Another AC anomaly?
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From: WM on 11 Dec 2009 10:15 On 11 Dez., 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <5333fb9a-1670-4fcc-85d3-25e75fb5b...(a)f16g2000yqm.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Dez., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Without the axiom of infinity omega would not be immediately existing. > > > So apparently there is a definition of omega without the axiom of > > > infinity. > > > Can you state that definition? > > > > Look into Cantor's papers. Look into my book. > > I have never seen there a proper definition of omega. > I knew you did not read it. Look at p. 93. There the natural numbers are constructed. On p. 86 all ordinals till eps^eps^eps^... are given. On p. 90 you see the axiom of infinity 7. > > > There are no concepts of mathematics without definitions. > > > > So? What is a set? > > Something that satisfies the axioms of ZF for instance. Is that a definition? But in case you shouldn't have been able to find a definition of actual infinity, here is more than that: omega + 1. > > If an infinite set exists as a limit, then it has gotten from the > > finite to the infinite one by one element. During this process there > > is no chance for any divergence between this set-function and its > > cardinality. > > If a function exists as a limit, then it has gotten from the finite to > the infinite one by one step. During this process there is no chance > of any divergence between the function and the integral. > > Now, what is wrong with that reasoning? > > Stronger: > lim(n -> oo) 1/n = 0 > 1/n > 0 I do not understand. Do you see a gap here? > > If a number exists as a limit, then it has gotten from the finite to the > infinite one by one step. During this process there is no chance of any > divergence between the element and the inequality. > > What is wrong with that reasoning? > > You are assuming that taking a limit is a final step in a sequence of steps. > In the definition I gave for the limit of a sequence of sets there is no > final step. I did not say that there is a final step. I say that there is no chance for a difference of lim card(S_n) and card(lim(S_n)) where lim means n --> oo. If, in your example you would claim that lim(1/n) = 0 and and 1/omega = 10, I would not accept such behaviour as mathematics (like your funny Sum n = 0 or Euler's -1/12). Regards, WM
From: WM on 11 Dec 2009 12:13 On 11 Dez., 03:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <fec95b83-39c5-4537-8cf7-b426b1779...(a)k17g2000yqh.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Dez., 16:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Before 1908 there was quite a lot of mathematics possible. > > > > > > Yes, and since than quite a lot of newer mathematics has been made > > > available. > > > > Most of it being rubbish. > > Nothing more than opinion while you have no idea what has been done in > mathematics since 1908. Algebraic number theory is rubbish? The answer is an explicit no. "Most" here concerns the magnitude of numbers involved. There was much ado about inaccessible cardinals. > > > > Moreover, before 1908 mathematicians did use concepts without actually > > > defining them, which is not so very good in my opinion. > > > > Cantor gave a definition of set. What is the present definition? > > Something that satisfies the axioms of ZF (when you are working within ZF). > It is similar to the concepts of group, ring and field. Something that > satisfies those axioms is such a thing. But I think you find all those > things rubbish. Why that? Group, ring and field are treated in my lessons. > > > > > There is not even one single infinite path! > > > > > > Eh? So there are no infinite paths in that tree? > > > > In fact no, but every path that you believe in is also in the tree, > > i.e., you will not be able to miss a path in the tree. > > I believe in infinite paths, you state they are not in the tree. So we > have a direct contradiction to your assertion. You believe in infinite paths. But you cannot name any digit that underpins your belief. Every digit that you name belongs to a finite path. Every digit that is on the diagonal of Canbtor's list is a member of a finite initial segment of a real number. You can only argue about such digits. And all of them (in form of bits) are present in my binary tree. > > > > > But there is every path > > > > which you believe to be an infinite path!! Which one is missing in > > > > your opinion? Do you see that 1/3 is there? > > > > > > If there are no infinite paths in that tree, 1/3 is not in that tree. > > > > 1/3 does not exist as a path. But everything you can ask for will be > > found in the tree. > > Everything of that kind is in the tree. > > This makes no sense. Every path in the tree (if all paths are finite) is > a rational with a power of 2 as the denominator. So 1/3 does not exist > as a path. In what way does it exist in the tree? It exists in that fundamentally arithmetical way: You can find every bit of it in my binary tree constructed from finite paths only. You will fail to point to a digit of 1/3 that is missing in my tree. Therefore I claim that every number that exists is in the tree. > > > Otherwise 1/3 would be a rational with a denominator that is a power of > > > 2 (each finite path defines such a number). > > > > > > > What node of pi is missing in the tree constructed by a countable > > > > number of finite paths (not even as a limit but by the axiom of > > > > infinity)? > > > > > > By the axiom of infinity there *are* infinite paths in that tree. So your > > > statement that there are none is a direct contradiction of the axiom of > > > infinity. > > > > Try to find something that exists in your opinion but that does not > > exist in the tree that I constructed. > > In what way do numbers like 1/3 exist in your tree? Not as a path, apparently, > but as something else. Isn't a path a sequence of nodes, is it? Everey node of 1/3 (that you can prove to belong to 1/3) is in the tree. > Similar for 'pi' and 'e'. Yes. Every digit is available on request. > So when you state that > the number of paths is countable that does not mean that the number of real > numbers is countable because there are apparently real numbers in your tree > without being a path. Wrong. Not only "apparantly" but provably (on request): Every digit of every real number that can be shown to exist exists in the tree. Or would you say that a number, every existing digit of which can be shown to exist in the tree too, is not in the tree as a path? Regards, WM
From: WM on 11 Dec 2009 12:16 On 11 Dez., 08:03, "K_h" <KHol...(a)SX729.com> wrote: > It depends on the context. When it comes to supertasks, > limsup={0,1} is basically useless. That is why those > definitions are not good, and invalid, for evaluating > supertasks -- in response to WM's supertask issues. Why do you think that the supertask of uniting all natural numbers is not invalid? Regards, WM
From: Marshall on 11 Dec 2009 13:53 On Dec 11, 9:13 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 11 Dez., 03:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > If there are no infinite paths in that tree, 1/3 is not in that tree. > > > > > > 1/3 does not exist as a path. But everything you can ask for will be > > > found in the tree. > > > Everything of that kind is in the tree. > > > This makes no sense. Every path in the tree (if all paths are finite) is > > a rational with a power of 2 as the denominator. So 1/3 does not exist > > as a path. In what way does it exist in the tree? > > It exists in that fundamentally arithmetical way: You can find every > bit of it in my binary tree constructed from finite paths only. You > will fail to point to a digit of 1/3 that is missing in my tree. > Therefore I claim that every number that exists is in the tree. This argument can be inverted to "prove" the existence of a natural number whose decimal expansion is an infinite string of 3s. The infinite-3 number exists in a fundamentally arithmetical way: you can find every digit of it in a preceding natural number constructed from finite string of 3s only. You will fail to point to a digit of ...333 that is missing in the natural numbers. Therefore you claim the naturals and the reals are just the same, but in the reverse direction, just as AP says they are. Marshall
From: WM on 11 Dec 2009 14:33
On 11 Dez., 19:53, Marshall <marshall.spi...(a)gmail.com> wrote: > On Dec 11, 9:13 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 11 Dez., 03:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > > If there are no infinite paths in that tree, 1/3 is not in that tree. > > > > > > > > 1/3 does not exist as a path. But everything you can ask for will be > > > > found in the tree. > > > > Everything of that kind is in the tree. > > > > This makes no sense. Every path in the tree (if all paths are finite) is > > > a rational with a power of 2 as the denominator. So 1/3 does not exist > > > as a path. In what way does it exist in the tree? > > > It exists in that fundamentally arithmetical way: You can find every > > bit of it in my binary tree constructed from finite paths only. You > > will fail to point to a digit of 1/3 that is missing in my tree. > > Therefore I claim that every number that exists is in the tree. > > This argument can be inverted to "prove" the existence of > a natural number whose decimal expansion is an infinite > string of 3s. No. Why should it? Of course the magnitude of natural numbers is not limited. For every number with n digits 333...333 there is another number with n^n digits. Nevertheless each one is finite. > The infinite-3 number exists in a fundamentally > arithmetical way: you can find every digit of it in a preceding > natural number constructed from finite string of 3s only. > You will fail to point to a digit of ...333 that is missing in > the natural numbers. Of course. That is because there is no digit missing in the natural numbers. > Therefore you claim the naturals and > the reals are just the same, but in the reverse direction, > just as AP says they are. They *are* just the same, because your argument that above procedure would prove an infinite string of 3's is wrong. There is neither a natural nor a rational with an infinite string of digits. To be able to determine every digit of a number you like does not imply that there is a number with a never ending sequence of digits. Why should it??? Because Cantor believed that God knows infinite strings? (He read it in civitate dei of St. Augustinus.) Or because Zermelo misunderstood Bolzana-Dedekind's definition of infinity? Are you really thinking that infinity comes into being because a Dr. Zermelo of Germany said so? Regards, WM |