Cantor Confusion
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From: Dik T. Winter on 27 Jan 2007 20:51 In article <1169822674.776815.294960(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 26 Jan., 13:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1169806783.611087.268...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > It has nothing to do with sophisticated models at all. They only have > > > to simulate the basic fact. A natural number in its most basic > > > representation, namely unary or unadic, simply is the superset of all > > > smaller numbers and a subset of all larger numbers. > > > > In that case you have first to *define* what you mean with "union of > > representations". There is no definition for it in mathematics. > > For natural numbers c is the same as <. 2 c 3 is the same as 2 < 3. > This is proven by II + I = III. Perhaps. You are focused at notation. Mathematics is *not* concerned with notation, it is concerned with the basics behind it. And in what way is III c IV c V ? > > > That is the origin > > > which has to be simulated by every correct theory of natural numbers. > > > II is a subset of III. > > > > By what definition? In what way are II and III sets? > > In that way which need no further declaratiojn. Put three nuts in your > hat. Then you can see the set. Yes, I have a set of three nuts. The set is a set of nuts. That does not mean that three is a set. Three is a set of what exactly? > > > This kowledge as to be taken into account when > > > denoting these numbers by 2 and 3. It has been done by the notation 2 < > > > 3. > > > > In the von Neumann model, indeed. a < b if a subset b. That is the > > definition of <. In the Peano axioms < can also be defined, using the > > successor function, but there it has not necessarily anything to do with > > subsetting. > > In unary representation it has. In other representations it has to be > defined. Oh. So you think that subsetting comes before ordering in unary, but it requires definition in other representations? Remarkable. Set theory and the Peano axioms are not interested in representations at all. They are only concerned with the abstract concept behind it. Moreover, when you want to apply set theory to unary representations you will get problems. Consider: 111 how do you propose to define it as a set? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 27 Jan 2007 21:34 In article <1169934207.714872.13260(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > I am interested in the fact that every set of natural numbers has a > finite maximum. What about the fact that it isn't a fact?
From: Dik T. Winter on 27 Jan 2007 21:31 In article <1169905223.354212.199580(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:21, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > A path-length which has no upper bound may be sufficent to be called > > > infinite. > > > > Sorry, a path-length is something fixed, so the wording is inadequate. > > But if you call a path-length infinite you do not satisty (1), because > > there you associate path-lengths with natural numbers. > > We can boil down the discussion about trees to the following simple > question, considering only one path, for instance the path p on the > outmost left hand side of the tree. This path p (in terms of nodes) is > the union of all paths of finite trees with length n, n in N. If you consider the union of paths as the union of sets of nodes, you are right. > Therefore all the path-*lengths* in the union are natural numbers. Yes, the cardinality of all those paths (as sets of nodes) is a natural number. > Notwithstanding the question whether there are infinitely many paths > in the union or not: If the union path p is infinite, Note that the union path is the union of sets of nodes. > then at least > one of the paths in the union must be infinite. Wrong. The union of a collection of finite sets can be infinite, and in this case is. > Is this so? No. You fail to see a crucial difference. Going back at natural numbers: union(n in N) {1, 2, 3, ..., n} is N but union(n in N) {{1, 2, 3, ..., n}} does not contain N. As your paths are equivalent to initial segments {1, 2, 3, ..., n} we find: union(n in N) p_n is p_oo but union(n in N) {p_n} does not contain p_oo. > > And, again, that is fundamentally wrong. It does not show anything of > > that kind. The union of all sets of paths from finite trees contains > > only finite paths. That is pretty basic set theory. If none of the > > sets used in the union contains an infinite path, there is also not an > > infinite path in their union. > > That is set theory. But it is wrong. The union of all finite paths of > length n is infinite, but not actually. It has no upper bound. A fundamentally misunderstanding. The union of a set of things is not the set of the union of things. > > > That is what I proved. In the union of the *sets* of paths in the > > > finite trees there is no infinite path. > > > > And that is what I am stating all along, but you are arguing against. > > I know that it is correct. But it implies that the union of all finite > paths is not infinite. This requirement can only be met by potential > infinity. You are wrong. There is no implication at all. Getting the union of *sets* of paths is not related to the union of paths at all. > > Indeed. That subset is empty. Because by your definitions *none* of the > > finite paths is a path in T(oo). Indeed, also, *none* of the paths in > > some T(m) is a path in T(n) when n > m. But, also by your definitions, > > T(oo) does contain paths. Or do you now claim that T(oo) does not contain > > paths at all? > > I claim that there is no actual infinity. T(oo) is a potentially > infinite tree (i.e. finite without an upper bound). And T(oo) does not contain any path? But how do you *define* "finite without an upper bound"? Can you come up with mathematical definitions of "actual infinitiy" and "potential infinity"? > > This reasoning > > is *exactly* the same as stating that N is finite. > > Correct! N is a potentially infinite set. No other form of infinity is > possible, as we observe in T(oo). I observe nothing of the sort. > > P_C is the set of finite sets P_C. > You mean finite sets P(k), I assume. Right. > > Now P_C is indeed a countable set. Its elements are P(1), P(2), etc. > We can form the union of all P(k) to get the set of all paths in all > finite trees. You can, indeed. > > A subset of P_C is something like { P(1), P(2) } with as elements > > *sets* of paths. However, P is a set with as elements paths. So P > > can *not* be a subset of P_C. > It is a subset of the union set P(1) U P(2) U ... Wrong, and that is not what you claimed. You claimed that P was a subset of P_C. You claimed: > 4) The set of paths in T(oo) is a subset of the countable set > of finite sets of all paths in the finite trees. P was the set of paths in T(oo). P(k) was the set of paths in the finite tree T(k). P_C was the set of finite sets P(k). But even this statement is indeed wrong: > It is a subset of the union set P(1) U P(2) U ... 0.010101... is a path in the complete tree, but is not a path in any of the P(k). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 27 Jan 2007 21:53 In article <1169934895.424017.240350(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 16:52, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > Please define path: > > We need only path-lengths n. > > > > > > > for instance the path p on the outmost left hand side of the tree.In the > > > sequence-of-nodes picture this is the path > > > > p = { (0, 2^i) | i e omega } > > Ok, but it is not necessary to consider the whole tree and to count > all its nodes and to sort out those belonging to the path. > > > The length of p_n is a number which has numbers (0 path-lengths) *as > elements*. > In the union of numbers, the numbers are elements. Every natural > number is a set the elements of which are all smaller natural > numbers. > n = {0,1,2,...,n-1} > The lengths of the paths in the union of path-lengths are represented > by natural numbers. But there can be paths in the union of all finite trees which are not paths in any finite tree, as paths are derived, not essentials to the tree structure. A path in a tree is is a maximal sequence of linked edges in that tree with each link connecting the child node of one edge to the parent of the next. Equivalently, a path in a tree is a maximal sequence of linked nodes in that tree with each successive node being the child of its immediate predecessor node. It is the maximality in its given tree that is essential for a path, as anything less is not a path in that tree. So that in the tree WM declaims as the union of all finite trees, such sequences of edges (or of nodes) cannot be finite and simultaneously be maximal. > > It is sufficient that the path with length n implies the existence of > the path with length n+1. If a union of paths must itself be a path then the above suffices to prove that there are paths which are not finite, and thus not represented by natural numbers. At least by the above maximality definition.
From: Virgil on 27 Jan 2007 21:55
In article <45bbd0a1$0$97250$892e7fe2(a)authen.yellow.readfreenews.net>, Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > > On 27 Jan., 16:52, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > [...] > > How can a union of finite numbers yield an infinite number? This is by > > definition impossible. > > The following applies: > > >> All in all you presented today next version of your 2003's theme > >> > >> X is not finite -> there must be an x in X which is infinite > >> > >> cast into paths. What comes next? > > > > Correct items will not change. > > Quantifier dyslexia? > > F. N. It appears that WM's misjudgements will not change either. |