Another AC anomaly?
in [Logic]
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From: WM on 18 Dec 2009 04:24 On 17 Dez., 22:52, Quaestor <quaes...(a)qqq.com> wrote: > > > > 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, ...}. > > > > As usual you confuse to be an element and to be a party of... > > > I am concerned with numbers. Your absurd distinctions are irrelevant. > > They are neither absurd nor irrelevant to mathematics. Ok, then you have a different idea of mathematics. In common mathematics, numbers can be defined by digit sequences. > > > Either N exists or not. If yes, then it is the union of all natural > > numbers as well as the union of all initial segments of the ordered > > set of all natural numbers as well as the union of all these and > > itself, because > > > {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...} > > > If you see that this equation is true, then you acknowledge my > > argument. > > On the contrary, that equation is true Fine, Virgil, that is one big step to comprehension. Regards, WM
From: Dik T. Winter on 18 Dec 2009 07:22 In article <SqSdnSUF1q8FobbWnZ2dnUVZ_uCdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > With the second option we > can restrict ourselves just to the wikipedia definitions and > define the sets n by: > > n = 0 = {} > n = 1 = {{}} > n = 2 = {{{}}} > ... > > Wikipedia gives liminf(n-->oo) n = 0. Only when we use the definition of limit on sequences of sets, not when we use the definition of limit on sequences of numbers. You have to distinguish the two and clearly state which one you are using. > So the notion of n > growing bigger, as one tends to the limit, is not embodied > here since |n| is 0,1,1,1,1... as one proceeds and is 0 in > the limiting case. Yes, as in many cases: lim | S_n | is not necessarily equal to |lim S_n|, whatever the definition of set limit. The same holds for limsup and liminf. -- 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 07:25 In article <eaCdnfgmKeFaorbWnZ2dnUVZ_hmdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > > Under the standard > > construction each natural is `accumulated' in the limit > > set, since each set persists beyond its introduction, and > > this preserves the notion of n growing bigger as one > > proceeds: |n| is 0,1,2,3,... and is N in the limiting > > case. > > Yo, it should be |N| in the limiting case! Eh? With your definition, lim {n} = {N} (as you stated earlier) and so lim n = N. But now you actually state (as |n| = n), lim n = |N|. Are N and |N| equal or not? -- 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 09:12 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/ |