Another AC anomaly?
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From: Dik T. Winter on 17 Dec 2009 06:49 In article <yrydnX_FwfczULTWnZ2dnUVZ_vmdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > lim(n ->oo) n = N is true for the standard definition of > natural numbers using just the wikipedia definitions for the > limit of a sequence of sets. But not with Zermelo's definition of the natural numbers. Nor when we take the natural numbers as embedded in the rational numbers. In my opinion, applying limits for sequences of sets when having sequences of numbers can be thoroughly misleading. -- 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: T.H. Ray on 16 Dec 2009 22:26 Jesse Hughes wrote > > > > 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 > Is {0,1} a set? Tom
From: Jesse F. Hughes on 17 Dec 2009 08:42 "T.H. Ray" <thray123(a)aol.com> writes: > Jesse Hughes wrote >> > >> > 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 >> > > Is {0,1} a set? Not sure why my name appears above. This exchange was between Dik Winter and WM. -- Jesse F. Hughes "Quincy, why should you not play with matches?" "Because... [pause] Aahhh! I'm on fire!!"
From: Dik T. Winter on 17 Dec 2009 08:44 In article <ba9f2c85-caa2-491d-9263-bea1bed3ab1e(a)p19g2000vbq.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > 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. And why, except by opinion? > > > 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. Aha, you are clearly a mindreader. Well, as far as I know mindreading is not part of mathematics. Anyhow, I can think of numbers larger than that path. But that is completely irrelevant. I am able to think about a set that contains all natural numbers, you apparently are not. > > > 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. Sorry, I have no knowledge of the bible. But live without that axiom when you can't stomach it. And do not attack mathematicians who live with that axiom. > > > 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? The infinite paths because you stated a priori that your tree did not contain infinite paths. So it is impossible to construct in your tree infinite paths by the axiom of infinity. You can construct infinite sequences of nodes, but as you stated *explicitly* that your tree did not contain infinite paths, those infinite sequences of nodes are apparently not paths within your terminology. > > > 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. Yup, you constructed a finite path, and that does not represent 1/3. > Then I appended an infinite tail. Whatever that may be, it is *not* a path according to your explicit statement that the tree did not contain infinite paths. > When the tree is completed, > then the tail becomes invisible, because every sequenece of nodes has > been constructed. But not every sequence of nodes is a path according to your explicit statements. > 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. Which of thos paths represents 1/3? (I may node that 1/3 does not have a last binary digit.) > > > 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. Yes, and by your statements a path is a finite sequence of nodes, so there are no infinite paths in your tree. > > > 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 is irrelevant. The path is missing. According to your explicit statements not every sequence of nodes is 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. Rubbish. You have stated, explicitly, that your tree contained only finite paths. So apparently there are sequences of nodes (the infinite sequences) that are not a path. > > > 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. That is irrelevant, it is not represented by a path because by your statements paths are finite. > > > 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... ? According to your definitions that is not part of a path. -- 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 17 Dec 2009 09:08
In article <6a57309a-a136-430c-a718-e38518c658bb(a)q16g2000vbc.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > 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. No, it is a matter of convention. In mathematics a, b, c, ..., z means a, b, c, continue this way until you reach z. But starting {1}, {1, 2}, {1, 2, 3} and going on you never reach {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. > > > > A sequence of paths is not a path. > > But a union of paths is. No. Suppose we have the paths 0.000 and 0.100, what is their union? And is it a path? > > > 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? Yes, but not with a step by step method that will ever be complete. > > > 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. Which mapping? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |