Another AC anomaly?
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From: T.H. Ray on 17 Dec 2009 21:57 Jesse Hughes wrote > "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. > Sorry. My error. Tom > -- > Jesse F. Hughes > > "Quincy, why should you not play with matches?" > "Because... [pause] Aahhh! I'm on fire!!"
From: Dik T. Winter on 18 Dec 2009 09:12 In article <a737fea3-6dd2-4a1d-9506-2df7cb6d412a(a)v30g2000yqm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Dez., 14:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > .com> WM <mueck...(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 axiomsbeing a ring is not rubbish? > > > > > > Yes, exactly that is true. > > > > And why, except by opinion? > > Look here: > http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb5da79d# What is the relation? > > 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. > > How do you know that without confirming it by thinking the last too? Why need I to think about a last one (which there isn't) to be able to think about a set that contains all natural numbers? Apparently you have some knowledge about how my mind works that I do not have. > > > > 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. > > To live with that axiom does not create uncountability. See the proof > here: > http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb5da79d# Where is the proof there? I see only you writing a bit of nonsense and two rebuttals. > > > 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. > > The axiom of infinity establishes the set N from finite numbers. It establishes the *existence* of a set N of finite numbers. > It establishes the infinite paths as well in my tree from finite > paths. No. That is impossible because you stated that the paths were finite. What it *does* establish is the extistence of a set P of finite paths. > > 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. > > They are not constructed, but it might happen that the come in by the > union of some finite paths: The sequence of finite paths 1, 11, > 111, ... may yield an infinite path 111... as a limit. How is it possible to yield an infinite path if by your definition paths are finite? That is like saying lim(n -> oo) is a natural number. > > > > 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. > > It 1/3 can be represented by a bit sequence, then it is represented in > my tree by the union of the node sets representing > 0.0, 0.01, 0.010, ... Right, but that is not a path by your definition. > > > > > 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. > > You have misread my construction. I use finite paths and then I append > an infinite tail. But that is not a path by your statement. -- 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 18 Dec 2009 09:19 In article <2cd3c08a-a072-4ef5-bc0b-f0aaa9126ea9(a)a21g2000yqc.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Dez., 15:08, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <6a57309a-a136-430c-a718-e38518c65...(a)q16g2000vbc.googlegroups= > .com> WM <mueck...(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, ...} > > That is true. Therefore it does not exist. That you can not get there step by step does not mean that it does not exist. > However, see Cantor, > collected works, p 445: > 0, 1, 2, 3, ... w_0, w_0 + 1, ..., gamma, ..., > He seems to reach far more. Right, he uses a convention that is no longer used. > > > > 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? > > No. But the union contains two paths. Wrong. If we look at the paths as sets, they are sets of nodes. Their union is a set of nodes, not a set of paths. And as a set of nodes we can form from them seven different paths. > And an infinite union of that > kind may contain the path 0.111... The union will contain a set of nodes. With the nodes we can form paths (but they are not elements of the union). But by your statements an infinite sequence is not a path, and so 0.111... is not a path. > > > > 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. > > The construction of the tree can be done within one step. Define: Let > there be every finite path. And every finite path is. The construction > of a path does not depend on a preceding step. It was in the construction *you* made. But what is the definition of path in this definition of the tree? > > > > > 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? > > Let every finite path of every infinite path be mapped on the elements > of omega. That was simple. By your statements infinite paths do not exist. But pray give such a mapping. Until now you have only asserted that such a mapping exists without showing that. -- 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 18 Dec 2009 09:22 In article <dd852490-e622-4bc4-b858-dfcc3b142f8c(a)l13g2000yqb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: .... > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > No. Suppose we have the paths 0.000 and 0.100, what is their union? > > > > And is it a path? > > > > > No. But the union contains two paths. And an infinite union of that > > > kind may contain the path 0.111... > > > > So according to WM a union of sets may contain an object not contained > > in any of the sets being unioned. > > For paths and initial segments to contain and to be is the same. Eh? paths contain nodes and initial segments contain numbers. But be aware that you use the word 'contains' with two different meanings at different times. > For paths and initial segments to contain and to be is the same. In > fact: > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > and also > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. In the mathematical sense the union contains numbers, not sets. > This is so according to set theory. Of course it is rubbish. Well, if you want to use terminology in a different meaning than standard, of course. -- 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 18 Dec 2009 09:26
In article <jrydnSyLVZ_6vbbWnZ2dnUVZ_vOdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:Kusoo8.xz(a)cwi.nl... .... > > Let's have some arbitrary object 'a' and the natural > > numbers. Create > > the sequence A_n where A_n = {a} and the sequence B_n > > where B_n = {n}. > > According to your definition: > > lim sup A_n = {a} > > and > > lim sup B_n = {N}. > > Now create the sequence C_n: C_2n = A_n, C_2n+1 = B_n. > > Again according > > to your definition: > > lim sup C_n = {a} > > which is not equal to union (lim sup A_n, lim sup B_n). > > This is a good example, thanks. Your theorem only applies > in special cases for the definition I have offered (although > my definition satisfies some different but interesting > theorems). Such as? Certainly not: limsup | S_n | = |limsup S_n| because see for that the sequence C_n above and limsup. Stranger, with your definition, lim C_n does exist and is equal to {a}, but lim B_n equals {N}, where B_n is a subsequence of C_n. Strange that an infinite subsequence can have a limit different from the limit of the original sequence. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |