Cantor Confusion
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From: Carsten Schultz on 25 Jan 2007 10:12 mueckenh(a)rz.fh-augsburg.de schrieb: > > On 25 Jan., 13:39, "William Hughes" <wpihug...(a)hotmail.com> wrote: > >>> 1) If we identify path-lengths with natural numbers, then there is no >>> infinite path in any of the finite trees, because there is no infinite >>> natural number. Therefore, also the union of all finite trees cannot >>> contain an infinite path. > >> Clearly the tha path-length of an infinite path cannot be a >> natural number. > > The path length is the number of nodes of the path. As long as the > nodes are enumerated by natural numbers, the length (distance from the > root node) is a natural number. This is just great. Of course this is true for WM, because he does not believe in infinite sets anyway, so in particular every set of natural numbers is finite. Good luck for those who think they can convince him of anything using mathematical arguments. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: mueckenh on 25 Jan 2007 10:13 On 25 Jan., 15:34, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 24 Jan., 21:05, Franziska Neugebauer > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > [...] > >> Using Virgil's trees we have the result that the induced std-tree > >> (M(P), E(P)) may be the same if different sets of paths P are given. > > > Then it is not what I am talking about.So what _are_ you then talking about? > > > If T(oo) = T concerning the set of nodes and edges,Axiom of Extensionality: Two sets a and b are equal if > > A c (c e a <-> c e b) > > There is no room in equality for "concerning" this or that. Tell it Virgil. But you are in error. The set {1, 0, 1} is identical to the set {0, 1, 1}, concerning the elements. Concerning the order of elements, both sets differ. > > > then they are identical with respect to paths too. There is no further > > parameter to fix. If you do not accept this identity, then further > > discussion is not meaningful. >Have you scrapped your plan to show a contradiction in contemporary set > theory? Why? > > >> [...] > > >> >> Since the union of V* is the std-union we know that that U V* > >> >> contains only "finite" paths. > > >> > Correct.This is correct _only_ for Virgil's trees. For std-trees > >> > there the > >> std-union is not a tree and for David's tree-union the tree-union is > >> not the std-union and the tree-union of all finite trees is the > >> binary tree G.[indentation correction:] > > >> > Hence 0.[10] is _not_ in UV*. > >> Using Virgil's trees this is correct. > >> Using David's tree-union this is not correct. > > > I do not intend to talk about this or that tree which may veil the > > facts.If you still want to show some contradiction in contemporary set theory > you have to commit to an appropriate formalization of what you > consider a tree. So I did. > > > In my proof there are two sorts of trees, solely defined by > > nodes and edges. Let the other trees grow, exist, contain, or whatever > > they like, when and where they can, if they can. > >> > It is valid for all existing natural numbers. Counting over to the > >> > infinite is nonsense.[indentation corrected] Counting occurs in the > >> > finite. > > > What you do _is_ "counting over to the infinite" > > > By no means! Every n in N is finite. >The set of finite trees V* = { T(n) | n e N } is not a finite set. > Induction over its members does not prove a property of V*. Then it does not consist of only finite levels. > This is quite the same as in the case of the infinite set > N = { n | n e N }. Induction over n for some property p(n) does not > prove p(N). Maybe you forgot that. No. All I require is that a property holds (or not) for all the natural numbers or all the levels or nodes or edges of trees which can be enumerated by natural numbers. > > >> when you claim that > >> induction proves existence of U V* where in fact is does only prove > >> existence of UV(n) A n e N. I am fully satisfied with the latter. I call it U T(n) for A n e N = T(oo). Nothings remains to be desired. Can you spot a difference to your "UV*"? Perhaps not a node but a path? Let it be. I am not interested in fairy tales. > > > Yes. > > >> V* is not a finite set, i.e. a set of the > >> form V(n). Hence induction does not "reach" V*. > > > Induction reaches and includes every tree which exists. > What *exists* in ZFC is defined by the axioms and the rules of > inference. If you still want to show a contradiction _in_ ZFC you must > adhere to that axioms and rules. Only if you find a difference between T(oo) and T. > > In ZFC/graph theory the tree T(oo) = G (the infinite binary tree) exists > but is only "reached" when using the non-std-union of David. When using > the std-union of Virgil, the _induced_ tree (M(P), E(P)) is "reached" > but the Virgil-tree (i.e. the union of all finite _paths_) is not > identical to the paths induces by the infinite binary tree T(oo)/G. I think this gives a delicious impression of the present state of mathematics. > > I must admit this is probably not simple enough at applied university > level but nonetheless valid. Better fitting to scientology or matheology? > > > As I said above: If you conclude from one and the same tree that there > > are different paths, > There exist different sets of paths (Virgil's trees) which *induce* > the same standard-tree (M(P), E(P)). There exist different systems of paths which may induce the same tree. A a simple finite example: 0.1 and 0.11 and 0.111 are paths which unioned result in the path 0.111. Of course you can also get to this same result when starting from the paths 0.1 and 0.111 or from the paths 0.11 and 0.111. That is clear. But once the union tree T(oo) is constructed, the set of path contained in it is uniquely defined. You always express very carefully that Virgls approach "leads to the same tree". Please state explicitly wether you are willing to assert that this same tree, T(oo), after construction, does contain different sets of paths, induced by the mental constitution of the observer? Regards, WM
From: William Hughes on 25 Jan 2007 10:19 On Jan 25, 9:45 am, mueck...(a)rz.fh-augsburg.de wrote: > On 25 Jan., 13:39, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > 1) If we identify path-lengths with natural numbers, then there is no > > > infinite path in any of the finite trees, because there is no infinite > > > natural number. Therefore, also the union of all finite trees cannot > > > contain an infinite path. > > Clearly the tha path-length of an infinite path cannot be a > > natural number.The path length is the number of nodes of the path. As long as the > nodes are enumerated by natural numbers, the length (distance from the > root node) is a natural number. > > > We are left with > > > > 2) If we identify path-length with sets of natural numbers, then there > > > are infinite paths in the union tree T(oo). > > >However, if L is a set of > > > limits L_k > > > L = { L_k | k in N } > > L is the set of finite limits. > > L is countable. Note that the set L does not > > contain an infinite limit.Every limit is an entity. > > > > > > and S is the set of corresponding sequences > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > then L cannot be larger than S, because there cannot be more limits > > > than sequences. > > Note that S contains only infinite sequences.Every sequence is in the union T(oo). > > > > > S can be and is larger than L. S is not countable. > That again is an unjustified and provably wrong assertion. > The sequences belong to the countable union of the finite trees. Take any sequence s in S. We know that each initial segment of a s belongs to a finite tree t(s). However there is no single finite tree t_D such that every initial segment of s belongs to t_D. s does not correspond to a single finite tree, s corresponds to a sequence of finite trees. > *Everything* in this union is countable. Yes, but s is a sequence of things from the union. The set of infinite sequences from a countable set is not countable. - William Hughes
From: mueckenh on 25 Jan 2007 10:50 Dik T. Winter schrieb: > > The tree is a collection of sets, namely the collection of levels L(n) > > as well as the collection of subtrees like T(n) = L(0) U L(1) U ... U > > L(n). The tree is also a set. Its elements are nodes. A path is a > > subset with the special property that it contains all nodes which go > > from top to the end or, in case there is no end, continue to go on. > > So now back to the largest paths again. In that case the union of the > sets of paths in two different trees is not the set of paths in the > union tree. If tree T1 contains the path {a, b, c} and T2 contains > the paths {a, b, c, d}, the union of T1 and T2 contains the path > {a, b, c, d} but the union of the set of paths in T1 and the set of > paths in T2 contains both {a, b, c} and {a, b, c, d}. Correct. But this union does not enter my proof other than by the theorem: The set of paths contained in T2 is but a subset of the *countable* union of the sets of paths in T1 and T2. That is important. But I do not define the union of trees by paths. > > > > > 3) The union of all finite trees includes the union of all nodes and, > > > > with it, the union of all such subsets which are paths (because every > > > > path is a well defined subset of the set of nodes if the structure of > > > > the tree is well defined). > > > > > > If a tree consists (amongst others) of the set of nodes in it, the union > > > of two trees indeed consists (amonst others) of the union of the sets of > > > nodes. > > > > I agree. But this union may not be intermingled with pair-axiom. The > > union of two or more paths is a set of nodes which can but need not be > > a path. > > But you are talking about the union of *sets* of paths. I do not use the union of sets of paths but the union of nodes. > > > In the tree T(1) > > a > > /\ > > b c > > There are two paths, namely p1 = {a, b} and p2 = {a, c}. By pair axiom > > we get a set {p1, p2} containing two paths as elements. The union p1 U > > p2 = {a, b, c}, however, is not a path. > > > > The tree T(0) = a contains only one path p(0) = a. The union of p(0) > > and p(1) is a path, namely p(1). > > But note that the union of the sets of paths in T(0) and T(1) is the set > of paths: {p0, p1, p2}. So the set of paths in a union is *not* the > union of the sets of paths. Correct! The set of paths in the union is a subset of the union of all sets of paths. But I do not use this latter union for my proof. > > > > Yes. But due to the schism "infinite set of finite numbers" there are > > two answers possible (but not more). > > 1) If we identify path-lengths with natural numbers, then there is no > > infinite path in any of the finite trees, because there is no infinite > > natural number. Therefore, also the union of all finite trees cannot > > contain an infinite path. Correct. Nevertheless there is no natural number which is larger than any path-length. > > 2) If we identify path-length with sets of natural numbers, then there > > are infinite paths in the union tree T(oo). However, if L is a set of > > (1) if you allow infinite paths (as you do) you can *not* identify > path-lengths with natural numbers. A path-length which has no upper bound may be sufficent to be called infinite. > (2) you can identify path-lengths with *ordinal* numbers (this also > allows for the zero-length path. However in some models the ordinal > numbers up to, and including, omega can be represented as sets of > natural numbers. So that is the way to go. First explain why path-length which has no upper bound is not to be called infinite. If you find a reasonable argument, then I will consider your suggestion. > > > However, if L is a set of > > limits L_k > > L = { L_k | k in N } > > and S is the set of corresponding sequences > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > This can not be S because you define S as a set of digits, not of sequences. You are in error. a_1k, a_2k, a_3k, ... is an element of S. Observe the symbol { | } above. > Properly done I get somethink like: > S_k = { a_1k, a_2k, a_3k, ... } with a in {0, 1} and k in N > S = { S_k | k in N } That would be a set of at most two bits. We require sequences a_1k, a_2k, a_3k, ... > > > then L cannot be larger than S, because there cannot be more limits > > than sequences. > > Yes. What is the relevance? If you have countably many sequences, you > have countably many limits. But you are thinking that you have > countably many sequences, and so have to prove that. As S above is > precisely the list from Cantor's proof, we know that S does not contain > *all* possible sequences. S above is in fact a list of sequences. The enumeration by N is justified by the fact that S is a subset of a countable set. T(oo) and all its constituents are at most countable. > > > Different limits for one sequence would mean a path > > splitting at a level n beyond any natural number, i.e., n not in N. > > There are no different limits for one sequence. Yes. Therefore the cardinality |R| of the limits R cannot surpass the cardinality |S| = |N| of the sequences S. > > > > But this is also wrong. > > > > No. When unioning the trees we union also their subsets, including > > their paths. > > What do you mean with this? The answer is probably "yes". But this > does *still* not make the union of the *sets* of paths the *set* of paths > of the union. Of course it does not do that. The *set* of paths of the union is a subset of the union of all *sets* of paths. The later, hoever, is countable. > > > A subset of a tree is a path if and only if all its elements (nodes) > > are connected by edges. > > The union of two paths is a path if and only if all its nodes are > > connected by edges. (See the examples above.) > > Yes, but this is all irrelevant for the *set* of paths. > > > The set of paths in T(oo) consists of unions of paths. > > Ah, perhaps yes, but badly worded. > > > It cannot > > contain more paths than were unioned, i.e.,countably many. > > But this is, eh, also correct. Still that makes it *not* the union > of countably many *sets* of paths. Bu it shows that the *set* of paths of the union is a proper subset of the union of all *sets* of paths. > > > the union of two or more paths are nodes. > > Eh? The union of two or more paths is a set of nodes. I was a bit slopy here. A set of nodes is correct. But it need not be a path. > > > > > 5) A countable union of countable sets is a countable set (according to > > > > ZF with AC). > > > > ==> The set of all path is countable. (==> The real numbers are > > > > countable.) > > > > > > The statement is correct, but what you state does follow, does not follow. > > > The set of all paths is *not* the countable union of countable sets of > > > paths. > > > > The set of all paths in T(oo) is not the union but only a subset of the > > countable union of countable sets of paths. > > No, you have not proven that. You use in the union the *sets* of paths in > the finite trees. As none of those sets contains an infinite path their > union also does not contain an infinite path. That is what I proved. In the union of the *sets* of paths in the finite trees there is no infinite path. Hence the subset which is in T(oo) also does not contain an infinite path. > > > > > 6) T(oo) = T contains only finite paths. > > > > 7) T(oo) = T contains all paths including all infinite paths. > > > > > > By your definition (paths are specific subsets of nodes), T contains > > > infinite paths. Conclusion (6) is false. On the other hand, the > > > union of the sets of paths contains only finite paths. And they are > > > countable indeed. > > > > If T(oo) = T concerning the set of nodes and edges, then they are > > identical with respect to paths to. > > I do not contradict that. T(oo) contains only finite path. > > Eh? You stated: > > > > 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. Yes. It is a subset. But I do not *define* the union of trees by their paths! > > Which is in itself nonsense. Which according to set theory is correct. > But if we elide "of finite sets" it makes sense, No. Below you talk about "finite trees". This implis finite paths and finite sets. > but is wrong. Also if we change "countable set" to "countable union" it makes > sense, but still is wrong. Note that: > the countable set of all paths in the finite trees > is identical to > the countable union of finite sets of all paths in the finite trees. Also this union is countable. > So both changes make it a sentence that can be interpreted, but is still > wrong. And in both cases you are actually talking about the union of > sets of paths. Of course I consider paths, because I wish to consider real numbers, but I do not *define* the union of trees by their paths! Regards, WM
From: mueckenh on 25 Jan 2007 10:56
On 25 Jan., 15:57, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: WM ] > | The union of two natural numbers is defined to be the larger one. > | This is a set theoretic union. > `---- > > So the union of two natural numbers is _not_ defined to be the larger > one but the "larger" of two distint naturals (sets) a and b is defined > to be a iff > > b c a > > or b iff > > b c a. > > So the order _is_ relevant to successfully prove you wrong. The union of two different natural numbers is defined to be the larger one. Two alternatives would be: The union of two different natural numbers is defined to be the smaller one. The union of two different natural numbers is undefined. Can you find more? Regards, WM |