Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Dik T. Winter on 16 Jan 2007 21:26 In article <ssXIUBVEGUrFFwxp(a)phoenixsystems.demon.co.uk> Andy Smith <Andy(a)phoenixsystems.co.uk> writes: .... > > That function cannot be made continuous at 0 at all. It is an essential > > discontinuity. > > > Doubtless you are right, it is very badly behaved, but I would like to > know the basis for the assertion. Simply because lim{x -> 0} sin(pi/x) does not exist. And a function is continuous only if the limit exists and if the limit is also the function value on the point. A fucntion is essentially discontinuous at some point if there is *no* value that makes it continuous. [ about exp(-1/x^2) ] > Yes, I agree, you can't define the derivatives, only the function value > at the point, my mistake. But I don't see the distinction between this > situation and sin(pi/x) - in one case, the function can apparently be > defined arbitrarily at x = 0, and in the other it has to be 0? No, in both cases you can set it arbitrarily. However, in the case of sin(pi/x) the limit to x = 0 does not exist. In the case of exp(-1/x^2) that limit *does* exist, and it is equal to 0. > If you > define exp(-1/x^2) to be other than 0 at x = 0, then the derivatives > will be discontinous (and infinite), No, when you define it as anything different from 0, the derivative is undefined at x = 0. > and such a defined function will > have no valid Taylor expansion, but so what? You are happy to introduce > a discontinuity and an arbitrary value at x=0 on sin(pi/x) ? Go ahead. But that does not make sin(pi/x) well defined at x = 0. There is *no* way to define that function for x = 0 so that the function is continuous at x = 0. The case is different for g(x) = exp(-1/x^2). Even when you define g(0) = 0, *all* the derivatives at x = 0 are 0. So there is no valid Taylor expansion around x = 0. Or perhaps it is valid, but all the coefficients are 0. To quote something you wrote earlier: > but you CAN define a function f by > e.g. Taylor expansion in x about some x0, such that f(x) = > exp(1/x^2) for all x!=0, There is no way to do that. If you do a Taylor expansion around some x0, that Taylor expansion will have a radius of convergence, i.e. a circle around x0 where it converges (assuming complex analysis). That circle never includes singular points of the function, and in this case the circle would be the circle around x0 that has the origin on its circumference. So the Taylor series would not work for -x0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 17 Jan 2007 02:24 >> >David Marcus wrote: > >If by "last" you mean "largest", then [0,1) does not have a last member. >More formally, for all x in [0,1), there exists y in [0,1) such that x >< y. A set T has a larget member if there exists x in T such that for >all y in T, y <= x. > Yes, OK, thanks. The [0,1) business was not a good example. -- 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: Andy Smith on 17 Jan 2007 02:36 David Marcus writes .... >> >It is true that k is not continuous, but it is a perfectly valid >> >function. Do you know the definition of "function"? >> >> Doubtless you lot have some legally watertight definition. As far as I >> am concerned, a function is a formula that provides a value for a given >> input (or inputs, in a multidimensional situation). > >Ah, you are at least a hundred years behind the times. No, a function is >most definitely not a formula. A function is a rule which assigns, to >each of certain real numbers, some other real number. For example, the >rule that assigns to each number a the number 0 if a is irrational and >the number 1 if a is rational is a function, but you will have a hard >time coming up with a formula (nor is a formula required). Thanks for that, but actually it appears "formula" has a technical meaning for you. rule and formula are synonyms for me. >> Well that is undoubtedly the answer to the question. Why doesn't the >> Taylor expansion work might be a better way of phrasing the question? > >An interesting question. It turns out that the behavior of the function >for complex values of x prevents the Tayor series from working for real >values. > OK, thanks, I will look that up I guess. Bit it is odd superficially in that e(-1/x^2) is super smooth with all its derivatives etc. > >You might enjoy looking at the book Calculus by Michael Spivak. I'm sure >that it explains Calculus in a way that is very different from how you >learned it. > Thanks I might well do that. -- Andy Smith
From: mueckenh on 17 Jan 2007 05:27 Han de Bruijn schrieb: > stephen(a)nomail.com wrote: > > > All you have proven is that you do not know what the Continuum Hypothesis is. > > LOL! Should I add a dozen smileys ? > :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) Add several dozens! The problem with your example is that set theorists distinguish very sharply between 2^omega and 2^aleph_0. On the other hand, they use omega instaed of aleph_0. Cantor already did so. There are some ways to show that there is no difference between 2^omega and 2^aleph_0. Compare my recent arguing which forces set theorists to assert that two binary trees with identical nodes contain different paths. This is obviously such a wishful thinking that I hope, most newcomers will soon recognize that set theory is as useless as an intellectual game as it is with respect to applications. Regards, WM
From: mueckenh on 17 Jan 2007 05:34
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > Franziska Neugebauer schrieb: > >> [...] > >> >> > Obviuosly yoou intermingle "unary" and "binary". > >> >> > >> >> Indeed I mean "binary represenation". To resume: If your set of > >> >> binary representations does not contain the representation of 1/3 > >> >> your proof is meaningless already before starting to write it > >> >> down. > >> > > >> > The representation of 1/3 is in the union of all finite trees. > >> > >> The path of 0.[01] is not a member of Sigma* with binary alphabet > >> Sigma = { 0, 1 }. > > > > But the path of 1/3 = 0.[01] is a member of a binary tree which > > contains all possible nodes --- infinitely many. > > 0.[01] (the alternating path) is a member of the infinite binary tree. > > > The union of all finite binary trees contians all nodes --- infinitely > > many. > > This is obviously not the union in the sense of set theory. Set > theoretically a tree is a directed graph which is defined as an ordered > pair (V, E). The union of two trees is in general not a tree at all. What I defined a sthe union of finite trees is the union in the sense of set theory. > > So obviously you mean a different tree representation in terms of sets > and/or a different definiens in the definition of "union". > > > Now apply your logic. > > To which definitons? I gave them. Look them up or leave it. I will not repreat them hre. > > >> > It is the limit of a set of finite paths like N is the limit of all > >> > of its finite initial segments. > >> > >> This should read: The binary representation of 1/3 is _not_ a member > >> of the set of all finite paths _as_ N is _not_ a member of the set of > >> all natural numbers. > > > > Correct. > > Aha. > > > One could add: Cantor's diagonal number is not a member of > > the set of all finite initial segments of Cantor's diagonal number. > > (And only such initial segments occur in Cantor's diagonal proof). > > Since no infinite string is a member of the set of all finite sequences > of strings this is trivially true. So what? The Equilateral Infinite Triangle (EIT) contains all lines which can be enumerated by natural numbers: 1 0.1 2 0.11 3 0.111 ............ In addition it contains the complete diagonal 0.111... (Otherwise, Cantor's diagonal proof would fail.) The union of all finite binary trees contains all levels which can be enumerated by natural numbers: 0 0. / \ 1 0 1 / \ / \ 2 0 1 0 1 ................................ But it does not contain the complete path 0.111... (Otherwise Cantor's proof of the uncountability of the real numbers was contradicted). Now, the path 0.111... in fact would look like a diagonal 0 0. 1 1 2 1 3 1 .................. Are the different notations "line" and "level" the reason for the difference? Regards, WM |