Cantor Confusion
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From: Franziska Neugebauer on 21 Jan 2007 05:58 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > William Hughes schrieb: [...] > I say: The union of indexes of paths {1} U {1,2} U {1,2,3} U ... is an > infinite index union {1,2,3,...} "{1} U {1, 2} U ..." means U {{1, ..., n} | n e N }. This set-union exists due to the axiom of union (in ZFC). It is valid that s = sup {{1, ..., n} | n e N} = {1, 2, ...} = N Note that N is not a member of s, N is not the maximum of {{1, ..., n} | n e N}. > corresponding to n infinite path. Where is your union of all finite trees? You have not yet shown: ,----[ <45af6ca0$0$97262$892e7fe2(a)authen.yellow.readfreenews.net> ] | 7. Now let V* denote the set of all finite trees { T(i) | i e N }. | U V is only defined for V having card(V) e N. Since V* does not | meet this requirement we (you?) have to define what | | U V* | | shall mean. The obvious definition | | "U V* = T(max(D(V*))" | | fails due to the reason that max(D(V*)) = max(omega) is not defined. | | The questions is: How do you define U V_omega? `---- F. N. -- xyz
From: Franziska Neugebauer on 21 Jan 2007 06:05 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> Whether a _meaning_ of >> >> >> >> T(1) U T(2) U T(3) U ... (inf-u) >> >> >> >> exists does _not_ depend on such question. It depends on a >> >> definition you have to provide. So please answer the question, >> >> what (inf-u) means and prove that it exists. >> >> >> >> > This is connected >> >> >> >> Whether that is "connected" is irrelevant to a possible definition >> >> of (inf-u). >> > >> > That is not a true statement. Nevertheless: >> > >> > Definition of the infinite union of trees by induction: If the >> > finite tree with n levels exists, the finite tree with n+1 levels >> > exist. The tree with 1 level exists. >> >> 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. > But if you, personally, think it is necessary, then simply use the > axiom of infinity. The axiom of inifinity in relation with the axiom of union does not apply here since: ,----[ <45af6ca0$0$97262$892e7fe2(a)authen.yellow.readfreenews.net> ] | 7. Now let V* denote the set of all finite trees { T(i) | i e N }. | U V is only defined for V having card(V) e N. Since V* does not | meet this requirement we (you?) have to define what | | U V* | | shall mean. The obvious definition | | "U V* = T(max(D(V*))" | | fails due to the reason that max(D(V*)) = max(omega) is not defined. | | The questions is: How do you define U V_omega? `---- > There is a set of all natural numbers, a set of all > levels, a set of all finite trees. Again: ,----[ <45af6ca0$0$97262$892e7fe2(a)authen.yellow.readfreenews.net> ] | 6. We now introduce the notion of the "union of a finite set of | trees" (sloppyly "union of trees") as: | | U V := T_1 U ... U T_n (n e N) | | Proof as an exercise that forall sets of finite trees V having card(V) | e N it holds that | | U V = T(max(D(V)). `---- But have asked for a definition of (inf-u). F. N. -- xyz
From: William Hughes on 21 Jan 2007 08:03 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > UT is not a tree. > > > > > > > > > > Call it as you like. Simply call it T1. (If I speak of "tree" blow, > > > simply read "eert".) It may be what you like. In any case it is the > > > same by nodes, edges > > > > Yes > > > > > and paths as T2. > > > > No. > > The directions of the paths are the same in T1 and T2. If you insist on > a difference, then it can only result from the length of the paths. Yes. T1 contains only finite paths. T2 contains only (potentially) infinite paths. > > > The paths in T1 are not determined by the nodes in T1. > > By its nodes and edges every path is determined. Yes, but the fact that these nodes and edges are in T1 does not mean the path determined by these nodes and edges in in T1. > > > The paths in T2 are determined by the nodes in T2. > > In fact? Why that? Because, T2 is the *tree* that contains the same nodes as the set of paths T1. T1 is not a tree (not every set of paths is a tree). > > > The paths > > in T1 are not the same as the paths in T2. > > > > T1 contains only finite paths. > > This is in fact the crucial question. If you are right, then I have > been working in vain for four years. If you are wrong, then thousands > of mathematician have been working in vain for 130 years. So let us > decide this question > > > T1 does not contain a single infinite > > path. The fact that there is an infinite path that can be contructed > > using the nodes from T1 is not relevant. The paths in T1 are > > not determined by the nodes in T1. > > What are the nodes and edges for, in your opinion? Given a set of nodes, you can create paths. However, the fact that a set of nodes M,is contained within a set of paths, S, does not mean that every path that can be made by the nodes M is in S. > Why do you think the EIT has an infinite diagonal? Because the diagonal of the EIT is definied as the (potentially) inifinite sequence consistting of the ends of the lines in the EIT. The diagonal of the EIT is not defined as the union of all the finite diagonals of finite initial seqments of the EIT. > I say: The union of indexes of paths {1} U {1,2} U {1,2,3} U ... is an > infinite index union {1,2,3,...} Correct and irrelevant. The union of indexes of paths is not the same thing as the union of paths. > corresponding to n infinite path. Incorrect. A path is not a set of indexes. A path contains a set of indexes. > If > you disagree, please prove your assertion. > Where is the end of one path of T1? > In the middle of the tree T2. T1 is not a tree. The paths in T1 do not have to "go to the bottom". Yes, given any path, P, in T1 you can find a node in T1 such that you can extend path P to path P1. This is irrelevant. There is nothing in the definition of T1 that says a path in T1 cannot be extended by a node in T1. > Please don't come up with answers like: "I don't know the end but I am > sure it exists." > A sequence is a function whose domain is either an initial segment of > natural numbers or all natural numbers N. A sequence whose domain is > some initial segment is called a finite sequence. > > In mathematics a sequence is proven finite if and only if there is a > natural number n_0 which not belonging to its domain. Otherwise a > sequence is infinite. > All irrelevant. A sequence is not its domain. If you have a set of sequences S, you can take the union of the domains, D, then construct T to be the set of sequences whose domain is contained in D. However, you cannot conclude that any sequence contained in T is contained in S. > > > > > > > An index denotes the level to which a node belongs. The union of all > > > indexes of nodes of finite paths is the union of all initial segments > > > of natural numbers > > > > {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... = {1, 2, 3, ...}. > > > > This is also the set of all last elements of the finite segments, i.e., > > > it is the set of all natural numbers N. This is the set of all indexes > > > - there is no one left out. With respect to this observation we > > > examine, for instance, all finite paths of the tree T1 which always > > > turn right: 0.1, 0.11, 0.111, ... If considered as sets of nodes, their > > > union is the infinite path representing the real number 0.111... = 1. > > > > No. Their union is a set of finite paths. > > The union of domains {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} > U ... is a domain {1, 2, 3, ...}. > Correct, but a domain is not a sequence. > > 0.111... is an > > (potentially) infinite path. 0.111... can be seen as a limit > > of a sequence of finite paths. It is not the union of a sequence > > of finite paths. Union and limit are not the same thing. > > (The union of {1/2},{3/4},{4/5} .... is not {1}). > > But the union of the intervals (0, n/(n+1)) has length 1. > No. A union of intervals is not an interval. The supremum of the intervals (0, n/(n+1)) has length 1. However, union and supremum are different things. > And the union of singletons {1} U {2} U {3} U... U {n} U ... is not an > infinite number omega? Yes it is. But as we are not discussing a union of singletons, this is irrelevant. > > Two trees which are identical by all nodes and edges are identical by > all paths. Correct, however, T1 is not a tree. > T1 as the countable union of all finite sets of finite paths contains > only a countable set of finite paths. Correct. > T1 = T2. ] No. T1 and T2 contain the same nodes, but T1 does not contain every path defined by its nodes. T1 and T2 do not contain the same paths. - William Hughes
From: Andy Smith on 21 Jan 2007 10:42 David Marcus writes >The less people know, the less they know what they know (and that's the >sane ones). > Personally I am with Rumsfield on that, I know what I know, and I know what I don't know but I know that there are things that I don't know that I don't know. -- Andy Smith
From: mueckenh on 21 Jan 2007 11:09
Virgil schrieb: > In article <1169332163.222236.89960(a)v45g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Andy Smith schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de writes > > > > > > > >Andy Smith schrieb: > > > > > > > >> I > > > >> > > > > The union of all finite binary trees contains all levels > > > >> > > > >which can be > > > >> > > > > enumerated by natural numbers: > > > >> > > > > > > > >> > > > > 0 0. > > > >> > > > > / \ > > > >> > > > > 1 0 1 > > > >> > > > > / \ / \ > > > >> > > > > 2 0 1 0 1 > > > >> > > > > ............................... > > > >> > > > > > > > >> > > > >> Out of interest, aren't the set of all numbers defined by the union of > > > >> all paths through a finite binary tree with N levels just all the > > > >> numbers addressed by the first N bits? If so, why do you bother with > > > >> the tree construction - does it have some special significance? > > > > > > > >The real numbers are represented as infinite paths in the "complete" > > > >infinite tree. Some even twice. > > > > > > > >The union of all finite trees is an infinite tree. > > > >Every finite tree contains only a finite set of paths. > > > >The countable union of all paths of the finite trees is therefore the > > > >countable union of all finite paths. > > > >The countable union of all finite paths is in the union of all finite > > > >trees. > > > >The "complete" tree containing all paths is identical to the union of > > > >al finite trees, with respect to nodes and edges. > > > >Identical trees cannot contain different sets of paths. > > > >Therefore, both trees contain the same set of paths. > > > >Therefore the "complete" set of all path is countable. > > > >Therefore the set of all real numbers is countable. > > > >Therefore ZFC is inconsistent. > > > > > > I would have said that the set of all paths in a finite tree of depth N > > > correspond 1:1 with the address range of N bits. > > > > You use N as a natural number? It is usual here to denote the set of > > all natural numbers by N, a natual number by n. > > > > > > An infinite tree corresponds to a number encoded in a countably infinite > > > set of bits. > > > > The nodes of the tree, yes. The edges of the tree too. The paths of the > > tree, no. > > An infinite tree corresponds fairly closely to the set of all real > numbers encodable in a countably infinite set of bits, except that some > numbers have dual encodings. So every number is represented by at least one path in the infinite tree? > > > > > > > Cantor's diagonalisation argument then applies. > > > > It does not apply to the paths of the tree with all nodes, because this > > tree contains the representations of all real numbers of the interval > > [0, 1]. > > That does not prevent it from applying, Of course, you may apply it, but in vain, because every real number is represented by at least one path in the infinite tree. > and, in fact, the > diagonalization argument applies perfectly to the set of all infinite > sequences of binary bits, which is where it was originally applied. So you can apply the argument *and* find a real number which is not represented by a path in the infinite tree? Regards, WM |