Cantor Confusion
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From: Andy Smith on 17 Jan 2007 17:13 David Marcus writes >I don't think "formula" has a precise meaning, but we generally use it >to mean some sort of expression, e.g., polynomials, trig functions, >exponential functions, a combination of these. If we define a function >using cases, e.g., > >f(x) = 17, if x < 0, > -3.14, if x > 0, > >then we usually don't consider this to be a formula. More formally, a >function is a set of ordered pairs such that if (a,b) and (c,d) are both >in the function and a = c, then b = d. Have you seen that definition? No, but (if you have the time) why is that useful? > >> >> 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. > >Yes. But, if you replace x by ix, then you get e^{1/x^2}. So, if you >come down the imaginary axis, the function isn't continuous. Yes, I can see that gets very nasty around z = 0. But, just because a function can be generalised to complex numbers, I don't see why one can't stick to a one-dimensional real representation if one wants to. > > >Besides discussing functions, Spivak also discusses why the Taylor >series of e^{-1/x^2} doesn't work. > Looking it up on Amazon now. Ty. -- Andy Smith
From: David Marcus on 17 Jan 2007 17:14 Han de Bruijn wrote: > Set theory is quite useful as a limited, relatively unimportant, part of > mathematics. You sound like Charlie-Boo. Are you related? -- David Marcus
From: David Marcus on 17 Jan 2007 17:35 Andy Smith wrote: > David Marcus writes > > >I don't think "formula" has a precise meaning, but we generally use it > >to mean some sort of expression, e.g., polynomials, trig functions, > >exponential functions, a combination of these. If we define a function > >using cases, e.g., > > > >f(x) = 17, if x < 0, > > -3.14, if x > 0, > > > >then we usually don't consider this to be a formula. More formally, a > >function is a set of ordered pairs such that if (a,b) and (c,d) are both > >in the function and a = c, then b = d. Have you seen that definition? > > No, but (if you have the time) why is that useful? It is more precise than saying "rule". I suggest reading Spivak. He has a lovely chapter where he discusses the definition of "function" and works up the final version in stages. > >> 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. > > > >Yes. But, if you replace x by ix, then you get e^{1/x^2}. So, if you > >come down the imaginary axis, the function isn't continuous. > > Yes, I can see that gets very nasty around z = 0. But, just because a > function can be generalised to complex numbers, I don't see why one > can't stick to a one-dimensional real representation if one wants to. Yes, you can. However, the Taylor expansion doesn't know you want to do this, and so has trouble because of the complex values. -- David Marcus
From: Virgil on 17 Jan 2007 17:43 In article <WYxG5Ixc+prFFwkG(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > David Marcus writes > > >I don't think "formula" has a precise meaning, but we generally use it > >to mean some sort of expression, e.g., polynomials, trig functions, > >exponential functions, a combination of these. If we define a function > >using cases, e.g., > > > >f(x) = 17, if x < 0, > > -3.14, if x > 0, > > > >then we usually don't consider this to be a formula. More formally, a > >function is a set of ordered pairs such that if (a,b) and (c,d) are both > >in the function and a = c, then b = d. Have you seen that definition? > > No, but (if you have the time) why is that useful? > > > >> >> 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. > > > >Yes. But, if you replace x by ix, then you get e^{1/x^2}. So, if you > >come down the imaginary axis, the function isn't continuous. > > Yes, I can see that gets very nasty around z = 0. But, just because a > function can be generalised to complex numbers, I don't see why one > can't stick to a one-dimensional real representation if one wants to. A power series series has a /circle/ of convergence whose radius often depends on a complex value of the variable for which the series diverges. > > > > > >Besides discussing functions, Spivak also discusses why the Taylor > >series of e^{-1/x^2} doesn't work. > > > Looking it up on Amazon now. Ty.
From: Dik T. Winter on 17 Jan 2007 19:52
In article <1169050591.632101.147680(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Ok. But this does not help either. The complete tree contains the > > path 0.0101010101...; none of the 'finite' trees contains that path. > > So that path is *not* in the union of the sets of paths of the finite > > trees, while it is in the union of trees. > > So you say 0.0101010101... is in the union of finite trees while ohers > say it is not? Probably they are using a slightly different definition for the union of trees. Because you at first did not define it at all, but only did use the term, it is no small wonder that people assume different things about what it means. In an earlier article I also gave my definition of what the union of trees could be and with that definition it was indeed not in the union of trees. That happens when you use terms without explicit definition. > It cannot be in the union of the finite trees, they say, because it is > not in one of the finite trees. They use something similar to the definition I gave previously where a tree is the triple [set of nodes, set of edges, set of paths], and in a union the sets are united. In that case it is indeed *not* in the union of trees. If as you (finally) defined, a tree consists only of the sets of nodes and edges, and a path in a tree is defined as a sequence of connected edges starting at the root of the tree, in that case the infinite tree contains the path 0.010101... . Further with the finite trees as you currently define them (with all having infinitely many nodes, so I the only finiteness is the number of splitting), the union of sets of paths in two finite trees is the set of paths in the union of those two trees. *But* that does not hold when you unite the whole collection of finite trees. > But *all* paths are in the union of the > finite trees, because it is simply silly to suspect different sets of > paths in a tree with all nodes. Eh? There is a whole collection of different sets of paths in a tree with all nodes (assuming the definition of tree as I stated above, that is the pair [set of nodes, set of edges]). Of course it is silly to suspect different complete sets of paths. *Assuming* you use the same definition of uniting trees. You never give proper definitions and that gives a lot of confusion. > Therefore the only conclusion is that > the path 0.010101... does not exist. Wrong. > > > The answer is: We can use sets of levels, sets of edges, sets of nodes > > > or sets of paths. Look at the "Trauerweide". > > > > Yes. But still the union of the sets of paths in the finite trees is not > > the set of paths in the complete tree. There are paths in the complete > > tree that are in *none* of the finite trees and so not in the union of > > those sets of paths. > > So you say. If you were right, there should be some indication for the > difference. > What makes the path 0.010101... be in a tree with all nodes but not in > a tree with all nodes? Again, I do not state that. What I state is that that path is (with the definitions above) in the complete tree. And I also state that that path is not in the union of the sets of paths of the finite trees (also again according to the above definition). And I say so (and am allowed to say so) because that path is in none of the finite trees. > > > > None. But all sets of diagonals of finite squares contain only finite > > > > diagonals, so there can not be an infinite diagonal in their union > > > > (which is a set of infinite size). > > > > > > So there cannot be an infinite diagonal at all? > > > > I do not state that. I state that in the union of the sets of diagonals > > of the finite squares only finite diagonals do exist (infinitely many > > of them). This does *not* prevent an infinite diagonal to exist. > > What can we do to switch its existence on and off? I do not do that. Where do I state that it does not exist? I only state that it exists but that it is not in the union of the sets of diagonals of finite squares. But again, all your arguing is lacking definitions. So perhaps we are talking at cross-purposes. Learn to once in a time to provide proper definitions. > > > 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.) > > > > Right. But the complete diagonal is not in the union of the sets of > > diagonals of the finite triangles. > > Where is it? Or is it nowhere? Why should it be in that union? What reasoning are you going to assume that it *must* be in that union? It is simply not in that union, and that is easily enough to prove. But it exists, but not in that union. > > > 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). > > > > It does contain that path (according to your definition above). > > Yes, that is obvious. But, for a moment, switch to the cut trees, > please. Does the union of all cut trees contain the paths 0.111...? You > would say no. And again, I would say yes. I even provided a definition of union of trees for that case and with that definition that was easily enough to prove. But then we have that the union of the sets of paths of two different finite trees is not the set of paths of the union of those two trees. > > But > > the union of the sets of paths in the finite trees is not the set of > > paths in the infinite trees. The first does not contain the path > > 0.01010101..., but it is in the second. > > If this path exists, then it must be constructible from all the nodes > present. There is a bit, 0 or 1, for every index n in N. That is not > sufficient? Not sufficient to be in the union of the sets of paths of the finite trees. The reason is that a path is in that union *if and only if* it is in one of the sets that are used to form the union. To get back to an earlier example, take the sets of rationals [1/2^n, 1 - 1/2^n]. Their union is (0, 1), not [0, 1]. And the reason is simply that binary 0.111... is not in the union while all of 0.1, 0.11, 0.111 etc., are. > > > > > There are more limits than sequences? > > > > > > > > Eh? Why do you think so? Each sequence has only a single limit (when the > > > > limit is properly defined and when it exists). > > > > > > Correct. > > > > So why your question? > > There are only countable many sequences. I disagree. There are uncountably many limits. You are trying to prove that there are countably many sequences and limits. If there are countably many sequences there are countably many limits, and the other way around. But until now you have failed to prove either the first or the other. Your proof of countably many sequences hinges on the assumption that the union of the sets of paths in the finite trees is the set of paths in the completed trees. But that one is false, as I did show above. > > > The countablity of all paths in the union of finite trees is a theorem > > > of set theory. > > > > No, it is not. Prove it. You use the *false* assumption that the > > union of the sets of paths in the finite trees is the set of paths > > in the complete tree. It is not because the first does not contain > > the path 0.01010101... . > > I use the fact that the union of all nodes contains the nion of all > paths. Now you are talking about unions of paths. The theorems of set theory are about unions of *sets* of paths. There is quite some difference. What is the union of all paths? How do you *define* the union of two paths? > I mean that in an infinite tree (= union of all finite trees) with all > nodes, there is a set of paths. Paths which are not therein, must > indicate why they are not therein - not be a spell but by a bit (or > node). Again, I never stated that there are paths missing from that set of paths. I only stated that there are paths that are not in the union of the sets of paths of the finite trees. And it is simple to state why, those are the paths that are not present in the finite trees. > > (1) and (2) together state that while all paths are different, there > > is *no* single edge that makes a single path stand out from all other > > paths. > > But paths which are not determined by the set of all nodes must stand > out by some indication, There are no such paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |