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From: David Marcus on 11 Jan 2007 16:13 Andy Smith wrote: > OK, thanks for that, content appreciated. > > I know I said I was calling it quits, but, once more into > the fray... > > What about the ordered set of the zero crossings of the > function: > > sin(exp(1/(x**2)) between say x = -1.0 and 1.0? > > a) is that set infinite (yes, for sure?) Yes. > b) does it have a well defined first and last member? If you mean smallest and largest, then yes. > So you can have an infinite set with a first and last member? Sure. Consider the following sets of real numbers: {x| 0 <= x <= 1} {x| 0 < x <= 1} {x| 0 < x < 1} -- David Marcus
From: mueckenh on 11 Jan 2007 16:23 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > [...] > > Conclusion: > > Every finite binary tree contains a finite set of path. > > True. > > > The countable union of finite sets is countable. > > True. The union of countably many countable sets is countable. Yes, that is with AC. But even without AC the countable union of finite sets is countable > > > The set of paths is countable. > > 1. The set of all rooted _finite_ paths in an infinite tree is > countable. The union of all finite paths is exactly this set of all > rooted finite paths. Yes, of infinitely many paths, each of which has infinitely many nodes. Why do you call the set {{1}, {1,2}, {1,2,3}, ...} which is equivalent and even isomorphic to the set{1, 2, 3, ...} = N finite (finite path)? The set of all natural numbers can be an infinite number omega, but the set of all initial segments of N cannot be an infinite segment? Even the diagonal of the EIT is no longer infinite, if it is inconvenient for set theory? > > 2. To "unary represent" every real in [0, 1] You cannot "unary represent" every real in [0, 1]. You can unary represent every natural number, and you can unary represent real numbers which contain only numerals a_n = const for every n in N (behind the decimal point), like 0.111 or 0.1111111. In addition you can unary represent omega or N = {1, 2, 3, ...}, namely as 0.111... = {0.1, 0.11, 0.111, ...}. Notice that this set does not contain infinite strings of 1. So you can unary represent an infinite string by an infinite set of finite strings. And now notice the important consequence: You can *binary represent* every real in [0, 1] by an infinite set of finite binary strings like 1/3 = 0.010101... {0.0, 0.01, 0.010, ...}. Therefore the union of all finite trees is the set of all infinite strings. > you also need _infinite_ > paths. (i.e. 1/3 is not represented as finite path and hence not by > any path in your union). Then the union of infinitely many finite initial segments {1, 2, 3, ...., n} cannot be an infinite set {1, 2, 3, ...}. > > > The set of real numbers in [0, 1] is countable. > > Non sequitur. The infinite paths of rational numbers are in the tree. At least {0.0, 0.00, 0.000, ...} = 0.000... = 0 and {0.1, 0.11, 0.111, ...} = 0.111... = 1 are in the union of all finite trees. But the paths of the irrational numbers are somewhat longer? Are you really arguing this way? (Nobody could contradict you. Belief in the infinite works just this way.) Regards, WM
From: David Marcus on 11 Jan 2007 16:27 Andy Smith wrote: > I was trying to get my head around Zeno's paradox - where > there is an infinite series, with no last element, which > sums to a finite amount (no problem so far for me), but the paradox > is made graphic by enumerating successive terms in amounts > of time that decrease pro-rata with the size of the terms. > Then the finite sum of the infinite series is achieved by > a process in a finite amount of time, also made up as an > infinite series in direct correspondence with the original > one. > > As I understand the professional answer kindly provided > here (several times) it is that we can't identify a last > instant of time, so there isn't a corresponding last jump. > So, no problem there then, other than in my head.... > > And again, by setting out an infinite series with a finite sum > twice, back to back, then it feels like one can have > a correspondence with a corresponding infinite sequence, > which definitely has a finite first and last element, > but an infinite number of terms in between. But that was > sat on hard too. The problem is your wording. Consider the sequence 1, -1/2, 1/4, -1/8, .... Then there is no last element to the sequence, but the set {1, -1/2, 1/4, -1/8, ...} certainly contains a smallest element and a largest elment. You have to remember that in mathematics, words (like "sequence") have precise technical meanings. You need to use the jargon correctly or people will correct you! > On the Peano thing, what happens if you remove the axiom > that says 0 is not the successor of any natural number? I'm not sure. I would think you'd have to modify the induction axiom too to get something sensible. > Then you get negative numbers included as natural numbers, > other properties unchanged for the positive numbers. > What does the e.g. binary representation of -1, -2 then > look like ? "-1" is the binary representation of -1. "-10" is the binary representation of -2. > A horror, with ...1111 as -1. Do -inf and > +inf converge on the same number? Are you asking about IEEE arithmetic or are you asking about how +oo and -oo are used in analysis? > Can one then just regard -1 > as the point that one counts round to eventually? Then > you could have a start and an end with an infinite number > of points in between, was how my uneducated thought > processes went, and then I wouldn't have to fret about > Zeno. I don't follow what is bothering you about Zeno. As for whether we can make the integers into a loop, the problem with such constructions is how to define the arithmetic operations and whether the resulting operations will have the properties that we expect or want them to. > I am not qualified to comment on that, don't really > believe it, and am liable to be called a troll (whatever > that is) for suggesting even the possibility of thinking of it.... A troll is someone who understands, but is pretending that they don't. If you really don't understand, but want to, then you are certainly not a troll. When I asked if you were trolling, I was asking a serious question. -- David Marcus
From: mueckenh on 11 Jan 2007 16:33 Franziska Neugebauer schrieb: > >> > This and your further questions are answerd by the following texts. > >> > >> I did not pose any questions. I have informed you about the fact that > >> there is no time and hence no temporal process _in_ math. > > > > Be informed then that mathematics is in time. > > From that it does not follow that there is a notion of time in > mathematics. A failure need not and will not persist for ever in mathematics. (From this you can see that time *is* in mathematics.) > > [Brouwer] maintains [...] > > Please tell us, whether you want to discuss views of > > [ ] Cantor > [ ] Brouwer > [ ] Mueckenheim > [ ] Cantor interpreted by Mueckenheim > [ ] Brouwer interpreted by Mueckenheim. > > If more than one applies please tag each sentence of yours accordingly. > I do not remember having posted interpretations, but only original quotes. I do not want to discuss pot. infinity with you. (Therefore sent Brouwer, Hilbert and Cantor to the front.) But the five alternatives do not differ at all with respect to the definition of the potentially infinite. Regards, WM
From: mueckenh on 11 Jan 2007 16:39
Franziska Neugebauer schrieb: > Virgil wrote: > > > In article <1168511743.391940.58070(a)77g2000hsv.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > [...] > >> Conclusion: Every finite binary tree contains a finite set of path. > >> The countable union of finite sets is countable. The set of paths is > >> countable. > > > > The set of /finite/ paths in the union is countable. > > But when one takes the union of sets of finite paths one only gets > > finite paths in that union. There are no infinite paths in that union. > > > > The same thing happens with ordinals. When one takes the union of all > > finite ordinals (like unary trees), there is no infinite ordinal IN > > that union > > Absolutely right. But an infinite ordinal is that union! Regards, WM |