Another AC anomaly?
in [Logic]
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From: WM on 16 Dec 2009 09:38 On 16 Dez., 03:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > 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? Yes, exactly that is true. > > 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. There is a finite path that contains larger numbers than you can ever think of. > > > 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. That is pure opinion, believd by the holy bible (Dominus regnabit in aeternum et ultra. [2. Buch Moses: Exodus 15 Vers 18]) or forced upon us by the men-made axiom of infinity. > > > 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. The tree contains all paths that can be constructed by nodes, using the axiom of infinity. Which one would be missing? > > > > > 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. It is. I constructed a finite path from the root node to each other node. Then I appended an infinite tail. When the tree is completed, then the tail becomes invisible, because every sequenece of nodes has been constructed. But if you don't believe me, then look at the tree: You can see and admire every node and its connection to the root and the continuing paths downwards. So let me know what you think is missing. > > > > 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. The tree is the union of all paths. There is no end. > > > 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'. Which node is missing? > > > > 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). You are wrong. The paths in the complete tree aer unions of all finite paths. > > > 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. Every real number is there, that can be represented by digits. > > > 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. Even if I appended the tail 010101... ? Regards, WM
From: WM on 16 Dec 2009 09:42 On 16 Dez., 03:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > {1} U {1, 2} U {1, 2, 3} U ... = ({1} U {1, 2} U {1, 2, 3} U ...) U {1, > > 2, 3, ...} That is a matter of taste. > > > 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. > > A sequence of paths is not a path. But a union of paths is. > > > Therefore: When all finite paths have been constructed within aleph_0 > > steps, then all paths have been constructed. > > Here, again, you err. You can not construct something in aleph_0 steps; you > will never complete your construction. You *cannot* get at aleph_0 step by > step. But you can make a bijection with all elements of omega? > > > This hold for every limit of every sequence of finite paths. > > A limit is not a step by step process. Then assume it is a mapping from omega. Regards, WM
From: WM on 16 Dec 2009 09:48 On 16 Dez., 04:56, "K_h" <KHol...(a)SX729.com> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > news:9904ae9d-de2c-4d5f-86da-be165e8e9d7e(a)p30g2000vbt.googlegroups.com... > On 15 Dez., 13:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> > wrote: > > > > > 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, > > ...} . > > All limit ordinals, X, satisfy UX=X. That means that the > union of the members of a limit ordinal is equal to the > ordinal itself. w=N is a limit ordinal so UN=N. So: > > {} U {0} U {0,1,} U {0,1 2,} U ... = {0,1,2,3, ...} > > is a true equation. Therefore 0.000..., as a path in the tree, is nothing but the union of its finite initial segments. pi is also nothing but the union of its finite initial segments. All finite initial segments form a countable set. Nothing else exists. Hence all paths in the tree form a countable set. Regards, WM pi
From: WM on 16 Dec 2009 10:35 On 16 Dez., 14:21, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <a586deed-42c7-4523-acb2-1567183f0...(a)g12g2000vbl.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 16 Dez., 03:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > > Without the axiom of infinity omega would not be immediatelry > > > > > > > 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. > > > > > Yes, so the definition uses the axiom of infinity. Without that axiom > > > omega would not be immediately existing, as I stated. > > > > No, you stated that you had never seen there a proper definition of > > omega. > > Look at the context. It started with my asking for a definition of omega > without the axiom of infinity. You state that there is such a definition > in your book. I find such a definition nowhere in your book. And the > 'proper' should be read in the context of 'without the axiom of infinity'.. The axiom has stolen the correct definition from Peano, (Zermelo says from Dedekind, Dedekind says from Bolzano) and has reversed its meaning from infinite to finished. That's all. > > > > > > > > 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. > > > > And omega is something that does never end. > > Whatever that may mean. The same as the axiom says. The axiom says, in addition, that this thing is a set, but as it is not said what a set is, this addition is idle. > > > > > 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. > > > > No, that's more than that. Omega already is actual infinity. And here > > are some other statements about actual infinity: > > So actual infinity is omega? In fact, it is said so. Sometimes it is carelessly used also for potential infinity. > > > Let us distinguish between the genetic, in the dictionary sense of > > pertaining to origins, and the formal. Numerals (terms containing > > only > > the unary function symbol S and the constant 0) are genetic; they are > > formed by human activity. All of mathematical activity is genetic, > > though the subject matter is formal. > > Numerals constitute a potential infinity. Given any numeraal, we can > > construct a new numeral by prefixing it with S. > > Now imagine this potential infinity to be completed. Imagine the > > inexhaustible process of constructing numerals somehow to have been > > finished, and call the result the set of all numbers, denoted by |N. > > But N is *not* defined as an inexhaustable process of constructing numerals > that somehow has been finished. The axiom of infinity state (together with > the actual definition) state that it does exist, not how it is created. Of course it is. If with n you can do n + 1, and if 0 is there, then you have omega. The axiom states that such a thing (it says set, but does not say what a set is) does exist. Regards, WM
From: Virgil on 16 Dec 2009 13:35
In article <ba9f2c85-caa2-491d-9263-bea1bed3ab1e(a)p19g2000vbq.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 16 Dez., 03:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > �> > > �> 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? > > Yes, exactly that is true. That WM thinks it true is enough to make serious mathematicians doubt it. > > > �> 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. > > There is a finite path that contains larger numbers than you can ever > think of. It is certainly true that an infinite path must contain such numbers, but one can always think of the last number of a finite increasing sequence. > > > > �> � � � 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. > > That is pure opinion it is certainly a better opinion than WM's contrary opinion. Which is all WM has in refutation. > > > > �> 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. > > The tree contains all paths that can be constructed by nodes, using > the axiom of infinity. Which one would be missing? All uncountably many endless ones. > > > > �> > �> 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. You will fail to point to a binary digit missing missing from {0,1} but you will not find 1/3 among them. > > �> 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. > > It is. I constructed a finite path from the root node to each other > node. Then I appended an infinite tail. As soon as you append such a tail, you no longer have a finite path. > When the tree is completed, > then the tail becomes invisible If it is there, then it is visible to eyes which can see. , because every sequenece of nodes has > been constructed. But if you don't believe me, then look at the tree: > You can see and admire every node and its connection to the root and > the continuing paths downwards. So let me know what you think is > missing. Common sense, for one thing. > > > > �> > 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. > > The tree is the union of all paths. There is no end. Your tree is nonsense. Here is a sensible model of a complete infinite binary tree: It starts with the set of naturals, N, with each natural being a node. For each node, n in N, its child nodes are 2*n and 2*n+1. A path in such a tree is a maximal subset containing 1 and for each n containing either 2*n or 2*n+1. but not both. > > > > �> � � � � � � � � � � � � � � � � � � � � �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'. > > Which node is missing? That question has no meaning, since in WM's trees there are no paths at all and thus no real numbers. > > > > �> > 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). > > You are wrong. The paths in the complete tree aer unions of all finite > paths. Then they are not finite. Either paths in the complete tree are infinite, possibly modeled as infinite sets of finite paths from finite sub-trees, or WM's tree is incomplete. Or both. > > > > �> � � � � � � � � � � � � � � � � � � � � � � � � � � � � Every digit of > > �> every real number that can be shown to exist exists in the tree. Every 'digit' in a binary tree is in {0,1}. So what? > > > > But not every real number is represented in the tree by a path. > > Every real number is there, that can be represented by digits. Not unless those digits are an infinite sequence of digits. > > > > �> 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. > > Even if I appended the tail 010101... ? Then you no longer have your alleged but phony finite path but an actually infinite path, which is what we have been saying all along. |