Cantor Confusion
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From: Andy Smith on 19 Jan 2007 21:17 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. It is not deliberate. It is a consequence of a misunderstanding on my part of something fundamental. 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. So, for example we can think of some representative points located in x 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 - 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? regards -- Andy Smith Phoenix Systems Mobile: +44 780 33 97 216 Tel: +44 208 549 8878 Fax: +44 208 287 9968 60 St Albans Road Kingston-upon-Thames Surrey KT2 5HH United Kingdom
From: Dik T. Winter on 19 Jan 2007 21:26 In article <rLq5NOUuSTsFFw6O(a)phoenixsystems.demon.co.uk> Andy Smith <Andy(a)phoenixsystems.co.uk> writes: .... > In this situation, > if we want to show that |x|^n converges to zero for all |x| < 1 as > n->oo, we can consider an eta>0, and ask, at a given value of n, what > range of |x| satisfies |x|^n <eta. Then we can then ask, as n->oo > whether we can reduce eta to 0, such that the range of |x| satisfying > |x|^n <eta is |x|<1. > > Is that dyslexic? 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 19 Jan 2007 21:30 In article <ec4EoOVn3UsFFwtu(a)phoenixsystems.demon.co.uk> Andy Smith <Andy(a)phoenixsystems.co.uk> writes: .... > > lim_{n->oo} |x|^n = 0. .... > OK, here is a suggested alternative argument to show that > lim_{n->oo} |x|^n != 0 for all |x| <0 Keep in mind the definition of limit. > 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. But that does not keep in mind how the limit is actually defined. See my previous article on this subject. > So lim_{n->oo} |x|^n cannot be 0 for all |x|<1 ? It is. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 19 Jan 2007 21:49 In article <JC1HI9.84(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <MPG.201861e1fc338409989b62(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: .... > > > I didn't receive it and I didn't enjoy it (how could the last happen if > > > the first did not happen?). > > > > It couldn't. Although, I guess you could have enjoyed not receiving it. > > Oh. We have in our librarys a few publications that prove the squaring of > a circle or the trisection of an angle. It actually is (in my opinion) > enjoyable reading. The reason is that it is like a puzzle: where do you > spot the first error. It is extremely like reading some of the articles > by JSH, when he tries to prove something. Archimedes Plutonium is a bit > more easy. I just received it today. At first glance it looks better than I thought it would (yes, I *did* consider contents). But I will read through it and come up with a review (but it will take some days). From a first view, the first half is about the development about the thought about infinities. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 19 Jan 2007 21:42
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.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |