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From: Dik T. Winter on 17 Dec 2009 09:15 In article <4c4737cb-07c4-4827-8957-ce2c7a678efa(a)p30g2000vbt.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 16 Dez., 14:21, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > 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. And what *is* that definition without the axiom of infinity? > > > > 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. The axiom does not talking about something that does not end. The axiom states: There exists an inductive set that is all. So what do you actually *mean* when you state that omega is something that does never end? It is similar to stating that 100 is something that never ends. > > > > 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. Who says so? And what is a definition of potential infinity? > > 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. No. > The axiom states that such a thing (it says set, but does not say what > a set is) does exist. It states that an inductive set exists. It does not state that it starts with 0, nor does it state that it consists only of the natural numbers. Suppose the successor function adds 1 to a number. In that case: {0, 1/2, 1, 3/2, 2, 5/2, ...} is an inductive set. When you start with 0 and successively apply the successor function you will *never* get at 1/2. -- 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 10:01 In article <876387t8ds.fsf(a)phiwumbda.org> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: .... > In general, the term limit is defined topologically: any topological > space comes with its own natural definition of limit. But it's not at > all clear to me whether "general set convergence" is a limit in this > particular sense. Is there a topology on Set so that the "natural" > definition of limit coincides with general set convergence? I do not think there is a proper topology, but I do not know. > If not, then I guess K_h has a point about the naturalness of this > definition of limit (and, conversely, if so, then K_h has no good > point at all). But here I think different. Within the definition presented there is a limit definition for sets that is derived from the underlying topology on the elements of those sets. At such it appears to me to be a natural definition, that is, an element is also element of the lim sup, it should be a limit point of one of the many pointwise sequences you can create when you take from each of the sets an element. lim inf consists of those elements that are limits of such pointwise sequences. And with the discrete topology on the elements we get just the definitions I quoted. So when we assume the discrete topology on N, we get lim sup(n -> oo) {n} = {} because there is no limit point. With the set "N union {oo}" with a topology based on delta(n, m) = |1/m - 1/n| we get a different result: lim sup(n -> oo) {n} = {oo}. K_h would like that if each S_n of a sequence is a set with a single element, that being a set exain (e.g. {X_n}) is that lim [sup|inf|] S_n = {lim [sup|inf|] X_n} but for that we need a more complicated (and in my opinion less natural) version of the limit of a sequence of sets, just because there is no proper topology on the collection of X_n. -- 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: Jesse F. Hughes on 17 Dec 2009 10:41 "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > Within the definition presented there is a limit definition for sets > that is derived from the underlying topology on the elements of > those sets. At such it appears to me to be a natural definition, > that is, an element is also element of the lim sup, it should be a > limit point of one of the many pointwise sequences you can create > when you take from each of the sets an element. lim inf consists of > those elements that are limits of such pointwise sequences. So, it's a kind of "lifting" of the definition of limit to sequences of subsets of a topological space. Yes, that's fairly natural, though it's still a different notion than a limit. -- Jesse F. Hughes "Mathematicians who read proofs of my results seem to basically lose some part of themselves, like it rips at their souls, and they are no longer quite right in the head." -- James S. Harris, Geek Cthulhu
From: WM on 17 Dec 2009 11:56 On 17 Dez., 14:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <ba9f2c85-caa2-491d-9263-bea1bed3a...(a)p19g2000vbq.googlegroups..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 axioms being > > > 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# > 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? > > > > > 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. To live with that axiom does not create uncountability. See the proof here: http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb5da79d# > > > > > 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. The axiom of infinity establishes the set N from finite numbers. It establishes the infinite paths as well in my tree from 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. However, there is never more than one limit of a sequence and every convergent sequence hat at least one finite term. > > > > > 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. 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, ... > > > 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. Regards, WM
From: WM on 17 Dec 2009 12:05
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. However, see Cantor, collected works, p 445: 0, 1, 2, 3, . . . w_0, w_0 + 1, . . ., gamma,. . ., He seems to reach far more. > > > > > 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? No. But the union contains two paths. And an infinite union of that kind may contain the path 0.111... > > > > > 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. 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. > > > > > 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. Regards, WM |