Cantor Confusion
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From: Andy Smith on 23 Jan 2007 05:45 Fuckwit <nomail(a)invalid.?.invalid> writes >> >> But, with respect, m'lud, can't I have as many bits as there are rows in >> the table? >> >Sure you can. So what? > >> >> [...] I am ... trying to clarify for myself why Cantor's line of >> argument is valid when the above isn't. >> >I repeat myself: > >"Cantor's argument doesn't work for natural numbers, because in this >case it's not guaranteed that the "diagonal" delivers a natural >number. (You might compare that with Cantor's argument concerning a >list of real numbers with decimal representation.)" > >To make a long story short: > >(a) List of (representations of) real numbers => diagonal delivers a >real number (which -as it turns out- is not in the list). > >(b) List of (representations of) natural numbers => diagonal doesn't >delivers a natural number. (Hence the diagonal argument does not >apply.) > >Clear enough? > Well I understand what you are saying, but, sorry to be dim, I still don't see it. There is no greatest natural number. So I can construct a finite ordered list of natural numbers, labelled 1 to n, and identify a number not in the list, and label that n+1. I can do that for all n. Cantor's hypothetical numbered list of the reals is also finite ? So Cantor's construction for a list with n elements just generates another real, which he can insert as the n+1 th row, and he can do that for all n? As I had previously understood it Cantor's argument relied on a hypothetically complete set of reals, with an actually infinite number of rows, and then showing that there was a real not included in the actually infinite list. But since you cannot have an actually infinite natural number, you cannot have an actually infinite list, so the argument is invalid? (I am sure that it isn't, but that is what I am trying to understand). -- Andy Smith
From: mueckenh on 23 Jan 2007 05:47 William Hughes schrieb: > > There is no path which ends at a node which you can specify. Every path > > of the union tree has maximum length. 0.11 is not a path in the tree of > > three or more levels. > > > Since T1 is not a tree, you cannot use a statement about the > properties of of a tree to say something about the properties > of T1. 0.11 is a path in T1. Whatever T1 may be called. 0.11 is, by definition, not a path in T1. > > > Define "the maximum set" as the union of > > > > {1}, {1,2}, {1,2,3}, ..., {1,2,3,...,n} > > > > "the maximum set" is different for every n. And for all n it is omega. > > > It is obviously nonsense to assert that every maximum set ends > > somewhere, because there is only one maximum set which ends at n. > > There are lots of different n so there are lots > of different maximum sets. Each different maximum > set ends some n. In the finte tree of n levels, but not in the union. In the union with n > 3 already, there is no path ending at n = 2. > > I do take it as the definition of a tree. However, as T1 is > not a tree, this definition is not relevant Why is T1 not a tree? Concerning nodes, edges and levels it is T2, as you agreed. > . > > > > > > More formally. T1 is the union of all finite paths. Let M be the > > > set of nodes in T1, (that is if a node m is in M, then there is a > > > path p(m) in T1, such that the node m is in p(m)). > > > > > > Let p1 be a (potentially) infinite path. Then > > > > > > Every node in p1 is contained in some path > > > in T1. > > > > The infinite union of all finite trees can only contain nodes which are > > in a path. Otherwise the nodes could not enter the union. Nodes are > > ends of paths of finit trees. > > And therefore every node in p1 is contained in some path in T1. > This is not enough to show that p1 is contained in T1. > > > > > > > However > > > > > > There is no path in T1 that contains every node in p1 > > > > Then not every node of T1 is in T1. > > No. Every node in p1 is contained in T1, however, you need more > than one path in T1 (indeed an infinite number of paths) to get > every node in p1. What would be necessary? We had an infinite number of paths in the finite trees. > > > There is no initial segment P of N which contains all natural numbers. > > Therefore the union of all P (which is N) does not exist? > > No. To have the union of all P to be N it is necessary that > for any natural number n, there is an initial segment p(n) > such that p(n) contains n. It is not necessary that p(n) > be the same segment for every n. But it cannot be avoided if all natural numbers are contained in he same set N. However, it may not be necessary that every n is contained in he same set N? > I have stated many times that the union of finite trees > contains the same nodes as the infinite tree. > The union of finite trees is not a tree. So you acknowledge the complete infinite tree as a tree and also every finite tree as a tree. Burt the union of all infinite trees is not a tree? Why. > The fact that > it contains the same nodes as the infinite tree does > not mean that it contains the same paths as the > infinite tree. Then you should be able to say what prevents some paths from being there where the corresponding nodes are available. I propose we should denote the paths (being absent) by Pnot and the corresponding nodes and edges (which would constitute the path Pnot if it were present) by Pyes. Then it will be easier to talk about the difference between Pnot and Pyes. Regards, WM
From: Franziska Neugebauer on 23 Jan 2007 05:50 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> >> By induction you only define (if at all) every _finite_ union >> >> >> >> >> >> T(1) U ... U T(n) >> >> > >> >> > That is completely sufficient as long as there is no upper >> >> > bound, but >> >> > n --> oo (potential infinity). >> >> >> >> Then I would like to have an explanation what >> >> >> >> T(1) U T(2) U ... >> >> >> >> shall mean in contrast to >> >> >> >> T(1) U T(2) U ... U T(n) n e N. >> > >> > There is no contrast but identity (for n --> oo). >> >> "M�ckenheim-Limes"? > > There is a tree T which contains the root node at level 0 and if it > contains the (finite tree T(n) down to) level n then it contains the > (finite tree T(n+1) down to) level n+1. You have to choose whether you want to see a tree as a) the standard tree, T = (M, E) b) the "Virgil"-tree. T = (set of paths having certain properties). Which definition do you choose? > The existence follows from the existence of the real numbers and all > their initial segments > > The union of all levels = the union of all finite trees = tree T. > >> >> U { T(i) | i e N } >> >> is not defined. >> >> > V* is not in the union (as omega is not in the union over all >> > natural numbers or over all finite intial segments of omega). >> >> I want you to define >> >> T(1) U T(2) U ... > > It is defined. It is the tree which contains the root node and if it > contains the tree with n levels, then it contains the tree with n+1 > levels. So you define it to be my G from <45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net> If so then the set of paths of G is cardinally greater than the std-union of sets of paths of the trees in the set of all finite trees V*. >> > > The union of all natural numbers does not include omega. >> > The union of all segments does not include it. >> > The union of all finite trees does not include it. >> >> I am asking for a definition of "the union of all fintie trees". >> You have not yet given one. > > Define the union of all segments of paths which always turn right. What is a segment of paths? > Define the union of all segments of paths which always turn left. > Define the union of all segments of paths which always alternate. > So you get 0.111... and 0.000... and 0.101010... > > Try to doubt the unions of all segments of any other infinite path. You have switched over to the Virgil-trees. In his view a tree is a set of paths. Since the union of V* is the std-union we know that that U V* contains only "finite" paths. Hence 0.[10] is _not_ in UV*. F. N. -- xyz
From: mueckenh on 23 Jan 2007 06:00 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> > Max n,m is not defined either, > >> Aha > >> > >> ,----[ WM in <1169111380.377993.67320(a)l53g2000cwa.googlegroups.com> ] > >> | The union of two finite trees T(m) and T(n) with m and n levels, > >> | respectively, where m < n, is the tree with n levels. > >> `---- > >> > >> So you mean m < n is not defined? Then it makes no sense at all to > >> write about trees? > >> > > Sorry, this should read: Max (n,m) is not defined *other* (than for > > finite m and n). > > The problem with defining the tree-union with the max-Function is that > max N is not defined. (N has no maximum). > > (If your N has a maximum, i.e. is finite, you leave the contemporary > set theory.) > > > The union of m and n is the maximum of both. > > Nevertheless the union of all natural numbers exists As well as the union of all levels L(n) of the tree, by the bijection L(n) <--> n. > > Your tree-union of two trees is _defined_ to be the tree having the > greatest of both trees depths. This is not a set-theoretical union. It is, because in the union every element appearing twice is eliminated. There are not two different roots or two different nodes 39 etc. Therefore it is exactly the set theoretic union. The union of two natural numbers is defined to be the larger one. This is a set theoretic union. > The only thing it has in common is the name "union" (equivocation). > > The extension of your tree-union to V* = { T(n) | n e N } fails as I > have pointed out already three times (or even more) because max (N) > does not exist. You can repeat it for 300 times without getting correct. > > > as well as the > > union of segments {1,...,n} and {1,..., m} and the infinite union of > > all segments. > > These unions are standard-unions of set theory, your tree-union is not. What about the union of levels? What about the union of initial segments of one path? > > Again. U V* is not defined. Again: I defined it so that everybody can constuct the finite union of finite trees. This construction does not come to an end. It is without end, i.e., infinite and can be described by U T(1) U T(2) U T(3) U ....., The "..." here are as appropriate as in 1,2,3,.... And even if this were never heard of in set theory. Why shouldn't we consider it as a new topic? Regards, WM
From: Franziska Neugebauer on 23 Jan 2007 06:06
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> > nevertheless the union of {1,...,n} and >> >> > {1,..., m} and the infinite union of all segments are defined. >> >> >> >> We are writing about trees. >> > >> > The trees have levels. For them we have the same as for the >> > initial segments given above. >> >> Again: Your notations >> >> T(1) U T(2) U ... >> >> and >> >> U {T(i) | i e N } >> >> are undefined. > > You are in error. The union of the trees T(n) and T(n+1) is defined. n > is a natural number. Therefore the union of all finite trees is > defined. You have misunderstood the induction principle. It is not made for "counting over to the infinite". If tree-union was the std-union of ZFC then by the axiom of union we can state that U V* is defined. But that requires a commitment to Virgil's definition of trees, which is not the standard-meaning of trees. > If you try to construct the tree with n levels, do you fail at some > number of levels? No. Therefore the union is defined for every n. But not for V*. Recall: V* = { T(n) | n e omega } is the infinite set of all finite binary trees. > More is not feasible. > >> >> >> >> >> | >> >> >> | The questions is: How do you define U V_omega? >> > >> > V_omega is not in the union. >> >> What is V_omega? > > The union of all finite trees. So you implicitly commit to Virgil's trees? I.e. a tree is a set of paths. >> > The union is only over omega elements, but that is not a problem >> > because it is covered by the axiom of infinity. >> >> Unions are covered by the axiom of union. But only if real >> set-theoretical unions are involved. >> >> It is an equivocation (fallacy) to claim that your tree-union (which >> selects the deepest out of two trees as "union") is a union. It has >> been explained by Virgil, WH and me that a set-theoretical union of >> trees is hardly a tree. Hence you are writing on undefined notations. >> > > It has been claimed, but falsely. If two trees, T(n) and T(m) are > idential down to level n but T(m) contains some moere levels, then the > union of both is T(m). Only under Virgil's definition this union is identical to the standard-set-theoretical notion of union. If so UV* is not identical to the full set of paths of the infinite binary tree G given by me in a elsewhere. Virgil's UV* only comprises finite paths. > Further, both Virgil and William understand that the union defined by > me is identical to te complete tree T as far as nodes and edges and > levels are concerned. Nodes, edges and levels. Shaken or stirred? > They merely doubt the identity of path due to some inexplicable > religious belief in a death religion. I must admit that great part of the discussion is to subtle for you. F. N. -- xyz |