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From: Dik T. Winter on 15 Dec 2009 08:01 In article <r_6dnUXDv4pMfbvWnZ2dnUVZ_tWdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:KunF0v.9y9(a)cwi.nl... .... > > By what definition is: > > lim(n -> oo) n = N? > > Any of the wikipedia definitions. Here is the proof again. .... > Theorem: > lim(n ->oo) n = N. Consider the naturals: > > S_0 = 0 = {} > S_1 = 1 = {0} .... This presupposes a particular construction for the natural number. There are other constructions that are consistent with ZF. Is the limit valid for all those possible models? -- 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 15 Dec 2009 16:26 On 15 Dez., 13:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > It is nonsense to define a pink unicorn. > > That statement is nonsense. Let pu be a pink unicorn is a proper definition. > However there is nothing that satisfies that definition. That is a matter of taste. > > > 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. > > As you have defined your tree as only containing finite paths it is trivial > to conclude that the infinite path 0.000... does not exist in your tree. > However, this does *not* show that N does not exist. If N does exist, then the tree containing all finite paths contains also infinite paths. > > > 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, ...} ? > > Ambiguous as stated. What is "..." standing for? Look two lines above: Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} hence {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...} . > Not for "go on in the > same way until", because when you go on in the same way you will never > get at {1, 2, 3, ...}. > > > If it is the same, then wie have a stop in transfinite counting. > > However, if we try to attach a meaning we find that it is {1, 2, 3, ...}. Correct. {1} U {1, 2} U {1, 2, 3} U ... = {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} There is no space in the binary tree to contain infinite paths in addition to the sequeneces of all finite paths. The sequence 0.0, 0.00, 0.000, ... when being completely constructed, is already the path 0.000... You cannot add it separately. Therefore: When all finite paths have been constructed within aleph_0 steps, then all paths have been constructed. There is no chance to construct the path 0.000... after all finite initial segments have been constructed. This hold for every limit of every sequence of finite paths. Regards, WM
From: Ross A. Finlayson on 15 Dec 2009 17:54 On Dec 15, 5:01 am, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <r_6dnUXDv4pMfbvWnZ2dnUVZ_tWdn...(a)giganews.com> "K_h" <KHol...(a)SX729.com> writes: > > > "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote in message > >news:KunF0v.9y9(a)cwi.nl... > ... > > > By what definition is: > > > lim(n -> oo) n = N? > > > > Any of the wikipedia definitions. Here is the proof again. > ... > > Theorem: > > lim(n ->oo) n = N. Consider the naturals: > > > > S_0 = 0 = {} > > S_1 = 1 = {0} > ... > This presupposes a particular construction for the natural number. There are > other constructions that are consistent with ZF. Is the limit valid for all > those possible models? > -- > dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ Half. That leads to then interesting expansions of the Factorial/Exponential Identity, about breakdown in asymptotics. Also it's half the other way as well, in the deterministic and well- founded. Regards, Ross F.
From: Dik T. Winter on 15 Dec 2009 21:15 In article <364fe723-5ae4-4ed5-9219-c6b3892b3df2(a)d21g2000yqn.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 11 Dez., 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > Without the axiom of infinity omega would not be immediately > > > > existinging. > > > > 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. I did 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. Yes, so the definition uses the axiom of infinity. Without that axiom omega would not be immediately existing, as I stated. But you thought there was a definition without that axiom. I still have not found such a definition. (The definition there is: omega = {0, 1, 2, 3, ...} and without the axiom of infinity it is not clear whether the right hand side is a set.) > > > > 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? That is not something unheard of. In mathematics a ring is something that satisfies the ring axioms, and that is pretty standard. > But in case you shouldn't have been able to find a definition of > actual infinity, here is more than that: omega + 1. Ok, so actual infinity now is omega + 1. In other articles you stated that it was a theory ("according to actual infinity") and various other things. > > > 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? Getting from something > 0 to something equal to 0. > > 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. Yes, you just say that. Without proof. > 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). What I state is: 1/n > 0 and lim(n -> oo) 1/n = 0 and you state that there is no chance for a difference of 1/n > 0 for all n and lim(n -> oo) 1/n = 0 About that "funny" sum, that "funny" sum has no definition, so it is allowed to give it a definition. But apparently you do not allow that. Yes, I know you dislike definitions and axioms, because they constrain you. -- 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: Dik T. Winter on 15 Dec 2009 21:31
In article <d5a795ed-90bd-43bd-af2f-dd4ab437aca8(a)d21g2000yqn.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > 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. That is a pretty strange usage of the word "most" in the context of "most newer mathematics". But apparently you think Graham's number rubbish. > > > > 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. You think that something that satisfies the ZF axioms being a collection of sets is rubbish, while something that satisfies the ring axioms being a ring is not rubbish? > > > > > 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. Right. But there is no finite path that contains them all. I believe in a path that contains them all, and that is an infinite path. > Every digit that is on the diagonal of Canbtor's list is a > member of a finite initial segment of a real number. Right, but there is no finite initial segment that contains them all. > You can only argue about such digits. And all of them (in form of > bits) are present in my binary tree. Right, but your tree does not contain infinite paths, as you explicitly stated. > > > 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. In that case you have a very strange notion of "existing in the tree". Apparently you do *not* mean "existing as a path". So when you say that the number of (finite) paths is countable, I agree, but 1/3 is not included in that, because it is not a path according to your statements. > > 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? Apparently not in your tree. In your tree a path is a finite sequence of nodes. > Everey node of 1/3 (that you > can prove to belong to 1/3) is in the tree. Right, but there is no path that denotes 1/3. > > Similar for 'pi' and 'e'. > > Yes. Every digit is available on request. Right, but there is no path that denotes either 'pi' or 'e'. > > 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): There is no proof needed. Apparently there are real numbers in your tree without being a path, because each path is finite (by your own definition). > Every digit of > every real number that can be shown to exist exists in the tree. But not every real number is represented in the tree by a path. > 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? Yes, by your own admissions. You state (explicitly) that every path is finite and it is easy to prove that every number that is represented by such a path is a rational number with a denominator that is a power of 2. So there are apparently real numbers of which every digit is in the tree that are not represented as a path, like 1/3. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |