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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
From: Marshall on 11 Dec 2009 22:49 On Dec 11, 6:39 pm, "K_h" <KHol...(a)SX729.com> wrote: > "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote in messagenews:KuHL7p.3zE(a)cwi.nl... > > > > > In article <5ZedndTmvoBac7zWnZ2dnUVZ_qOdn...(a)giganews.com> > > "K_h" <KHol...(a)SX729.com> writes: > > > "Dik T. Winter" <Dik.Win...(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. It certainly appears to me that the definitions quoted give lim(n -> oo){n} = {}. One of us is misreading it. Marshall
From: WM on 12 Dec 2009 04:32
On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > 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. Every digit of 1/3 that you can handle, I can handle too. Only the claim "all digits" is incorrect. In my mathematics, this is seen, in yours it is not. > > 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. Consider how a union of paths is counted (I copy from another posting, therefore the quotation symbols): > > > {1} > > {1, 2} > > {1, 2, 3} > > _________ > > {1, 2, 3} > > ======== > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > _________ > > {1, 2, 3, ...} > > ========= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > _________ > > {1, 2, 3, ...} > > ========= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > _____________ > > {1, 2, 3, ..., a} > > ============ > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ______________ > > {1, 2, 3, ..., a, aa} > > ============= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ... > > __________________ > > {1, 2, 3, ..., a, aa, ...} > > ================= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ... > > {1, 2, 3, ..., a, aa, ...} > > __________________ > > {1, 2, 3, ..., a, aa, ...} > > ================= Contrary to the usual 1, 2, 3, ..., w, w+1, w+2, ..., w+w, w+w+1, ... this leads to the following sequence which stops from time to time (at limit ordinal paths): {1} {1, 2} {1, 2, 3} .... {1, 2, 3, ...} {1, 2, 3, ...} {1, 2, 3, ..., a} {1, 2, 3, ..., a, aa} {1, 2, 3, ..., a, aa, aaa} .... {1, 2, 3, ..., a, aa, aaa, ...} {1, 2, 3, ..., a, aa, aaa, ...} {1, 2, 3, ..., a, aa, aaa, ..., b} {1, 2, 3, ..., a, aa, aaa, ..., b, bb} {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb} .... {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb, ...} {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb, ...} .... The first double-meaning, at {1, 2, 3, ...}, is responsible for the fact that the infinite binary tree can be constructed completely from finite paths (hence contains only countably many paths) and simultaneously is said to contain uncountably many infinite paths. Regards, WM |