Cantor Confusion
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From: mueckenh on 17 Oct 2006 06:21 David Marcus schrieb: > stephen(a)nomail.com wrote: > > Consider the twin slit experiment. Is the fact that none of > > the following accurately describe the situation an inconsistency? > > a) the photon goes through one slit > > b) the photon goes through both slits > > c) the photon goes through neither slit > > In Bohmian Mechanics (and similar theories), the photon goes through > only one slit. Physicists could learn something about logical thinking > from mathematicians. LOL. Physicist know already that something which cannot be known does not exist. Mathematicians have to learn it. Some will achieve it, but certainly not all. Bohmian Mechanics allows for action at a distance. Without that unrelativistic assumption case (b) is correct. But if no case were correct, then no case would exist. That is the same as with actual infinity. Regards. WM
From: mueckenh on 17 Oct 2006 06:22 William Hughes schrieb: > mueck...(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > One cannot compute a list of all computable numbers. By this > > > > definition, > > > > (1) the computable numbers are uncountable. > > > > (2) There is no question, that the computable numbers form a countable > > > > set. > > > > This is a contradiction. It is not necessary to come up with a list of > > > > all computable numbers. > > > > > > Nope. You are mixing two approaches.. > > > > > > A set X is countable if there exists a surjective function,f, > > > from the natural numbers to X > > > > > > There are two possibilities > > > > No. > > > > > > A: you allow arbitrary functions f > > > > > > B: you allow only computable functions f > > > > What is a function which is not computable? > > > > > > In case A, the computable reals are countable [ but you > > > also have arbitrary reals, and the set of reals (computable > > > and arbitrary) is uncountable.] > > > > > > In case B. the computable reals are not countable > > > (i.e. there is no list of computable reals) > > > > > > You cannot arrive at a contradiction by taking one result from > > > case B (the computable reals are uncountable) > > > and one result from case A > > > (the computable reals are countable). > > > > I take not at all results from your mysterious cases. > > I see: Every list of computable numbers supplies a diagonal number > > which is computable but not contained in the list. > > And I see: Every list of real numbers supplies a diagonal number which > > is not contained in the list. > > There is absolutely no difference. > > > > If real numbers are also computable numbers there is absoluetly > no difference. If real numbers contain both computable > real numbers and non-computable real numbers then > there is a difference. But in either case > > Every list of real numbers supplies a diagonal number which > is not contained in the list. > > so any way you cut it there is no complete list of real numbers. And there is no complete list of computable numbers. But they are countable. Hence the diagonal argument does not prove anything. Regards, WM
From: Randy Poe on 17 Oct 2006 06:36 mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > According to the ZFC system: The vase is empty at noon, because all > > > > > natural numbers left it before noon. > > > > > By means of the ZFC system we can formulate sequences and their limits > > > > > in mathematical language. From this it follows that lim {n-->oo} n > 1. > > > > > And from this it follows that the vase is not empty at noon. > > > > > > > > By what axiom do you conclude that the limit as t increases towards noon > > > > of any function and the value of that function at noon must be the same? > > > > > > By that or those axiom(s) which lead(s) to the result lim {t-->oo} 1/t > > > = 0. > > > > Here is your theorem: Let f(x) be any function f:R->R. Then > > lim(x->0-) f(x) = f(0). That is, the limit of f(x) as x approaches > > 0 from the left is f(0). > > > > Can you show me how the axiom(s) you describe prove > > that theorem? > > > > Can you then show me how the theorem applies to this > > function? f(x) = 1 if x<0, f(x) = -1 if x>=0. > > If there is no stepwise continuity in f(t) = n, can you show me why the > set of balls/numbers removed from the vase is containing all natural > numbers at noon after the number of transactions t --> oo? Simple. Because n being a natural number => there is a removal time t_n < noon. Therefore every natural is a member of the set of balls removed before noon. What I don't understand is how anyone can think there are balls with removal times which are still in the vase. - Randy
From: Dave Seaman on 17 Oct 2006 07:39 On Tue, 17 Oct 2006 02:48:55 GMT, Dik T. Winter wrote: > In article <1161019210.489120.290720(a)e3g2000cwe.googlegroups.com> "MoeBlee" <jazzmobe(a)hotmail.com> writes: > > > > Set theory has advanced since Cantor. The best source to learn about > > current set theoretic definitions of 'infinite' is not Cantor. > Indeed. There are some statements in that work that are false according to > current set theory (as I already did notice before). The works are a good > starting point when you want to talk about the history of set theory, but > that is all. > One of the most serious errors can he found in the statement that > "to count sets of first cardinality you need ordinals of the second > class" Are you sure you are quoting him correctly? Cantor did say (in fact, this is a section heading in boldface): The Power of the Second Number-Class is equal to the Second Greatest Transfinite Cardinal Number Aleph-One. I am quoting from a Dover edition of "Contributions to the Founding of the Theory of Transfinite Numbers" (ISBN 0-486-60045-9), and that is the title for section 16 on page 169. > where (I may have first and second wrong), first cardinality means (in > current terminology) aleph-0 and second class ordinals means (in current > terminology) omega and larger (until omega^omega or somesuch). The > error is of course that there is a set of cardinality aleph-0 that can > be counted using finite ordinals only (the natural numbers). No, the second number class according to Cantor is the set of all ordinals having cardinality aleph_0. So the smallest is omega, but it goes well beyond omega^omega, all the way up to (but not including) omega_1. And he is quite explicit about stating that omega is the smallest member of the second number class. > It is exactly the same error that is present in Tony's presentations and in > many of Wolfgang's presentations. In spite of Wolfgang's protestations, > the contents of the set N represent a potential infinity while the size > represents an actual infinity. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: William Hughes on 17 Oct 2006 07:41
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueck...(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > One cannot compute a list of all computable numbers. By this > > > > > definition, > > > > > (1) the computable numbers are uncountable. > > > > > (2) There is no question, that the computable numbers form a countable > > > > > set. > > > > > This is a contradiction. It is not necessary to come up with a list of > > > > > all computable numbers. > > > > > > > > Nope. You are mixing two approaches.. > > > > > > > > A set X is countable if there exists a surjective function,f, > > > > from the natural numbers to X > > > > > > > > There are two possibilities > > > > > > No. > > > > > > > > A: you allow arbitrary functions f > > > > > > > > B: you allow only computable functions f > > > > > > What is a function which is not computable? > > > > > > > > In case A, the computable reals are countable [ but you > > > > also have arbitrary reals, and the set of reals (computable > > > > and arbitrary) is uncountable.] > > > > > > > > In case B. the computable reals are not countable > > > > (i.e. there is no list of computable reals) > > > > > > > > You cannot arrive at a contradiction by taking one result from > > > > case B (the computable reals are uncountable) > > > > and one result from case A > > > > (the computable reals are countable). > > > > > > I take not at all results from your mysterious cases. > > > I see: Every list of computable numbers supplies a diagonal number > > > which is computable but not contained in the list. > > > And I see: Every list of real numbers supplies a diagonal number which > > > is not contained in the list. > > > There is absolutely no difference. > > > > > > > If real numbers are also computable numbers there is absoluetly > > no difference. If real numbers contain both computable > > real numbers and non-computable real numbers then > > there is a difference. But in either case > > > > Every list of real numbers supplies a diagonal number which > > is not contained in the list. > > > > so any way you cut it there is no complete list of real numbers. > > And there is no complete list of computable numbers. But they are > countable. Hence the diagonal argument does not prove anything. > You cannot say that the computable numbers are countable unless you use a definiton of countable that includes non-computable functions. (The usual argument, that each computable number is given by a finite string and there are only a countable number of finite strings, does not work here. Yes, there is a computable function, f_1, from the natural numbers to the set of finite strings but it is not true that therefore given B a subset of the finite strings that there is a computable function, f_2, from the natural numbers to the set B.) - William Hughes - William Hughes |