Cantor Confusion
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From: Andy Smith on 19 Jan 2007 22:34 Dik T. Winter writes > >Perhaps, but I do not know. To show that |x|^n converges to 0 for all >|x| < 1 it is sufficient to get the definition of limit from the cupboard: > lim{n -> oo} f(n) = L iff for all eps > 0 there can be found an n0 such > that for all n > n0, |f(n) - L| < eps. >So assume some eps < 1. Take n0 = ceil[log eps / log |x|]. We have eps and >|x| < 1, so log eps and log |x| < 0 and the quotient is > 0, so n0 > 0. Now >set n some integer > n0, we have: > n > log eps / log |x| > n.log |x| < log eps (because log |x| < 0) > exp(n.log |x|) < exp(log eps) (because exp is strictly increasing) > |x|^n < eps >as required. So given an arbitrary eps we can find an n0, and that proves >that the limit is what it should be. Thank you. Helpful. Do you see any possible difficulties with the above as eps->0 and |x|->1 ? -- Andy Smith
From: David Marcus on 19 Jan 2007 22:42 Andy Smith wrote: > David Marcus writes > >> > > >> > >> OK, here is a suggested alternative argument to show that > >> lim_{n->oo} |x|^n != 0 for all |x| <0 > >> > >> We can consider the mapping from x to y_n given by y_n = |x|^n. The > >> effect of the exponentiation is to 'stretch' [0,1] non linearly, so that > >> , depending on the size of n, progressively more points uniformly spaced > >> originally in [0,1] are concentrated in the bottom of [0,1] in y_n. But, > >> that doesn't matter - we have plenty of reals. There are as many reals > >> in any interval of y_n in [0,1] as that of x. > >> > >> So it doesn't matter how large n becomes - y_n has the same "density" of > >> reals as x. And that must also apply at the limit n->oo - if it were > >> true that you could 'stretch' the reals by such a power law, so as to > >> map all the reals in 0,1 to either 0 or 1, this would be tantamount to > >> saying that the reals are not continuous on the line - the reals are > >> infinitely expandable? > >> > >> So lim_{n->oo} |x|^n cannot be 0 for all |x|<1 ? > > > >Did you read what I wrote? Doesn't seem like it. I'll try again. The > >sentence > > > > lim_{n->oo} |x|^n = 0, for |x| < 1 > > > >means > > > > lim_{n->oo} |0|^n = 0, > > lim_{n->oo} |0.1|^n = 0, > > lim_{n->oo} |0.2|^n = 0, > > lim_{n->oo} |-0.278|^n = 0, > > lim_{n->oo} |0.78|^n = 0, > > lim_{n->oo} |sqrt(2)/2|^n = 0, > > etc. > > > >That's all it means. It does not mean > > > > lim_{n->oo} sup_{|x|<1} |x|^n. > > > >When we want to say, "lim_{n->oo} sup_{|x|<1} |x|^n", we say, "lim_{n-> > >oo} sup_{|x|<1} |x|^n". We don't say, "lim_{n->oo} |x|^n = 0, for |x| < > >1", because the latter means that lim_{n->oo} |x|^n = 0 for whatever > >value of x we happen to try as long as |x| < 1. The only way that > > > > lim_{n->oo} |x|^n = 0, for |x| < 1 > > > >can be false, is if there is a real number y such that |y| < 1 and > > > > lim_{n->oo} |y|^n != 0. > > > >Is there such a y? > > I can see that I am irritating you. No, you aren't. > It is not deliberate. It is a > consequence of a misunderstanding on my part of something fundamental. I'm sure! > You say, is there such a y? Well, if we consider the mapping > w = Lim n->oo x^n, 0<=x<=1 , we can consider that as the limit of a > succession of mappings. This is simply a language problem. The meaning of what is written is not what you think. It is possible to write what you are thinking: see below. An analogy would be if I tell you what the word "cat" means and you ask why can't a cat be the animal that barks. It could be, but it isn't. We have another name for that animal. > So, for example we can think of some > representative points located in x You can't say "in x". x is a number. We don't know exactly which, but it is still a number. So, just as you can't say "in 7", you can't say "in x". "x" is a name for a number, the number x. Similarly, "7" is a name for the number 7. > e.g x - 0,1/4,1/2,1 on [0,1]. After > x^2 we get > 0,1/16,1/4,1 and after x^n these points map to 0,1/(4^n),(1/2^n), 1. > But we still have a continuum of points in [0,1] after n iterations, and > each of these map back to an original location in x. And that has to be > true after n->oo There is no "after n -> oo". > - the line is infinitely elastic. Otherwise you could > define a space between two reals in x? > > There is then an inverse mapping from the points in w in [0,1] back to > their original locations in x? And these would all describe the y that > you request? Name a number y with |y| < 1 that will work. You can't. The following is from Chapter 24 of the book "Calculus" by Spivak. "Let {f_n} be a sequence of functions defined on the set A. Let f be a function which is also defined on A. Then f is called the uniform limit of {f_n} on A if for every e > 0 there is some [natural number] N such that for all x in A, if n > N, then |f(x) - f_n(x)| < e. "We also say that {f_n} converges uniformly to f on A, or that f_n approaches f uniformly on A. "As a contrast to this definition, if we only know that f(x) = lim_{n->oo} f_n(x) for each x in A, then we say that {f_n} converges pointwise to f on A. Clearly, uniform convergence implies pointwise convergence (but not conversely!)." Another way to write that {f_n} converges uniformly to f on A is lim_{n->oo} sup_{x in A} |f_n(x) - f(x)| = 0. Let f_n(x) = x^n. Let f(x) = 0. Then {f_n} converges pointwise to f on [0,1), but not uniformly. -- David Marcus
From: Andy Smith on 19 Jan 2007 22:46 In message <JC5BJn.8C1(a)cwi.nl>, Dik T. Winter <Dik.Winter(a)cwi.nl> writes >In article <ZNAVsEWGCVsFFwdL(a)phoenixsystems.demon.co.uk> Andy Smith ><Andy(a)phoenixsystems.co.uk> writes: >... > > >Why? If m is 10^90, is it hard to talk about (10^90)! ? > > > > > Some time back you showed me that sin(pi/x) has a fundamental > > discontinuity at x=0. > >Yes, that is fundamental. > > > Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can > > you explain, please? > >That function has a whole lot of problematical points. It is not even >defined for irrational x. So there are a whole lot of discontinuities. >But they are not fundamental. Defining the function: > f(x) = lim{m -> oo} cos(m!.pi.x) when x is rational and > f(x) = 1 when x is irrational >solves them all and leaves a continuous function. So the discontinuities >are not fundamental. (Unless someone is able to show that for some x, >lim{m -> oo} cos(m!.pi.x) is defined.) Yes, thanks. But actually the whole point of all of this discussion was a throwaway remark by David Marcus to the effect that finding a formula f(x) to generate 1 if x was rational, and 0 if x was irrational would be difficult (for me), someone generated the smarty formula, and I said subsequently that I might have been able to generate it, but would not be able to rigorously justify it. ... Anyway, redefining lim{m -> oo} cos(m!.pi.x) as a function of rationality of x rather defeats the purpose of that exercise... -- Andy Smith
From: David Marcus on 19 Jan 2007 22:51 Andy Smith wrote: > David Marcus writes > >Andy Smith wrote: > >> > >> Some time back you showed me that sin(pi/x) has a fundamental > >> discontinuity at x=0. > > > >We said two things. Division by zero is not defined. So, the expression > >sin(pi/x) is not defined for x = 0. We also said that the function > > > > f(x) = sin(pi/x), x != 0 > > > >can not be extended to a continuous function that is defined on all of > >R. > > > >> Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can > >> you explain, please? > > > >You don't evaluate limits by taking the value you are approaching and > >sticking it in the expression. And, oo isn't a number, so it wouldn't > >even make sense in this case. cos(m! pi x) is defined for all real > >numbers x and all nonnegative integers m. So, it isn't any harder to > >consider the limit as m -> oo as it would be to consider the limit as x > >-> oo. > > If x is irrational cos(m! pi x) does not have a limit as m->oo, it is > undefined? Correct. It does not have a limit. So, if x is irrational, then lim_{m->oo) cos(m! pi x) is undefined. If that is what you meant a few posts back, then you were correct. Sorry if I said otherwise. -- David Marcus
From: David Marcus on 19 Jan 2007 22:53
Andy Smith wrote: > Dik T. Winter writes > > > >Perhaps, but I do not know. To show that |x|^n converges to 0 for all > >|x| < 1 it is sufficient to get the definition of limit from the cupboard: > > lim{n -> oo} f(n) = L iff for all eps > 0 there can be found an n0 such > > that for all n > n0, |f(n) - L| < eps. > >So assume some eps < 1. Take n0 = ceil[log eps / log |x|]. We have eps and > >|x| < 1, so log eps and log |x| < 0 and the quotient is > 0, so n0 > 0. Now > >set n some integer > n0, we have: > > n > log eps / log |x| > > n.log |x| < log eps (because log |x| < 0) > > exp(n.log |x|) < exp(log eps) (because exp is strictly increasing) > > |x|^n < eps > >as required. So given an arbitrary eps we can find an n0, and that proves > >that the limit is what it should be. > > Thank you. Helpful. > > Do you see any possible difficulties with the above as eps->0 and |x|->1 > ? I don't think your question makes sense. First you pick x. Then you pick an eps. Then you have to find n0. -- David Marcus |