Cantor Confusion
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From: David Marcus on 14 Jan 2007 15:28 Franziska Neugebauer wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > << snip >> > Without notice you paste from a different post. This is considered bad > practice. It is originally from > <45a6c9da$0$97224$892e7fe2(a)authen.yellow.readfreenews.net>. I would have thought it was also bad practice to post nonsense for two years. Ah, well. -- David Marcus
From: Andy Smith on 14 Jan 2007 17:58 Either David Marcus, or Dave Seaman, probably both plus others have told me that I was wrong to assume that an infinite series needs to have a last member. e.g. 1/2+1/4 + .. and I do see that now. But can you have an infinite series that does have a last term? The point about the sin(pi/x) zero crossings that I was feeling towards was that: IF you can have an infinite ordered sequence, such as the set of zero crossings ordered on their x coordinate, and IF you can identify a last one such as at x=0 in [-1,0], then can't one assert that: - all the crossings exist - there is then a countably infinite series where each term is the difference in x coordinate of successive zero crossings - that series has a last member - it sums to 1. ? And if you say that we can define sin(pi/x) = 0 at x =0, can't we do that? Not wishing to start another rabbit running on sin(pi/x, it is very badly behaved around 0, with an infinite number of zero crossings in ANY finite e around 0. I would see that as good evidence for ihe existance of something infinitessimal i.e. a number that you need to add an infinity of to get something finite. Otherwise it is hard to see how the function sin(pi/x) ever gets away from x=0 ... but that is probably not a very mathematical way of expressing it... > I'm not using a Web interface. Try using a non-Web > interface. > Such as e.g. ? (excuse the ignorance) --- andy
From: Andy Smith on 14 Jan 2007 19:07 The sin(pi/x) zeros thing is a red herring, and I am being unecessarily dim. I can choose any N reals in the closed interval [0,1] a(0) to a(N-1) and order them such that a(k)>a(k-1) for all k >0, k<N. I can define a series: S = (a1-a0) + (a2-a1) + ... + (a(N-1) - a(N-2)) which necessarily integrates to a(N-1)-a(0) = 1 If I could choose a (countably) infinite N, then I can have an infinite series with a defined last term. But, you will say, I cannot let N become (countably) infinite, because there will be terms in the series that I cannot identify? i.e. the kth term in the sequence: 0,1/2,3/4,..((2^k-1)/2^k) can be made arbitrarily close to 1, but cannot be 1. But even if I define a sequence 0,1/2,3/4,..((2^k-1)/2^k),..,1 by including the additional element I can't say that the corresponding series S = 1/2 + 1/4 + ... = 1 has a last term in it, presumably because I can't define it?
From: Dave Seaman on 15 Jan 2007 07:28 On Mon, 15 Jan 2007 03:58:33 EST, Andy Smith wrote: > Either David Marcus, or Dave Seaman, probably both plus others > have told me that I was wrong to assume that an infinite > series needs to have a last member. e.g. 1/2+1/4 + .. and I > do see that now. But can you have an infinite series that does > have a last term? Let me clarify a bit. First of all, what I was saying is that an infinite ordered set may have a last member. A *series* is something rather different. It probably makes more sense in the present context to speak of a *sequence* rather than a *series*. Usually the word "sequence" means a mapping whose domain is the natural numbers, and in that sense a sequence never has a last member. However, the word "sequence" also has a more general meaning. If alpha is any ordinal, then an alpha-sequence is a mapping defined on alpha. An ordinary sequence in the sense of the previous paragraph is thus an w-sequence. A sequence in this more general sense therefore has a last member whenever the domain is a successor ordinal. In particular, an (w+1)-sequence is a mapping f: w+1 -> X for some set X, and such a sequence does indeed have a last element, namely f(w). > The point about the sin(pi/x) zero crossings that I was > feeling towards was that: > IF you can have an infinite > ordered sequence, such as the set of zero crossings > ordered on their x coordinate, > and IF you can identify > a last one such as at x=0 in [-1,0], > then can't one assert that: > - all the crossings exist > - there is then a countably infinite series where each term is the difference > in x coordinate of successive zero crossings > - that series has a last member > - it sums to 1. ? The set of crossings is not a series, or even a sequence in the sense that I have explained. It is an ordered set, but not well ordered. > And if you say that we can define sin(pi/x) = 0 at x =0, > can't we do that? A function can be defined any way you want. > Not wishing to start another rabbit running on sin(pi/x, > it is very badly behaved around 0, with an infinite > number of zero crossings in ANY finite e around 0. I would > see that as good evidence for ihe existance of something > infinitessimal i.e. a number that you need to add an infinity of > to get something finite. Otherwise it is hard to see how the function sin(pi/x) ever gets away from x=0 ... but > that is probably not a very mathematical way of expressing it... Each of the crossings is at some finite x. No infinitesimals are involved. >> I'm not using a Web interface. Try using a non-Web >> interface. > Such as e.g. ? (excuse the ignorance) I wasn't the one who made that comment, and I am not aware of the context. There are lots of dedicated newsreaders available, some of which offer a GUI interface. I use slrn myself, which has a text-based interface. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dave Seaman on 15 Jan 2007 07:41
On Mon, 15 Jan 2007 05:07:08 EST, Andy Smith wrote: > The sin(pi/x) zeros thing is a red herring, and I am > being unecessarily dim. > I can choose any N reals in the closed interval [0,1] > a(0) to a(N-1) and order them such that a(k)>a(k-1) for > all k >0, k<N. > I can define a series: > S = (a1-a0) + (a2-a1) + ... + (a(N-1) - a(N-2)) > which necessarily integrates to a(N-1)-a(0) = 1 Why does a(N-1)-a(0) = 1? > If I could choose a (countably) infinite N, then I can have an infinite series with a defined last term. Suppose a(k) = 1 - 1/2^k for each natural number k, and suppose we set a(w) = 1. Then what is your "series"? For each finite k > 1 you have a term a(k) - a(k-1), but you don't have a "last" term because w has no predecessor. > But, you will say, I cannot let N become (countably) > infinite, because there will be terms in the series > that I cannot identify? Each point can be identified, but your difference-of-successive-points scheme breaks down. > i.e. the kth term in the sequence: > 0,1/2,3/4,..((2^k-1)/2^k) can be made arbitrarily > close to 1, but cannot be 1. But even if I define > a sequence > 0,1/2,3/4,..((2^k-1)/2^k),..,1 > by including the additional element > I can't say that the corresponding series > S = 1/2 + 1/4 + ... = 1 > has a last term in it, presumably because I can't > define it? There's no problem with the last member of the sequence (1), but you can't define a difference of successive terms at that point. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/> |