Cantor Confusion
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From: mueckenh on 9 Feb 2007 02:30 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. Then substitute P for S to have a brief notation. Regards, WM
From: mueckenh on 9 Feb 2007 02:39 On 8 Feb., 15:00, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 8 Feb., 11:05, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > [...] > >> >> Who has told you that? Whatever, Mückenheim axiom 2: > > >> >> "Numbers cannot exist without name" > > >> > If you take all names (in the widest sense, including symbols and > >> > notations and defining equations) from a number and attach it to > >> > another number, what remains with the first? > > >> >> Let's call that numbers "named numbers". If the length of names > >> >> "must" be finite one can easily prove, that any set of named > >> >> numbers has cardinality <= card(omega). > > >> > Of course. That is not new to me. > > >> It seems new to you that in real mathematics numbers do not > >> necessarily have a name to exist. > > > What do those nameless numbers need to exist? > > Wrong question. You need Mückenheim axiom 2 to erase those innocent > inhabitants of Cantor's paradise. Wrong questions are possible? How can we show that a number without name does exist? > > > How can we distinguish them from nonexisting numbers? > > Good grief! First you terminate their existence and afterwards you > complain about their non-existence. I did not terminate their existence. I am looking for a sign of life to localize them and possibly help them to acquire a more solid form of being. Regards, WM Regards, WM
From: Ross A. Finlayson on 9 Feb 2007 02:41 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. Then substitute P for S to > have a brief notation. > > Regards, WM Hi Wolfgang, I think you, too, should support having sci.math.infinity, and instead post your topics there, or on sci.math as I don't care. Please do so. Thank you, Ross F.
From: mueckenh on 9 Feb 2007 02:59 On 8 Feb., 20:29, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > On Feb 8, 1:25 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 7 Feb., 21:24, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > > > > On Feb 7, 4:46 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > There is an > > > > > > axiom which requires the existence of the non existing and seems to > > > > > > make some people happy > > > > > > If there is such an axiom anywhere, it is only an axiom in WM's system, > > > > > not in anyone else's. > > > > > > > (like the axiom which requires the finity of > > > > > > the infinite). > > > > > > Any such axiom exists only in WM's system, and not in anyone else's. > > > > > Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: > > > > "Foundations of Set Theory", 2nd edn., North Holland, Amsterdam > > > > (1984): "Intuitionists reject the very notion of an arbitrary > > > > sequence of integers, as denoting something finished and definite as > > > > illegitimate. Such a sequence is considered to be a growing object > > > > only and not a finished one." > > > > > Who considers it a finished one? > > > > Which particular formal intuitionistic axiomatization do you have in > > > mind? (There are intuitionistic formal systems, but I don't know in > > > particular which one you have in mind so that we can evaluate whatever > > > it is you think holds in or about them.) > > > The question here is not at all what these or those intuitionists > > think, but what "others" think, according to the opinion of Frenkel et > > al. > > > Intuitionists reject N as something finished. Who considers N as > > something finished? > > "Finished" is informal. It expresses exactly the state of art: "There is" > Meanwhile, ExAy(yex <-> y is a natural number) > (where 'is a natural number' is here a rendering of a certain defined > predicate symbol) is a theorem of many recursive axiomatizations. Spare this. I know it. But I understand that finished infinity cannot be swallowed other than formal, i.e., without thinking and with eyes wide shut. > > Meanwhile, you still have not stated from what system of logic and > what mathematical axioms your own mathematical conclusions are > supposed to be derived from. MatheRealism. > Or perhaps (?) you said that you reject > formal axiomatization? Okay, then I don't see what objective basis you > offer for determining which mathematical propositions have been > established and which have not. Reality. > "The finity of the infinite" is whose expression? Cantor's, Hilbert's, Fraenkel's etc. (Informal, of course.) > If 'finity' (as > opposed to 'is finite') and 'the infinite' (as opposed to 'is > infinite') are supposed to be about formal set theory, then what are > the set theoretic definitions and what specfic axiom do you have in > mind and how do you think it states such a requirement? The axiom of > infinity? It states that there exists a set that has 0 as a member and > is closed under successorship. Nothing about 'the finity of the > infinite'. "There exists" means "it is ready, complete(d), finished". But better ask Fraenkel et al. or Hilbert or other mathematicians who know about the things they talk about. (Levy for instance is still alive and can be contacted.) Regards, WM
From: mueckenh on 9 Feb 2007 03:44
On 8 Feb., 14:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170854542.370817.5...(a)a34g2000cwb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 7 Feb., 04:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > No. A finite tree, chain and path are all sets of nodes (by your own > > > definition). The elements in the union are *not* the trees, chains > > > or paths, but the nodes. > > > > Does not matter. The nodes are finite elements. The sets of nodes are > > finite too. > > The sets are finite, unless you unite *all* trees, or *all* chains, or > *all* paths. In that case the union is the set containing *all* nodes. And all subsets of the set of nodes which are nodes of paths. > > > > > Every element is a set. There are only sets in ZFC. > > > > > > 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. > 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. > > > > > > The union of sets of elements does > > > > > *not* contain the union of elements. > > > > > > > > The tree contains it. > > > > > > Yes. So what? > > > > Is your argument about infinite sets of finite sets of nodes which > > cannot contain infinite paths now removed? > > 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 "..." > 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. > > > > 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. > 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. > > > > 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. > > > > > The set of paths contained in the finite tree T(0), the set of paths > > > > contained in the finite tree T(1), ..., the set of paths contained in > > > > the finite tree T(n), and so on. > > > > > > In what way does the union of p(0), p(1), p(2), ..., contain p(oo)? It > > > does not contain it as an element, but it *is* the union. > > > > 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. > > > > > Every path is a subsets of the union, for example {0, 2, 6} c {0, 1, > > > > 2, 3, 4, 5, 6}. Which path do you miss? > > > > > > None. So every path is a subset of the union of the union of the sets of > > > finite paths. That is *not* what you did state. You stated that every > > > path is an element of the union of the sets of finite paths. > > > > 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. > And as the set of all paths is a > subset of that powerset you can not prove this way that that set is > countable. > (And (1), (2) and (3) are not really needed, they merely > serve as obfuscation.) > > > > > > Yes. Infinite unions can give finite things. But that is not a > > > > > contradiction. > > > > > > > > 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. My example shows that a union need not be infinite. > > > > So P(oo) is not the union of all sets of finite paths? But P(oo) is the > > > set of the union of all finite paths? There is a difference. > > > > It is or is not in the tree. That is important. > > 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. > 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. > > > > > That set is a tree but contains the paths as subsets. > > > > > > Yes. But it is not a path. > > > > What is this remark good for? > > 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. > 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. Regards, WM |