1, q2, q3 as columns is that, because S{-1} translates from quenya to sindarin: the 1-st column is how you write q1 in sindarin; the 2-nd column is how you write q2 in sindarin, and the 3rd column is how you write q3 in sindarin. What does S*A*S{-1} do? S{-1} only 'understands' quenya, and when you give it a vector in quenya (the new base), it translates it into the old base (sindarin). Then you give this sindarin vector to A, who promptly tells you, in sindarin, what happens when you apply f to it. Then S takes that sindarin vector and translates it into the new base (quenya). In summary: you give the entire thing a vector in quenya (the new base), and it responds with the value of f at that vector, also expressed in the new base (quenya). If you only do AS{-1}, then you have to give the vector in quenya (the only thing S{-1} understands as inputs); then S{-1} translates it into sindarin, which A understands; then A tells you, in sindarin again, what f does to that vector. So AS{-1} takes as an input a vector v expressed in terms of Q, [v]_Q, and returns the vector f(v) expressed in terms of the standard basis. If you do SA, then you have to give it a vector in sindarin (the only thing A understands); then A tells you what f of that vector is in sindarin, and then S, which understands sindarin, translates it into quenya (the new base). So SA takes as input a vector v expressed in terms of the standard basis, and returns the fector f(v) expressed in terms of Q, [f(v)]_Q. Your B was given by B = SAS{-1}, which takes a vector v expressed in terms of Q as an input, [v]_Q, and returns the vector f(v) expressed in terms of Q as an output, [f(v)]_Q. -- === Subject: Re: Still problems when using change of base matrix days. My association with the Department is that of an alumnus. >>If I had a vector blop expressed in Q I would need to left multiply it >>with S(-1) to get it in the old base and then I could left multiply blop >>with A to get f(blop) in the old base.....hope I am finally getting this >>right. >Yes! Ehr... Caveat: If you have a vector blop expressed in terms of Q, then you would need to left multiply it by S{-1} to get it in the old base, which would yield a vector, let's call it, flop. Then you left multiply ->flop<- (not the original blop) by A, to get f(blop) in the old base. For example: recalling that your map f is defined as having the matrix A = ( 1 -2 -1 ) ( 2 0 2 ) ( 0 1 1 ) as the coordinate matrix with respect to the standard basis, and that Q = {q1,q2,q3} = {(0,4,1), (3,2,-1), (1,1,-1) } suppose you are told that you have the vector v=[(1,2,3)]_Q; that is, the vector which has coordinate vector (1,2,3) with respect to Q. This is the vector which, in terms of the standard basis, is: 1*q1 + 2*q2 + 3*q3 = (0,4,1) + (6,4,-2) + (3,3,-3) = (9,11,-4). If you then left multiply by A, you will get (9+22, -18-4, -9+22-4) = (31,-22,9), which is f(v) expressed in terms of the new base. You can do all of this by taking (1,2,3), left multiplying by S{-1} (which will give you (9,11,-4)), and then left multiplying by A (which will give you (31,-22,9)). If you then further left multiply by S, you will get f(v) expressed in terms of Q. But you have to remember that (1,2,3) does not represent the vector (1,2,3) as we usually think about it, but is instead the name of the vector (9,11,-4) when written in terms of the basis Q. -- === Subject: Re: Still problems when using change of base matrix > No. The old base is the standard base. q1 is ALREADY expressed in > the standard base. You don't have to change it before applying A. >>god damn this is driving me nuts :-) I think I see your point. q1 are as >>you >>said expressed in the old base. What I think I am missing (and >>constantly >>seems to think q1 is) is a vector that is expressed in the base Q >>={q1,q2,q3}. >>If I had a vector blop expressed in Q I would need to left multiply it >>with S(-1) to get it in the old base and then I could left multiply blop >>with A to get f(blop) in the old base.....hope I am finally getting this >>right. > Yes! > I think your confusion stems from the fact that q1 is not expressed > in terms of Q, it is expressed in terms of the old base. The > expression of q1 in terms of Q is just (1,0,0) (add one times the > first vector of Q, plus zero times the second vector of Q, plus zero > times the third vector of Q, with Q= {q1, q2, q3}). > Perhaps it might help you to think of it this way: you have two > languages, sindarin (the old/standard base), and quenya (the new > base Q). > Matrix A only understands sindarin, and resplies in sindarin. Matrix S > translates from sindarin to quenya; matrix S{-1} translates from > quenya to sindarin. > The reason why S{-1} has the vectors q1, q2, q3 as columns is that, > because S{-1} translates from quenya to sindarin: the 1-st column is > how you write q1 in sindarin; the 2-nd column is how you write q2 in > sindarin, and the 3rd column is how you write q3 in sindarin. > What does S*A*S{-1} do? S{-1} only 'understands' quenya, and when > you give it a vector in quenya (the new base), it translates it into > the old base (sindarin). Then you give this sindarin vector to A, who > promptly tells you, in sindarin, what happens when you apply f to > it. Then S takes that sindarin vector and translates it into the new > base (quenya). In summary: you give the entire thing a vector in > quenya (the new base), and it responds with the value of f at that > vector, also expressed in the new base (quenya). > If you only do AS{-1}, then you have to give the vector in quenya > (the only thing S{-1} understands as inputs); then S{-1} translates > it into sindarin, which A understands; then A tells you, in sindarin > again, what f does to that vector. So AS{-1} takes as an input a > vector v expressed in terms of Q, [v]_Q, and returns the vector f(v) > expressed in terms of the standard basis. > If you do SA, then you have to give it a vector in sindarin (the only > thing A understands); then A tells you what f of that vector is in > sindarin, and then S, which understands sindarin, translates it into > quenya (the new base). So SA takes as input a vector v expressed in > terms of the standard basis, and returns the fector f(v) expressed in > terms of Q, [f(v)]_Q. > Your B was given by B = SAS{-1}, which takes a vector v expressed in > terms of Q as an input, [v]_Q, and returns the vector f(v) expressed > in terms of Q as an output, [f(v)]_Q. the change of base matrix. I will keep this post for future use! === === Subject: Re: I have just rediscovered Fermat's proof of the Last Theorem This reminds me of a talk that Joseph Gallian gave about cracking >driver's licenses. He started with the Minnesota DLs, as a recreation, >and then moved on to other states, and even went so far as to write >several states and ask for information about them. (This was pre-9/11.) >One state -- I don't remember which -- said they could tell whether his >guess was correct, but they wouldn't for security reasons. >>What is there to crack about driver's licenses? > In some states, nothing; they're assigned in sequential order. But in > states like Minnesota, the driver's license depends only on your name > and your birthday. Right. For example, in Minnesota, Michigan, and Maryland the license number is a function of the last, first, and middle names and the date of birth. New Jersey uses eye color, too. There's a nice summary of the scheme that some US states use in Joseph A. Gallian, _Mathematics Magazine_, vol. 64, no. 1, Feb, 1991.=== Subject: Re: Questions about the marble problem. [...] >*>That they give the same answer (in any particular case) does not >make >*>them equivalent--what makes them equivalent is that the first is a >*>standard, mathematically precise way to restate the second. You >*>recall the definition of pointwise convergence, don't you? In this >*>example, the definition of pointwise convergence can be stated as: >The set of marbles in the urn _converges pointwise_ to the set S >iff >*>for every marble m in the system, the state of m is constant after >*>steps (where n may depend on m), and this constant state is in >iff m >*>is an element of S. >You are claiming precisely that The set of marbles in the urn at >time >*>infinity is equal to S iff the set of marbles in the urn convereges >*>pointwise to S, are you not? >Please do reread this section and reply to it. Ok. >I think it encapuslates >the heart of our disagreement. I don't. >> Not that limits have anything to do with it, but even >> if limits _did_ have something to do with it, why >> in the world would anyone assume Axiom B was true? >> The cardinality of a set S is simply _not_ a continuous >> function of S, in any sense that I'm aware of. So >> why would we assume that in this problem the cardinality >> _is_ continuous? We're not assuming *anything* is continuous in Axiom B. The state of >the urn at infinity simply hasn't been defined, so we're just defining >it. Do you know what the word continuous means? >Er, I'm taking a graduate course in analysis at present. I am well >>aware of the meanings of all the terms at play here. Looking at this >>again, I suppose my Axiom B could be regarded as extending the >>definition of the number of balls in the urn at time s to infinity by >>assuming the continuity of the counting function, >> That's progress. >> Now, ignoring for a second that the question really has nothing >> to do with convergence at all, and pretending that it does. >> The state of the system at a given time is determined by a certain >> set, namely the set of marbles in the vase at that time. There >> are various notions one could give for the convergence of a >> squence of sets. >> Question (which I believe I've asked before, by the way): >> Can you give _one_ definition of convergence of a sequence >> of sets with the property that the cardinality of the set >> is a continuous function of the set? >You haven't asked it in this thread, I don't think. See below for an >example. >> If not then B is simply unreasonable - the cardinality of >> a set is simply not a continuous function of the set, so >> there is simply no reason one would assume that B holds. >Whether the cardinality is a continuous function depends on the >topology we put on the sets. I state below a definition that makes it >continuous. In this particular example, I don't think there's any >(reasonable) topology that would do that, but that doesn't mean we >can't talk about the number of balls without specifying the set of >balls. Oh for heaven's sake, I give up. The number of balls in the vase means something other than the cardinality of the set of balls in the vase? Yes, we can make any definition we want. As of now the answer to the original question is Pennsylvania. By definition. >> It seems to me that the people who _do_ assume B are doing >> so because the fact that they have little experience >> dealing with infinite sets leads them to assume that >> card(S) _is_ a continuous function of S. Of course the >> people assuming this don't have a clear notion of >> convergence of sets in mind in the first place. >> What _is_ a notion of convergence of sets that makes >> card(S) a continuous function of S? >How about this: A sequence {S_n} of sets converges to the set S iff >there is some natural number N such that for all n > N, S_n = S. >Rather trivial, but the cardinality is certainly continuous under this >definition. Right. Now give a definition of convergence for sets under which cardinality is continuous, and which also has some relevance to the question. >>But compare the case >>where we have only two balls and we swap them at each step: certainly >>there is a ball in the urn at infinity, even though we can't name it >>because it's been swapped infinitely many times. >> That's nonsense. The answer to is there a ball in the vase >> at infinity is undefined. The difference is that here >> we are performing infinitely many operations on each ball; >> there's simply no such thing as the position of the ball >> after we perform all those operations. In other words, the position of each ball diverges pointwise. But even >though we can't say where either ball is at infinity, it would seem >silly to say that the urn doensn't contain a ball of some sort at >infinity. If you say so. In fact in _this_ question the state of the > vase after infinitely many steps _is_ undefined - claiming > anything else seems very silly to me. >Just because we can't say what's in the vase doesn't mean we can't say >>*anything* about it. Perhaps you recall that in measure and >>integration theory, many functions are only defined up to a set of >>measure zero. >> These analogies are irrelevant, and if they were relevant they >> would hold no water. Here, ignoring the irrelevance and pretending >> that this has something to do with that, the reason the analogy >> holds no water is that on a discrete space with counting measure >> there are no nulls sets except the empty set. >Er, it's an *analogy*. I wasn't trying to apply it to this problem, >just giving an example of a well-defined mathematical object that >doesn't specify values at points. >> In particular, we can't specify a *single value* of a >>function defined in this way (at least in the case of Lebesgue >>measure). Would you say that this makes these functions (and therefore >>most of modern analysis) illegitimate? >> Of course not. I don't see the point to any of these analogies. >May I recall that *you* were the first one to bring up analogies to the >real number system? Just as an example of a similar question where the answer is not the limit. > I happen to find analogies generally helpful for >clarification, but if you don't, that's fine with me. >> But since you _insist_ on analogies, here's an analogy for you, >> regarding the second scenario where you say that there clearly >> is a ball in the vase at time infinity, we just can't say which >> one: >> Say X = {0,1}. If n is even say f_n(0) = 0, f_n(1) = 1. If n >> is odd say f_n(0) = 1, f_n(1) = 0. Presumably you'd claim that >> f = lim f_n exists, and that f(x) = 1 for some x in X, although >> we can't say which x that is? >> No, of course you wouldn't say that. And no, that doesn't >> follow from what you said about that second situation. >> But it does show that _if_ we're reasoning by analogy with >> measureable functions then no, we can't say that at infinity >> the vase contains a ball. >I don't think you've thought about this carefully. Would you agree >that the integrals of the sequence {f_n} converge under counting >measure to 1? Because that is precisely what I mean when I say there >is one ball left at infinity. The integrals can converge, even when >the functions do not. Uh, the person who hasn't thought about this is you. The limit of the integrals is 1. But the limit of the integrals is not the integral of the limit. The question was how many balls remain in the vase. If we are talking about limits, that's a question about the integral of the limit, not about the limit of the integral. Confusing the two is exactly how people arrive at the erroneous answers they've been giving. >>Also, has it ocurred to you >>that undefined means we haven't defined it yet? We are free to >>define the state at infinity to be anything we want, as long as our >>definition is internally consistent. >> So if I wish I can simply define the state of the vase at noon >> to have exactly 42 balls in it, and saying 42 in reply to the >> original question makes perfect sense. Fine. >Precisely--until we make some definition, we can't say anything. The >number 42 seems rather arbitrary to me, but if you've got a good reason >for it, I'm willing to listen :-) >> Does it really seem unreasonable >>to say that if the urn contains exactly one ball at every finite step, >>it contains exactly one ball at infinity, when no other information is >>given? >> Yes, it seems like the silliest thing I've ever heard. >> This seems to relate to your put in the marble and do nothing >>example. >> Huh? You're not putting in a marble and doing nothing, you're >> putting it in and taking it out again, infinitely many times. >> Seems like a teensy difference there. >Think about this. Suppose I tell you nothing about the scenario except >that there is a marble in the vase at every step s. You would not, >given only this information, be willing to say that there is a marble >in the bag at infinity? This is precisely the situation of putting it >in and doing nothing. >Not wrong, however, since we haven't defined the state at >infinity. For that matter, if we wanted to be really silly, we could >label the balls with the numbers 1 and 2, and claim that at infinity >the urn contains a ball labelled 1.5, or perhaps a mango or a pear. >The state at infinity isn't defined until we define it. > As opposed to the current question, where only two >> operations are performed on each marble. So the position of each ball converges pointwise to out, which we agree on. We agree that each marble is outside the vase, but somehow > one of us cannot get from each marble is outside the vase > to there are no marbles in the vase. >This is a deduction, which while entirely indisputable at any finite >>step, is not so evident at infinity. >> That's bizarre. _At time infinity_ each marble is not in the vase, >> because each one has been removed. We can't get from at time infinity >> each marble is not in the vase to at time infinity there are no >> marbles in the vase? >As I have stated before, the assumption that allows us to make this >deduction is known in mathematical circles as pointwise convergence, >and while it is obvious for any finite time, it requires an explicit >assumption at infinity. >> Or are we assuming that at time infinity a marble which was previously >> not in the vase jumps back in for some reason? >No, I have stated numerous times that we have no magic marbles--the >state of each individual marble is well-defined by the problem >statement. What we do have, and what requires us to be careful, is a >magical vase which can hold infinitely many balls (and for that matter, >a magical way of moving these balls at infinite speed). No, those bits of magic are not sufficient for what you're claiming. You also need balls appearing that were never inserted. Since you agree that each ball that was inserted was later removed but insist that it's reasonable to say that the number of balls remaining is non-zero. >>The formula (# of balls = >>9*number of steps) is also indisputable for any finite number of steps, >>but you seem unwilling to consider the possiblilty that we can extend >>*this* obvious truth to infinity. >> That's because the cardinality of a set is not a continuous function >> of the set. (Unless you have some notion of convergence of a sequence >> of sets that I've never heard of.) >But to talk about something being a continuous function of the set, >we must put some sort of topology on the sets. I provided above a >topology in which the cardinality *is* continuous, so your blanket >statement here is unwarranted. And of course, as I stated above, the >limit of the cardinalities may exist even when the limit of the sets >does not, and it is a perfectly well-defined mathematical object, even >in these cases. What do you have against defining the number of balls >at infinity to be the limit of the cardinalities (if this limit >exists)? This seems to me to be just as reasonable as defining the >number of balls at infinity to be the cardinality of the limiting set >(if one exists). The number of this and the number of that is the cardinality of a certain set. The question is about how many balls are in the vase at noon. > === Subject: Re: Questions about the marble problem. I think perhaps it is time to end this discussion, since it seems to be diverging to pointlessness (assuming it wasn't there to begin with). I'm sure that both of us have better things to do, involving actual mathematics, rather than wrangling over poorly-stated mental excercises. Therefore I am making a few final comments and leaving it at that; you may take the last word or not, as you wish. > The number of this and the number of that is the cardinality of > a certain set. The question is about how many balls are in the > vase at noon. In the mathematical nomenclature *I* have learned, the concept of number is a good deal broader than the notion of cardinality. In particular, I am familiar with such objects as negative numbers, rational numbers,irrational numbers, imaginary numbers, and even infinite ordinal numbers. Merely asking the question how many does not, in my opinion, suffice to make clear that the number in question is the cardinality of some set. Moreover, even if we grant that the answer must be the cardinality of some set, it is not obvious that this set must be the pointwise limit. Recall the swap at each step example. The pointwise limit doesn't exist, but we *could* if we wanted to, take the set at infinity to be the limsup or liminf, which always exist. (Under the reasoning the ball was in the urn at an arbitrarily late time, so it's in at infinity or similarly, replacing out with in.) Perhaps I'm just a victim of too much mathematical training, but I'm not willing to accept that the one and only possible limit of a sequence of sets is the pointwise limit. Finally, in the context of analysis (which is, more likely than not, the context in which this sort of problem comes up) the answer to a question like how much or how many is usually thought of in terms of integrals. The integral of the limit may not equal the limit of the integrals, but that does not mean the limit of the integrals is not the answer to our question. (I am working right now, as a matter of fact on a homework problem which asks for the latter, rather than the former!) That's all. === Subject: limits of measurable functions How do I prove that for a sequence fn of measurable functions, the set X={x: lim fn(x) eixsts} is measurable. Just some hints or steps towards the proof would be great. === Subject: Re: limits of measurable functions >How do I prove that for a sequence fn of measurable functions, the set >X={x: lim fn(x) eixsts} is measurable. Just some hints or steps towards >the proof would be great. Hint: A sequence of numbers converges iff it is Cauchy. Write f_n(x) is Cauchy in terms of epsilon and n's (but use only countably many values of epsilon), and write your set in terms of unions and intersections of sets that you know are measurable. === Subject: Re: Hermitian matrix >Are hermitian matrices only used in regard to complex numbers? I'm not sure what you mean by used in regard to. Hermitian matrices are the natural generalization to complex scalars of real symmetric matrices. Any real symmetric matrix is a hermitian matrix. Most of the well-known facts about real symmetric matrices are actually facts about hermitian matrices. === Subject: Re: Hermitian matrix So maybe you can help me with this idea... Supposed A is an element of C(mxm matrix) is hermitian, or in the real case, symmetric. a) describe a strategy of symmetric pivoting to preserve the hermitian structure while still leading to a unit lower-triangular matrix with entries |l_ij|<=1. Does this involve row interchanges and column interchanges? b) what is the form of the matrix factorization computed by the strategy? c) what is its asymptotic operation count? === Subject: Hermitian matrix Supposed A is an element of C(mxm matrix) is hermitian, or in the real case, symmetric. a) describe a strategy of symmetric pivoting to preserve the hermitian structure while still leading to a unit lower-triangular matrix with entries |l_ij|<=1. b) what is the form of the matrix factorization computed by the strategy? c) what is its asymptotic operation count? === Subject: Re: countable? the complement of the cantor set on [0,1] >> im trying to show that the complement of the cantor set is countable. >> Good luck. >> i >> have >> solved the recursion >> to find 2k -1 is the number of intervals (in the complement of the >> cantor >> set) >> how do i show this is countable? >> Is an interval countable? >> also i am trouble showing that if there are two differnet cantor sets >> on [0,1] thats there exists a bijection between the complements >> It's easier to show that the cardinalities are the same than it is to produce >> an actual bijection. > Doesn't an order preserving bijection of the endpoints of the intervals > making up the Cantor sets lead fairly easily to a bijection of the sets > themselves? Isn't each point in a Cantor set the limit of a sequence of > these endpoints? > it might even be possible to do this so it is continuous over intervals > containing the Cantor sets by making it piecewise linear on the > compliments of the cantor sets. I don't see how that's easier than observing that any subset of the reals that contains an open interval must have cardinality c. But the OP has stated that this was not the question that was intended. The intended question was how to show that the complement of C is a countable union of open intervals. That's easy, since each of those intervals contains a rational. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: A stupid post about Intel's latest computer chip ( s) Forth-style model, separating the return stack from the data stack >> allows much shallower stacks than a C-style stack frame requires. > i don't see how this affects the space needed for a stack. > it's just partitioned. >> Does C require a combine stack? It is a common implementation, but >> I don't believe it is required. > no. in fact, the return address can be passed in a register. > yeah? how would recursion work? This thread is full of interesting information about register only based recursion and interupt service design methodologies. Maybe read the 1802, micro-thread. My peers are many. ( Also, maybe check the main-thread(s) and re-check all the groups because of broken usenet group linkage from a casterated Followup-To, comp.lang.forth, comp.lang.java.machine, sci.electronics.design, sometimes used by jerky trolls.) > as for FORTH, doesn't it require two memory fetches for each interpreter > cycle? > if the top end of the stacks are stored in special processor registers > that needn't be. > I've heard of forth implementations that run in machine code - define > something and the definition is stored as machine code that executes the > operations in the definition. would that be a compiler, or is it still an > interpreter? > Bye. > ANSI Forth, which includes all extensions, is an interpeter, however, that does not prevent a fully ANSI Forth compliante native code compiler the ability to perform dynamic linkage of a native machine code, thereby, extending the ANSI Forth interpreter with a native code subroutine ( or an inline native code macro expansion), whether, 1) machine forth ( for example, 0< AND XOR DROP OVER DUP @ ! 2* 2/ >R R> INVERT + JUMP_IF_ZERO JUMP CALL RETURN LITERAL ) or 2) ancient, mostly register based, arcitectures, such as Intel. ( a similar example, would include thousands of Intel native machine code statements) ( ..., or 3) my core parallel stack machine engine suggestion, ( , maybe a hundred native codes, at most, for core VLIW, but could also, in theory, generate machine forth as a virtual machine code, like the JVM but where the target could look like this, URL, ) Both some LISPs and some FORTHs are able to perform dynamic linkage of native machine code and extend their interpreter language vocabulary(ies), somewhat alternatively, Kawa Scheme does this in Java byte code. mawcowboy === Subject: Re: A stupid post about Intel's latest computer chip ( s) even if an architecture defines trap-only registers, interrupts must >be disabled for interrupt sservice within the architecture to avoid >multi-trap collisions. > I'd consider that [disabling interrupts to handle each > interrupt] to be a big-time design bug. It doesn't deal > with interrupts that have a higher priority; if an interrupt > has a higher priority you have to catch it, not ignore it, let > alone disable it. Let me try that again It seems accidental CarriageReturn = SendMessageNow. Some architectures only save one prior level of context. E.g. a RISC processor would just copy the old program counter and processor status value into a specific register pair, disable interrupts and jam the interrupt address into the PC. The interrupt routine preamble saves the prior state and reenables. The RCA 1802 COSMAC, circa 1975, it was 5V CMOS (4000 series and TTL compatible), with the power consumption directly proportional to clock speed. You could stop the clock and draw only uAmps. The architecture had its own special brand of weirdness. It had no stack NOR a formal program counter! My memory may be a bit off, but IIRC... It was 16 bit address, 8 bit data, with an 8 register bank. There was a Program Counter Pointer (PCPR) register that indicated which of the general registers was the program counter. On interrupt, the PCPR was incremented and the next general register was used as a program counter. It was up to the interrupt routine to save the state as it required and manipulate the registers and PCPR if it didn't want to use up all the general registers saving prior program counters. Eric === Subject: convergence problem I am trying to prove the following question: Assume X_1, X_2, ... are i.i.d. random variables. Find necessary and sufficient conditions for X_n/n converges to 0 almost surely. I have no idea how to start this question. Could someone give me a === Subject: Re: convergence problem >I am trying to prove the following question: >Assume X_1, X_2, ... are i.i.d. random variables. Find necessary and >sufficient conditions for X_n/n converges to 0 almost surely. >I have no idea how to start this question. Could someone give me a Hint: the Borel-Cantelli lemmas will be useful. === Subject: Re: Is sci.math on the slide? Baez | >| There are only 4 balls inside the cube. The others are 8 at the corners | >| and 6 in the centres of the faces. The lines connected to those 4 do | >| correspond to bonds, the others don't. Each is at the centre of a | >| tetrahedron formed by the 4 balls/atoms it is bonded to, one corner ball | >| and three face-centre ones. Much clearer pictures can be found at | >| http://phycomp.technion.ac.il/~nika/diamond structure.html | >| | >Each dark blue corner ball has 4 links. | >Each interior pale blue ball has 4 links. | >Each green face-centred ball has 2 links, leaving 12 links open | >to connect to 12 cubes. Every atom connects to four other | >atoms. Four atoms form the vertices of a tetrahedron. | >Therefore the space is tesselated with tetrahedra. | | I am not sure what you mean by links here. There are indeed four lines | linked to each dark blue ball in the diagram referenced above, but only | one represents a chemical bond; the others are just to show the | underlying cubical structure. The four linked balls do, of course, form | a tetrahedron, but it is not regular, the corner ball is not inside it, | and it overlaps some of the other tetrahedrons. If you mean it has 1 | bond within the cube, and 3 to balls in other cubes then you are | correct. | | If you throw in all the tetrahedrons formed by these sets of 4 atoms | linked to a single atom they overlap, as each is in the centre of one | tetrahedron and a vertex of 4 others, (but they still don't fill space). | If you take just the ones with a pale blue ball at the centre and dark | blue or green balls at the vertices (or vice versa) you get a lovely | array of regular tetrahedra, connected at each vertex, but otherwise not | touching. There are large spaces between them, they are not a | tessellation. These spaces cannot be filled in with more regular | tetrahedra. (They can of course be filled with irregular tetrahedra, but | that's not the issue.) | | >Take a large tetrahedron. Cut the top off at half the height. | >That leaves a triangular base. | > / | > /A A is a tetrahedron. | > | | | > cut 1. | > | | | > / R R = remainder. | > /B R cut 2. | > R/ /C cut 3. | >Cut 4 is behind the view. D is a tetrahedron. | >What is R after cut 4? An inverted tetrahedron. | | No. Its base is a triangle, not a point. Okay, yes, you are correct. No fooling you, is there? Its upper face is also a triangle. That leaves three side faces extending down to three side faces. Five faces altogether is not an octohedron :-) | Round the middle you're left Can't have any rounding, sorry. You can cut a plane, but no rounding allowed. Isn't this fun, trying to communicate on Usenet? I'm so used to the other person being an idiot that it just isn't possible for me to be one as well, so you won't mind if I make ridiculous statements to try to wriggle out of it, will you? | with one triangle from each side (the bit left in your diagram of cut | 3), and a triangle where each of B, C and D attached, and on top you | have the triangle where A attached. 8 sides, all triangles, it's an | octahedron. | -- | I'm retired now, but I used to work in flight simulation. Most modern simulator platforms are mounted on 6 extendable hydraulic legs, coupled to a triangle at the base and a reversed triangle under the platform, making an octohedron. http://www.cadsoft.de/~kls/fltsim/platform.jpg Needless to say I was fully aware the whole time of the octohedral interior of the tetrahedron, but please look at the thread title. I was doing a Baez. When someone intelligent such as yourself broke in, that kinda spoilt the fun. Baez is a typical know-it-all idiot that cannot think, the thread crossed into sci.physics.relativity and Einstein's idiocy which Baez and his ilk propagate is no different to me claiming that an octogon is made up of four tetrahedra. But I admit you caught me. Neither Baez nor Tom Roberts who idolizes Baez have the courage to do that. They'd prefer to go on believing in nonsense for the rest of eternity and continue to spread malicious lies to young, unsuspecting students for the benefit of their own glory. 137. Baez Oct 3, 5:50 pm hide options Local: Mon, Oct 3 2005 5:50 pm === Subject: Re: Baez's stupidity. Reply | Reply to Author | Forward | Print | Individual Message | Show original | Report Abuse >In >http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/doppler.gif >Baez has constructed 16 arrows in the left diagram and 32 in the right >diagram. This sentence is false. If you could read, you'd know why: I may not be able to read, but I can count my fingers and toes. Baez, Ph.D, cannot count. Heaven only knows where he got his degree but it wasn't from my Alma Mater. author Local: Sat, Sep 17 2005 6:57 pm === Subject: Re: Does the 'Curvature of Spacetime' cause gravity? Yes, tests of strong fields are few and far between, but there are some: the binary pulsars, and observations of accretion disks near black holes I've asked Roberts Ph.D. for the right ascension and declination of the the accretion disk near the black hole he's OBSERVED. I want to see it also. My apologies for jerking your chain. You didn't deserve it. === Subject: Re: Falling in curves | > A free-falling fart follows a curved path through spaceTIME. http://www.freefarts.com Androcles. === Subject: Re: Falling in curves Einstein found it. But he did not > find the cause of not falling in curves. > You can freefall in a straight line. > Just drop something and it travels a > straight line. > A cannonball falls along a parabola (approximately) -- that is a curve. > It speeds up rather than > curves. > Yes, of course, RELATIVE TO COORDINATES FIXED ON THE EARTHS SURFACE. In > GR this is explained by the acceleration of those coordinates. > A free-falling object follows a curved path through spaceTIME. > The speedup mechanism is what Einstein > missed. The theory of gravity is > incomplete without it. > Einstein got it right. You did not. You never will until you actually > LEARN something about physics. > A free-falling object follows a curved path through spaceTIME. > Well, in a sense, a free-falling object is one that > *doesn't* follow a curved path through space-time; a free Yes, of course. And a geodesic is the appropriate generalization of straight line into a curved manifold. I was implicitly using the context of the original poster, and its implicit reference to coordinates fixed on the surface of the earth. Tom Roberts tjroberts@lucent.com === Subject: Re: Falling in curves > A free-falling object follows a curved path through spaceTIME. >> Well, in a sense, a free-falling object is one that >> *doesn't* follow a curved path through space-time; a free > I was implicitly using the context of the original poster, and its > implicit reference to coordinates fixed on the surface of the earth. Right: that makes more sense. I was a bit puzzled about why you would have written that. === Subject: Maximization Problem Let's say that I want to find f[x] such that Integral[f[x]*Log[f[x],{x,0,1}] is a max and f[0] = f[1] = 0, Integral[f[x],{x,0,1}] = 1 and f[x] >= 0 for all 0<=x<=1. How would I go about doing this? Lagrangian multipliers only seem to yeild a constant function - although such a function does maximize the above integral, it does not do so subject to the constraints I set up. How can I include said constraints in my maximization? === Subject: Polar-Decomposition vs. Procrustres rotation Cc: ukintzel@aol.com, arndt@jjj.de why I got a different solution from that of the author see http://www.jjj.de/fxt/fxtbook.pdf.gz J.9arg Arndt Chap 15.5 The subject is also dealt in http://opus.kobv.de/tuberlin/volltexte/2005/972/pdf/kintzel_ulric.pdf of Ulric Kintzel, but I haven't read that long text to its end yet. [28] set listing=on Given Matrix is A A : | 1.00 1.00 0.75 | | -0.50 1.50 1.00 | | 0.75 0.50 -1.00 | A = E1 * B1, where B1 is symmetric and E1 is orthogonal [29] b1 = sqrt(a'*a) b1 : | 1.30 0.24 -0.23 | | 0.24 1.77 0.55 | | -0.23 0.55 1.48 | [30] e1 = a*inv(b1) e1 : | 0.80 0.29 0.52 | | -0.49 0.82 0.29 | | 0.34 0.49 -0.80 | [31] chk =e1'*a // should be B1 chk : | 1.30 0.24 -0.23 | | 0.24 1.77 0.55 | | -0.23 0.55 1.48 | [32] a_chk = e1*b1 // should be A a_chk : | 1.00 1.00 0.75 | | -0.50 1.50 1.00 | | 0.75 0.50 -1.00 | With my own implementation of procrustres-rotation I get a different, but also valid result: a procrustes rotation to unity-matrix A = E * B, where B is symmetric and E is orthogonal [33] e = gettrans(a',procrustres,einh(a)) // finds a column rotation matrix for A' approximating unity-matrix e : | 0.08 0.95 -0.29 | | 0.11 0.28 0.95 | | 0.99 -0.11 -0.08 | [34] b = e'*a b : | 0.77 0.73 -0.83 | | 0.73 1.32 1.10 | | -0.83 1.10 0.82 | [35] chk = e'*a // should be B chk : | 0.77 0.73 -0.83 | | 0.73 1.32 1.10 | | -0.83 1.10 0.82 | [36] a_chk = e*b // should be A a_chk : | 1.00 1.00 0.75 | | -0.50 1.50 1.00 | | 0.75 0.50 -1.00 | I also could find some different solutions by performing row rotations. Until now I thought my procrustres - method was giving a unique result... Also the deviation of B from identity-matrix is worse than the standard-polar-method. Does someone have an idea? Gottfried Helms === Subject: Re: Geometry Problem > _________________ > A / B > / > M /____P_______________N > / > / > D/__________Q_______________C Given: > Trapezoid ABCD > M is midpoint on leg AD > N is midpoint on leg BC Prove: MN is parallel to CD Draw APQ parallel to BNC,P is midpoint of AQ.... > Why, exactly? I thought of a similar approach, but realized that > it didn't work very smoothly. > Ken Pledger. > .... Yes, I see all that. But just after you've drawn APQ parallel to BNC, you've shown _neither_ that MP is parallel to DQ (which is what sudden claim that P is the mid-point of AQ. Ken Pledger. === Subject: Re: New packings of unit squares in squares > I'm not sure that Prai Jei thought they'd be simple. > Surely he just wanted precise expressions, simple or not. I do not understand Prai Jei's reaction now, but anyhow he did ask an interesting question und you gave an interesting answer. === Subject: Re: New packings of unit squares in squares message <20051025004458.819$BG@newsreader.com>: > Packing 29 unit squares in a square of side length s = 5.9465+ > The link throws a 403 forbidden error when I try it > For comparison, the previous best known packing of 29 squares, due > to Bidwell in 1997, having s = 5.965 approx, can be seen at Erich's > Packing Center . ISTR a debate about how the stars on the US flag would be rearranged if ever more states joined the Union. (I'm a British citizen so that debate was rather academic for me.) The layouts of squares shown for 51 (but with the central strip moved up to a more symmetrical position) and 52 would seem to answer that question for those cases, but the layout for 53 looks just *horrible*. -- The internet is missing, I think I've deleted it. Interchange the alphabetic letter groups to reply === Subject: Re: New packings of unit squares in squares >> Could you please clarify what the different s values are, e.g. cube >> root of 215 or solution of x3 - x = 200 or whatever. >> What makes you think there should be such a simple expression? > I'm not sure that Prai Jei thought they'd be simple. Surely he just > wanted precise expressions, simple or not. > For n = 29, s = Root[51 - 146*#1 + 71*#12 - 14*#13 + #14 & , 2] > which is notation for a specific root (the second one in Mathematica's > ordering) of the polynomial 51 - 146*x + 71*x2 - 14*x3 + x4. > Of course this could also be expressed in radicals, if desired. The next > ones are messier. > For n = 55, s = Root[3263162 - 5430600*#1 + 3533776*#12 - 1132594*#13 > + 186713*#14 - 15166*#15 + 481*#16 & , 2]. > For n = 71, s = Root[-563063 + 1228316*#1 - 784922*#12 + 211944*#13 > - 21677*#14 - 1064*#15 + 438*#16 - 36*#17 + #18 & , 6]. > David I didn't recognise them as square roots, cube roots or anything simple like that so I guessed they would be something a bit more involved. But I knew that there must be some rationale behind the values. Or should that be irrationale? :) Yes I think it should, and I'll submit that to the neologisms section on the langmaker website http://www.langmaker.com -- The internet is missing, I think I've deleted it. Interchange the alphabetic letter groups to reply === Subject: Re: infinity I suspect that the sweepers at the local animal center where one might take the kids may disagree with you -- but it depends on the animal. :-) I doubt lizards shed many follicles thereof, for example -- though polar bears might clog the drains of their swimming pools. As for transfinite cardinals -- those are as mythical as numbers, which can't be caught in anything even remotely resembling a cage, butterfly net, or sack, but it's fairly obvious that there are an indefinite (and infinite!) number of transfinites, starting with card(N) and continuing with P(N), P(P(N)), P(P(P(N))), ... . And that's probably not all of them, either; there might be ones in the middle, for all I know. (Did we solve card(R) = card(P(N)) yet?) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: infinity ... >> [Alrecht] are claiming that one of the members of the infinite set is >> equal to the size of the set. This is not obvious to us because it >> contradicts proven set theory. It is your responsibility to prove >> that your claim is true, and that standard set theory is wrong. >> Saying that your claim is obvious is not a proof. >> His claim is obvious based on the diagram he offered, which graphically >> shows the element values along one side of a growing square, and the >> element count along the other side. They are obviously, graphically, >> inductively, always equal. One is not infinite while the other is finite. >> It's impossible. > Here's his second diagram: >> # O O O O O O O O O ... 1 >> # # O O O O O O O O ... 2 >> # # # O O O O O O O ... 3 >> # # # # O O O O O O ... 4 >> # # # # # O O O O O ... 5 >> # # # # # # O O O O ... 6 >> # # # # # # # O O O ... 7 >> # # # # # # # # O O ... 8 >> # # # # # # # # # O ... 9 >> . . . . . . . . . . . > Every natural is represented by a horizontal string of #'s. The vertical > string of zeroes directly to the right of the last # in each natural denotes > the size of the set through that natural. With the addition of each natural, > both the horizontal number of #'s and the verticle number of 0's are > incremented in tandem, so this equality is preserved. The slope of -1 shows > that there is a 1-1 correspondence between element value and set size. > This is a good graphical illustration of what I have been trying to say > regarding the naturals. You only have an infinite number of them when you > allow infinite values for them. > If that's the case, at what point do the naturals on the right side > stop being finite and start being infinite? > All I see is an infinite list of rows, where each row is labeled by > a finite natural k and contains k leading # marks and an infinite > number of trailing O marks. > column contains a finite number of O marks on top and an infinite > number of # marks on the bottom. > But why can't there be an infinite number of finite naturals? > Do you have some kind of proof of this? You have never offered one. > You've offered plenty of statements about unidentifiable largest > naturals and set ranges and unit infinities, but you've never > actually pinpointed how you get from the finites to the infinites. Think about the numbers which follows in horizontal direction as sets of Os as natural numbers and the numbers in vertical direction which are sets of #s as cardinal numbers. Or think of the example with the infinite triangle it might be clearer to you since it is very impressive. I understand, it could not be easy to end indoctrinated thinking, and it is not done in one day. AS === Subject: Re: infinity > In general: when you biject to a series you are implying Ordinal > similarity. > WRONG! > Bijection of itself says nothing more that a one to one matching of > members of the sets. Order is irrelevant to the bijection, even when the > bijection is based on the ordering of members of the sets being bijected. > Fine then biject the series S as I defined it in a previous post > when as all (n,0)pairs being placed befor all (n,1) pairs. > S={ (0,0),(1,0),(2,0),.........,(0,1),(1,1),(2,1),..........} > in this order and then I want you to tell me how the field of that > series would equal to the cardinality of N. > Zuhair Cardinality depends only on existence of injections, surjections and bijections between SETS, and sets are not intrinsically ordered, even though their representations may be. presented its members is irrelelvant, I am free to reorder it as I please, as long as I am only considering set properties not order properties of the set S. I do as follows: (S,<) = {(0,0); (0,1),(1,0); (0,2),(1,1),(2,0); ...}, i.e., ordering so that (x,y) < (u,v) when (a) x + u < u + v, or (b) should x + y = u + v, then when x < u. With this ordering, the set is both well ordered and order isorphic to the naturals N = {0,1,2,...}, so this order isomrphism of ordered sets produces a bijection of sets. Now if Zuhair wants an order isomorphism between N and S with the natural order on N and HIS order on S, that is not possible === Subject: Re: infinity ... You misinterpret totally when you say, I think there must be an > infinite natural number. I don't think so. I only argue that, if there > are infinite sets, there must be infinite natural numbers (since nat. > numbers are sets). > OK. Make the substitutions, natural numbers = Greeks, sets = mortals > and > infintite = German [1] > Then the statment if there are infinite sets, there must be infinite > natural numbers (since nat. numbers are sets) becomes if there are > German mortals, there must be German Greeks (since Greeks are > mortals). > Aristotle must be rolling in his grave. About the sillyness a man could shows like you do? Don't forget Einstein: The universe and the stupidity of men are infinite. About the universe I'm not really shure. > -William Hughes > [1] Deutchland, Deutchland, ueber alles [2] William Hughes before the fuehrer AS === Subject: Re: infinity > I presume by now you know the definition of a bijection from > set A to set B. > 1. For every element of B, there is some element of A mapped > to that element of B (surjection). > 2. No two elements of A are mapped to the same element of B > (injection). > I think I should elaborate on this notion bijection which is repeatedly > used > in this group. Bijection is used as similarity criterion, when two > things are > bijected then they are similar in what they are bijected. That is garbled. A bijection between {square, circle, elephant} and {lettuce, tomato, blue} does not establish similarity in the elements. It only means elements of one set can be used as labels for elements of the other. It is used as a definition of cardinality. > I want only to speak alittle bit about what cardinality mean. Since clearly you do not share our view of what cardinality means, then what you are going to be speaking on is your private definition of cardinality. > Cardinality is an index of comparison between the multiplicities of > discriminated collection of items . > Now for example { Lion , elephant , dog} this is a discriminated > collection of items > usually called set which is a group of things fulfilling a specific > propositional > function and having no specific relation between them other than > discrimination. > when we say the cardinality of { Lion , elephant,dog} = x then this > means also > that it is the same for any arrangment of that set No, we don't mean that. That is something which can be proven for finite sets, but there is nothing in the definition of cardinality which implies that to be true in general. > In order to determin similarity of cardinality we use the concept of > one-one corresponce. We define cardinality in terms of bijections. One-to-one correspondence (injection) is necessary but not sufficient for a map to be a bijection. > Before we go to one-one corresponce we should know what is one-one > relation > A relation is said to be one-one when,if x has the relation in > question to y, no other term x' has the same relation to y, This is more or less the definition of injection. It is often given in this form: f(x) = f(x') implies x = x'. > and x does > not have the same relation to any term y' other than y. Not everyone uses that as part of the definition of one-to-one correspondence. Because there are two separate conventions for the meaning of that phrase, the term bijection is preferable, having only one interpretation. > if we have X={a,b,c}, Y={e,f,g} and their is one-one relation between > all their members then this is one-one corresponce ( bijection) > Now the above is cardinal bijection and it implies similarity in > cardinality. It is the DEFINITION of EQUAL cardinality. > However their is another kind of bijection No, there isn't. > For example:A= 4,5,6 bijected to B=1,2,3 , this kind of bijection is > not cardinal bijection. No, a bijective map between these two sets is precisely the same as one between the above two sets: it is an injection and a surjection. What possible difference do you see? > For example if a,b,c is bijected to 1,2,3 this means that a,b,c are of > the same sequence as 1,2,3 What the heck does of the same sequence mean? > It DOESN'T imply that a,b,c are of the same cardinality By definition, it means that {a,b,c} and {1,2,3} have the same cardinality. Far enough. This is already seriously wrong-headed. === Subject: Re: ring/field/algebra >> We can, but we do not have to. >So, up to Sylow's Theorems OK? > As it should have been obvious to anybody without a personal agenda, > that statement was meant to be hyperboly and over the top. The > original poster was asking wether one could and whether one should > include redunt axioms into the definition of groups and rings. Yes, > one can. One can include any theorem deducible from axioms into the > axioms and obtain a new axiom set that is logically equivalent to the > original one. But in fact, MY point when I made that statement (as was > evidently clear) was that the fact that we ->can<- include such > statements does not, in any way, mean that one ->should<- include such > statements. That it was completely unnecessary to include the > statement of closure in the specific definition being discussed (that > which appeared in MathWorld), given the CONTEXT of that definition. I would always oppose adding unnecessary structures on a whim. > [.snip.] >>The axioms for a system are meant to be the shortest list of >>statements that makes it into that system. >> There are other considerations besides brevity and independence that >> often go into a choice of which axioms to use in in an axiom system. >> Other things being equal, both brevity and independence in an axiom >> system are to be desired, but other things are rarely equal. >Yes I meant an independent list. And in fact it is a good idea to use >the best fit in that if there is an independent list of axioms that is >better than yet equivalent to the 'shortest' list that is fine by me. > Pretty much all groups in ring theory include commutativity of > addition, and many include the existence of a multiplicative > identity. Yet, from the existence of a multiplicative identity, and > the distributive laws, commutativity can be derived. Yet I do not see > people agitating for the removal of commutativity from the list of > axioms of ring theory. You do appear to have got me there. But remember what you call rings today were once rings with 1. In other words the rings were there and requirement for a multiplicative identity was added on to make the study of rings smooth. Now we have chosen an add-on condition as an axiom. That is bound to have some effect. Now why have they not shortened the list? Expediency, I suppose or may be laziness. There could be other reasons, like having to jilt the notations '+'and '0'. Not wanting to sever the ring connection etc. But this cannot be compared with what you said. >This is how the usual definition of a group was kept as standard. But >threatening to include up to Sylow's Theorems in the group axioms, >without legitimate cause! > You misunderstood the context of that statement. Par for your course, > apparently. (I almost took your word for it, and I am sure most of your readers must have thought the diagram came from where you said it did.) For the readers' information a better diagram and much more on Schreier and pre-Schreier rings can be found at http://www.lohar.com/mit.html My comments are at the tail-end of the following pasted material. (I am particularly angry about that whole affair because you guys splashed my name across the internet so much that it probably made my (ex) colleagues so uncomfortable that they had to pass a rule that no one can work as a visitor for more than three years. What two paper Professor wants to have a visiting assistant professor who gets mentioned on the net? Now, I lost my job and did not get a well-deserved citation and that makes me angry.) - Hide quoted text - - Show quoted text - > [.snip.] >>It's actually a little bit better than just the existence of a GCD: each >>finitely generated ideal in the algebraic integers is principal. So not >>only do elements a and b have a GCD, but it can be written in the form >>ua+vb for algebraic integers u and v. It suffices for this to be true for >>ideals generated by two elements. Rings having this property are called >>Bezout. I believe some rings have GCDs without being Bezout. Being Bezout >>is partway toward having unique factorization: any two factorizations can >>be refined into ones which only differ by units. >The property you mention here, any two factorizations can be refined >into equivalent ones, is actually a bit weaker than the Bezout >property, and were called the pre-Schreier property by >P.M. Cohn. People more familiar with universal algebra/lattice theory >may know it as the Riesz Interpolation Property on Ideals. Every >Bezout domain is pre-Schreier (in fact, every Bezout domain is a >Schreier domain, meaning that in addition to being pre-Schreier, it is >also integrally closed in its field of fractions), but not every >pre-Schreier domain is a Bezout domain; for example, any UFD is a >Schreier domain, so Z[x] is a Schreier domain, but not a Bezout >domain. >P.M. Cohn proved that a domain is a UFD if and only if it satisfies >the Schreier property AND every element may be factored into >irreducibles in at least one way. > [.snip.] >Bill posted several references to papers dealing with this, which were >a great help to me personally in revising the Gauss's Lemma >manuscript to include those results. One of the papers which covers >this and more is >D.D. Anderson and R.O. Quintero, Some generalizations of >GCD-domains. In Factorization in integral domains (Iowa City, >1996) , Lecture Notes in Pure and Applied Mathematics 189, Dekker >1997, pp. 189-195, MR 98i:13001. >A table of implications which appears there includes the following >(but I have left out several other intermediate terms): >PID -> UFD > | | > V V >Bezout-> GCD -> Schreier -> Pre-Schreier -> AP >(AP means atoms are primes, that is, every irreducible satisfies >the prime divisor property: x|ay -> x|a or x|y). >None of the arrows are reversible, so you are correct that there are >GCD domains which are not Bezout, and not UFDs either (to get rid of >the trivial examples like Z[x]). >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magi...@math.berkeley.edu How about proving your statement: PID -> UFD > | | > V V >Bezout-> GCD -> Schreier -> Pre-Schreier -> AP Using the Anderson-Quintero paper: D.D. Anderson and R.O. Quintero, Some generalizations of >GCD-domains. In Factorization in integral domains (Iowa City, >1996) , Lecture Notes in Pure and Applied Mathematics 189, Dekker >1997, pp. 189-195, MR 98i:13001? The only thing that I could find from that paper is something like: GCD property-->PSP property-->GL property-->AP property Apart from the GCD propert none of the other properties implies Schreier property! > -- === Subject: Re: Pure mathematics as intellectual dead end He is merely saying > that mathematicians should have a preference for applications, that > mathematics divorced from application leads to useless > self-involvement. The mathematics useful for applications necessarily has empirical content.