Another AC anomaly?
in [Logic]
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From: Virgil on 11 Dec 2009 15:53 In article <364fe723-5ae4-4ed5-9219-c6b3892b3df2(a)d21g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > If, in your example you would claim that lim(1/n) = 0 and and 1/omega > = 10 That seems to be much more like something that WM would claim than something Dik would claim.
From: Virgil on 11 Dec 2009 16:22 In article <d5a795ed-90bd-43bd-af2f-dd4ab437aca8(a)d21g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. There has been much ado about all sorts of other things since 1908, enough so that what has been said about cardinals is far from being "most" of it. > > > > �> > 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. Are they treated as badly as sets? > > > > > �> > �> 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 belongs to infinitely many finite path-segments, so why not to any infinite path? To get an infinite tree one must have something being infinite, so why does WM claim that there can there be infinitely many paths in the tee but not infinitely many nodes in any path? How is one such actual infiniteness any more actual than any other? > Every digit that is on the diagonal of Canbtor's list is a > member of a finite initial segment of a real number. Actually, any such list is not Cantor's but is presented to him by a challenger, and he is challenged to find an unlisted sequence, which he can always do, by applying a simple rule of construction. > > You can only argue about such digits. On the contrary, we can argue about a good deal more, e.g., sequences of digits. > And all of them (in form of bits) are present in my binary tree. But your tree has no paths (MAXIMAL sequences of bits with each bit having exactly one child bit in the same path). > > > > �> > �> � � � � � � � � � � � � � � � � � 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. I can find every letter in any book in an alphabet, but that does not make any alphabet contain the book. > You > will fail to point to a digit of 1/3 that is missing in my tree. You will fail to point to any letter of 'Hamlet' that is missing in my alphabet. > Therefore I claim that every number that exists is in the tree. Therefore I claim that everything in 'Hamlet' is in my alphabet. > > Isn't a path a sequence of nodes, is it? In any tree, if a path has a last node, then that node must be a leaf node having no child nodes. > 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. Every letter in 'Hamlet' is in my alphabet. > > 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? 'Hamlet' is not "in" any alphabet, even though every letter in it is in the alphabet. It is not the individual letters so much as the sequence of them that creates 'Hamlet'. Unless every sequence of digits in real number is in a path, that path does not represent that number, and possibly not even when they are.
From: Virgil on 11 Dec 2009 16:34 In article <dad673b7-b69b-411a-abef-7c3d4e3bb999(a)a32g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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? Why shouldn't it? It would make at least as much sense as your arguments. > 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. But then, since all your paths are finite, they do not represent all reals, which can have non-finite representations. > > > 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. If the string of binary digits's in a path has a last position, then it does not represent 1/3. Then 1/3 can only be represented by an actually infinite set of such strings, at which point actual infiniteness is established and all of WM's objections to it fail. > There is neither a > natural nor a rational with an infinite string of digits. There is no finite string of binary digit representing 1/3. Or any rational not of form m/2^n for m an integer and n a natural. > 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. I like 1/3 which, in any base not divisible by 3, requires a never ending sequence of digits to represent exactly. > > Why should it??? Because Cantor believed that God knows infinite > strings? (He read it in civitate dei of St. Augustinus.) Because it does. > > 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? Does WM think he is more in touch with God than any of the many mathematical geniuses whom he trashes here?
From: Marshall on 11 Dec 2009 20:01 On Dec 11, 11:33 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > 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. If your math can't handle 1/3, it is exceedingly weak. A third grader can do as much. You propose various severe restrictions on math and say that everyone else ought to adopt them, but you supply no motivation for doing so. > 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? I don't do ancestor worship. I have never read anything of Cantor's or Zermelo's and can't see any reason to; their work has been improved upon since. If I were a doctor I wouldn't be reading nineteenth century medical papers. The answer to your "why" question is because many good and useful results come about if I do. For example, I can handle decimal expansions of fractions such as 1/3. I know you don't like many of those results, however, coming up with results that are palatable to some guy I never met is not a goal of mine. To get any attention to your demands, you need to supply some motivation for people to listen. Showing a contradiction would qualify, but it's been well established that you don't know how to do that. There are, on the one hand, contradictions of the form A ^ ~A, and on the other hand results that WM doesn't like. I know you don't distinguish between these two things, but most people do, and attend only to the first. Until you also become able to make the distinction, no one has any motivation to listen to you. Marshall
From: K_h on 11 Dec 2009 21:39
"Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuHL7p.3zE(a)cwi.nl... > 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} = {} The only way lim(n ->oo){n}={} is if limsup and liminf both equal {}. If limsup and liminf are different then the limit does not exist and cannot equal an existing set like {}. Since the empty set exists, and since you are claiming that lim(n ->oo){n}={}, you need to show that limsup and liminf are both {} by the definition you are using. WM attributed this definition to you: - Given a sequence of sets S_n then: - lim sup{n -> oo} S_n contains those elements that occur in infinitely many S_n. - lim inf{n -> oo} S_n contains those elements that occur in all S_n from a certain S_n (which can be different for each element). - lim{n -> oo} S_n exists whenever lim sup and lim inf are equal. All the definitions on the wikipedia pages, and the above definition WM attributes to you, give lim(n ->oo){n}=N. Consider the naturals: S_0 = 0 = {} S_1 = 1 = {0} S_2 = 2 = {0,1} S_3 = 3 = {0,1,2} S_4 = 4 = {0,1,2,3} S_5 = 5 = {0,1,2,3,4} .... First, let's consider the definition WM attributes to you: - lim sup{n -> oo} S_n contains those elements that occur in infinitely many S_n. * Every natural, n, occurs infinitely many times after S_n so limsup=N. - lim inf{n -> oo} S_n contains those elements that occur in all S_n from a certain S_n (which can be different for each element). * Every natural, n, occurs in each S_m after S_n so liminf=N. - lim{n -> oo} S_n exists whenever lim sup and lim inf are equal. * limsup=liminf=N and so the limit exists and is N. On the wikipedia page, apply the definition under the section titled "General set convergence" and apply both definitions under the section titled "special case: discrete metric" to the above S_n and you will find that the limit is N in all cases. So, in all cases we get lim|S_n|=|limS_n|. My previous post already has the proof for wikipedia's last definition: liminf S_n=\/(n=0, oo)[/\(m=n, oo) S_m] (n-->oo) limsup S_n=/\(n=0, oo)[\/(m=n, oo) S_m] (n-->oo) In fact, the wikipedia page says the definitions are equivalent in certain cases: "The difference between the two definitions involves the topology (i.e., how to quantify separation) is defined. In fact, the second definition is identical to the first when the discrete metric is used to induce the topology on X." k |