Another AC anomaly?
in [Logic]
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From: Dik T. Winter on 11 Dec 2009 06:57 In article <5ZedndTmvoBac7zWnZ2dnUVZ_qOdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:KuFy3L.Cxt(a)cwi.nl... .... > > Have a look at <http://en.wikipedia.org/wiki/Lim_inf> in > > the section > > titled "Special case: dicrete metric". An example is > > given with the > > sequence {0}, {1}, {0}, {1}, ... > > where lim sup is {0, 1} and lim inf is {}. > > > > Moreover, in what way can a definition be invalid? > > 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. I am not discussing supertasks, nor is WM here. The question is simply whether it is possible that: lim | S_n | != | lim S_n | with some form of limit. And by the definitions I gave (and which you also will find on the wikipedia page above): lim(n -> oo) {n} = {} -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 11 Dec 2009 09:58 On 11 Dez., 03:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <89fb6e91-b6b1-4926-afca-820492e3c...(a)r24g2000yqd.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 10 Dez., 15:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Have a look at <http://en.wikipedia.org/wiki/Lim_inf> in the section > > > titled "Special case: dicrete metric". =A0An example is given with the > > > sequence {0}, {1}, {0}, {1}, ... > > > where lim sup is {0, 1} and lim inf is {}. > > > > > > Moreover, in what way can a definition be invalid? > > > > It can be nonsense like the definition: Let N be the set of all > > natural numbers. > > In what way is it nonsense? Either that set does exist or it does not exist. > If it does exist there is indeed such a set, if it does not exist there is > no set satisfying the definition. In both cases the definition is not > nonsense in itself. It is nonsense to define a pink unicorn. The set N does not exist as the union of its finite initial segments. This is shown by the (not existing) path 0.000... in the binary tree. Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}. What then is {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} ? If it is the same, then wie have a stop in transfinite counting. If it is not the same, what is it? > > But apparently you are of the opinion that you are only allowed to define > things that do exist. Most essential things in mathematics exist without definitions and, above all, without axioms. Regards, WM
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 |