Cantor Confusion
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From: Franziska Neugebauer on 9 Feb 2007 04:58 mueckenh(a)rz.fh-augsburg.de wrote: > On 8 Feb., 14:43, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 8 Feb., 11:02, Franziska Neugebauer <Franziska- >> > Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> >> This is not true. A tree *then* is the structure plus the set of >> >> paths and/or nodes. You persistently try to vaporize the essential >> >> constituents of a tree until it fits your preconception. >> >> > What is wrong with my use of trees, except that it is inconvenient >> > if one is a bit clumsy in adapting new ideas? >> >> ,----[ WN in <1170852009.005792.107...(a)p10g2000cwp.googlegroups.com> >> ] >> | The tree is a set of nodes. [(TISON)] >> `---- >> >> (This is not Virgil's definition that a tree is a set of paths. >> This is not a definition of tree at all since tree structure is not >> taken into account. A set of nodes is simply a set of nodes.) >> >> But let us /assume/ that a tree is a set of nodes. Then we read >> >> ,----[ contd. ] >> | It contains paths as subsets. >> `---- >> >> This is wrong and remains wrong under your definitions of path: >> >> ,----[ WM in <1170923642.541518.218...(a)h3g2000cwc.googlegroups.com> ] >> | A path is an ordered set of nodes. [(*)] >> `---- > > Instead of a path P consider the set S of nodes K which belong to a > path P. Do your calculation and arguing. Then substitute P for S to > have a brief notation. > >> Assume that the tree is the set of nodes S and there is a path p' = >> (S', <). Your claim (*) is >> >> p' c S >> -> A x(x e p' -> x e S) >> -> (S' e p' -> S' e S) & (< e p' -> < e S) >> -> S' e S & < e S >> -> false & false >> -> false >> >> Hence your "new ideas" are plain wrong. > > Instead of a path P consider the set S of nodes K which belong to that > path P. Do your calculation and arguing. You should change the wording of your definitions/conjectures first. > Then substitute P for S to have a brief notation. Fallacy of composition. F. N. -- xyz
From: Franziska Neugebauer on 9 Feb 2007 05:09 mueckenh(a)rz.fh-augsburg.de wrote: > I am about to show that infinity does not exist. Good luck! > So I will not create sci.math.infinity. Surely not. F. N. -- xyz
From: Dik T. Winter on 9 Feb 2007 08:34 In article <1171010690.654184.223540(a)a34g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 8 Feb., 14:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > In some models of ZFC. You can not prove that every element is > > > > a set based on ZFC only. > > > > > > It is simply a definition. > > > > Yes, you can not prove it. > > You cannot disprove it. As I have not seen a definition yet that makes it true. > > So a statement like "every element is a set > > in ZFC" is false. > > It is used by most authors of text books on set theory. Talk to them > about their errors. I am no interested in what they use to denote the > elements (or sets) of their erroneous theory. But they give a proper foundation for that statement. You do not. You pull that statement out of thin air. > > No. I still argue that p(0) U p(1) U p(2) U ... *is* p(oo), it does > > not contain p(oo). > > Fine. I agree. For me it is unimportant how you think p(oo) is > established. Ayhow we can say p(oo) is the limit of the union p(0) U > p(1) U p(2) U ..., which is expressed by the three "..." Limits are not involved here. The union of a collection of sets is defined without using limits, and it does not matter whether the collection if finite, countable or uncountable. In set theory limits are not defined. > > Moreover, when P(i) is the set of paths at a finite > > level, P(0) U P(1) U P(2) U ... does *not* contain p(oo). Because, for > > this union to contain p(oo) at least one of the constituent sets P(i) > > *must* contain p(oo). But the infinite tree *does* contain p(oo). > > In short P(oo) is *not* a subset of P(0) U P(1) U P(2) ... . That is > > my objection against your argument. When you unite sets of paths you > > do *not* unite paths. > > But your argument is that the union P(0) U P(1) U P(2) U ... > establishes the existience of P(oo) while the union of P(0) U P(1) U > P(2) U ... U (some other paths) does not establish the existence of > P(oo). This is inconceivable. Read what I wrote. P(i) (uppercase P) is a set of paths, p(i) (lowercase p) is a path. (This is from your notation from an earlier article.) I state: (1) p(0) U p(1) U p(2) U ... = p(oo) (2) P(0) U P(1) U P(2) U ... != P(oo) see the difference? The union of paths establishes the infinite path. The union of *sets* of paths does not establish the set of infinite paths. > > > > How do you find that > > > > {p(0), p(1), p(2), ...} = p(oo)? > > > > on the left there is a set of sets of nodes, on the right there is > > > > a set of nodes. How can they be equal? > > > > > > Because the tree contains the union U{p(0), p(1), p(2), ...} = p(oo). > > > > You are still maintaining that {p(0), p(1), p(2), ...} = p(oo)? > > I leave it to you how the union of all finite paths p(n) belonging to > p(oo) establishes the exisence of p(oo). I am not at all interested in > the mechansm how p(oo) appears. You say that it appears. That is > enough. You state that {p(0), p(1), p(2), ...} = p(oo). I wonder how you can state that when on the left the elements of the set are paths, while on the right the elements are nodes. > > So that > > a set of sets of nodes is a set of nodes? Or a set of paths is a path? > > That does not matter the least. The union of all paths belonging to > p(oo) establishes the existence of p(oo). The union of those and some > more paths does not establish this existence. This is a contradiction. > However you need it in order to save set theory. But you stated that {p(0), p(1), p(2), ...} = p(oo). I am asking you to either retract that statement or to prove it. > > > > I agree that > > > > p(0) U p(1) U p(2) U ... = p(oo), > > > > (allowing finite paths in p(oo)), but neither is equal to > > > > {p(0), p(1), p(2), ...} > > > > It is already not true for the finite case, so I wonder why it must be > > > > true in the infinite case. > > > > > > The tree makes the union! > > > > How can a set of paths be a path, when a path is defined as a set of nodes? > > The union of sets of nodes is a set of nodes. What is your problem? > The union of paths (in same direction) is a path. How can a a "set of paths" be a path? What you are actually stating here is: { {0}, {0, 1}, {0, 1, 2} } = {0, 1, 2} How is that possible? > > > It is the union. Just in the same way T(1), T(2), T(3), ... is the > > > union T(oo). > > > > Pray define "contain". Up to now I have seen you use it in two ways: > > (1) a contains b if b is an element of a > > (2) a contains b if b is a (possibly improper) subset of a > > Both is correct and can be applied. But that makes a confusing discussion. > > > 1) The union of finite trees contains the union of finite paths. > > > 2) The union of finite trees is the whole tree. > > > 3) The whole tree contains all paths. > > > 4) The union of finite paths contains all paths. > > > 5) This union is countable. > > > > The union of finite paths is indeed a countable set of nodes. It contains > > all paths as subsets. But the set of subsets of that set of nodes is > > the powerset, which is *not* countable. > > Wrong, the uncountable power set contains all combinations of nodes. > The finite paths, however, are countable. Therefore are sequences are > countable. Therefore their limits are countable too. Prove it. > > > > > It shows clearly that not the number of elements is decisive > > > > > but their sizes. > > > > > > > > Prove it. > > > > > > The infinite union of sets {1} is a finite set. > > > > The infinite union of the sets {1}, {2}, {3}, is not a finite set. > > That is your assertion. You know that it does not hold wihout the > axiom of infinity. Yes. > My example shows that a union need not be infinite. But I did *not* object to that. I only stated that an infinite union *can* be infinite, regardless of the size of the sets united. > > But you want to show that P(oo) is countable by showing that it is the > > union of finite sets. > > But if it is the set of the unions of finite > > paths, it is *not* the union of finite sets. > > p(oo) is a set. This set is the union of finite paths. This is a union > of finite sets. Again, you do not want to prove countably of p(oo) (which is a path), but of P(oo), which is a set of paths. (Note the difference between lowercase and uppercase, something you introduced in an earlier article.) > > It is a set of unions. > > Above you say: p(0) U p(1) U p(2) U ... = p(oo). > This is a union of finite sets. Now again, lowercase p. Please pay attention. I am not talking here about paths (lowercase p) but about sets of paths (uppercase P). Pray properly distinguish *paths* and *sets of paths*. You want to prove something about a *set of paths*, not about a *path*. > > Because P(oo) can not be the set of unions of finite paths. > > P(oo) is as a subset in the union of finite paths which is the set of > all nodes of the tree. It is not a subset of the set of nodes. P(oo) is a *set* of paths, and so its elements are paths, not nodes. Each subset of the set of all nodes is a set of nodes, not paths. > > Because > > most unions of finite paths are not paths, and so are not in P(oo). > > The unions which are not paths (and which probably would imply > uncountability) are not in the tree and, threfore, do not exist in our > problem. But those paths which are necessary to establish P(oo) are in > the tree. And THEIR union does exist. But it is not about THEIR union. It is about the union of *sets of paths*. If we define (note: uppercase P) P(i) as the set of paths in the finite i-th tree, we have: P(oo) not subset P(0) U P(1) U P(2) U ... because P(oo) contains only infinite paths, while the union on the right contains only finite paths, namely *all* paths of finite length. Your continuous mixing the discussion of paths with the discussion of sets of paths is quite confusing. So much so, that it confuses yourself. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 9 Feb 2007 08:53 In article <1171011850.731985.236670(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 8 Feb., 13:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > Why not. IIIII can be abbreviated by 5 or V or your proposal or ... > > > > Again, still no definition. So IIIIII can be abbreviated 5I, I5, VI or > > IV? And IIIIIIIIII can be abbreviated 55 or VV? > > That is a matter of convention. Therefore the unary represantation, > which is not a matter of convention, is preferable. Also unary representation is a matter of convention in my opinion. > > > > No. Numbers are abstract entities. Their representations are > > > > concretisations of those abstract entities. > > > > > > Where and what are these entities? > > > > Abstract entities. > > Where are they? Further your answer does not distinguish between > numbers and impressions of colour or beauty. If you say a number > exists, then you must be able to distinguish it from *everything* else > (including other numbers. This is certainly not done by saying: > abstract entity. But the distinctions can be performed. I know precisely how to compare the abstract entities behind sqrt(2) and sqrt(3). > > > You have to use some agrrement if you want to use abbreviations. For > > > IIII you need no agreement. > > > > Oh. I think one is needed. > > I believe that without any other cultural contact two persons could > start communicating by unary numbers. Yes, your belief. > > > If you use "nat�rliche Zahl" or "N" does not matter. You use something > > > not yet defined (and never defined by Peano). > > > > You apparently do not know how a recursive definition works. (1), (2) > > (when properly corrected) define the natural numbers. (3) defines the > > set of natural numbers. > > I know it. Cantor already used it (see p. 128 of my book). That does > not mean that it is good. > (1), (2) have not to be corrected (they appear in several text books). > Your assertion that " x is in N" and "x is a natural number" were > different is wrong. Both has to be defined. But, in (1) you state "is a natural number", in (2) you use "is in N", without stating anywhere in that article that the two things mean the same. So the combination of those two is *not* a proper definition. > > > N is only an abbrevation of set of natural numbers. > > > > An undefined abbreviation. When doing definitions you should do it the > > correct way. See: > > We are going to define natural numbers: > > (1) 1 is a natural number > > (2) if a is a natural number, the successor of a is also a natural > > number > > These two together define the natural numbers. > > Not yet. According to your definition also -7 and pi could be natural > numbers. How do you come to that conclusion? At which step is either -7 or pi generated by (2)? > But we could write your two axioms also as: > > We are going to define N: > > (1) 1 is in N > > (2) if a is in N, the successor of a is also in N. > > These two together define N. Right. But not as you did: (1) 1 is a natural number (2) if a is in N, the successor of a is also in N. These two together define the natural numbers. By this definition only 1 is a natural number, because you can not even start with statement (2). > > > Correct. There is no infinite line existing. All we assume is a line > > > longer than any line we have measured yet. > > > > Such lines also do not have physical existence. Each and every physical > > line has a width smaller than anything measured yet. And I do not think > > that physical lines are really straight either. > > The physical existence of a line is "a measurable distance" between > two points. The points exists as sets of coordinates. Oh. What is a circle? What is a parabola? Anyhow, what you do is not plane geometry, but analytic geometry. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: G. Frege on 9 Feb 2007 08:56
On Fri, 9 Feb 2007 13:34:36 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >>> >>> So a statement like "every element is a set in ZFC" is false. >>> Actually, the common chant (concerning ZFC) is: "Everything is a set." It's even possible to "prove" that. ;-) Define: set(x) =df Ey(y e x) v x = 0. (This definition would make sense if there were urelements in our theory. Yes, I know, there are no urelements in ZFC. :-) Then the following theorem holds: Ax set(x). ;-) >> >> It is used by most authors of text books on set theory. Talk to them >> about their errors. I am no interested in what they use to denote the >> elements (or sets) of their erroneous theory. >> > But they give a proper foundation for that statement. You do not. You > pull that statement out of thin air. > Though -I think you will admit that- this doesn't make WM's claim _false_. Right? His claim might be formalized the following way: Ax(Ey(x e y) -> set(x)). "For every x: If x is an element (of some y) then x is a set." Since in ZFC we have Ax set(x) WM's claim is trivially true. No? (On the other hand it _would be_ false if there were urelements in ZFC. But since there are no ...) F. -- E-mail: info<at>simple-line<dot>de |