Cantor Confusion
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From: G. Frege on 23 Jan 2007 07:51 On Tue, 23 Jan 2007 12:33:06 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >> >> All natural numbers are "finite", but (still) there are infinitely >> many of them (in the set IN). >> > Yes = simple lay definition of infinity, infinity meaning without end ? > I guess that's ok. "Without end" is a good picture (I think), especially concerning the (set of) natural numbers. >> >> Try to get that straight: Even though IN is a infinite set, each and >> any natural number in IN is finite. (The latter claim makes sense when >> we construct the natural numbers as certain finite sets.) >> >> So if we consider a sequence of real numbers (i.e. your "list"), there >> may be infinitely many of them in the "list" considered (of course). >> >> a_1 >> a_2 >> a_3 >> : >> >> where a_i e IR for any i e {1, 2, 3, ...}. >> > whereas an infinite [sequence] of bits (... a_3 a_2 a_1) > does not represent a natural number. > Bits? Well, let's consider digits. If you consider in infinite string of (non zero) digits, this string does not represent a natural number. With other word's, the infinite string (sequence of digits) ...321319879482374 doesn't represent a natural number. BUT the infinite string (sequence of digits) 0.473284978913121... represents a real number. You see, there's a certain difference here. (Indeed, that's the difference which makes the difference. :-) > > - you can consider an actually infinite set but no > actually infinite member of it. A key distinction. > Sorry, can't follow you here. But it seems to me that you are going astray, again. :-) > > So Cantor's argument doesn't (bla bla) > I was right. :-) F. -- E-mail: info<at>simple-line<dot>de
From: Dave Seaman on 23 Jan 2007 07:56 On Tue, 23 Jan 2007 10:45:22 GMT, Andy Smith wrote: > 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 ? Definitely not. The list is countably infinite. To be precise, Cantor's list is a mapping f: N -> R. The whole point is to show that no such f is a surjection, thus implying that R is uncountable. If you replace N by some finite set, then the most you can hope to prove is that R is not finite. We can do better than that. > 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. Although Cantor's proof is sometimes presented as an indirect proof (assume f is a surjection and then derive a contradiction), there is no need to do it that way. All we need is to assume f: N -> R is an arbitrary mapping, and then show it is not a surjection. That is a direct proof, not an indirect one. > 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). N is an infinite set, but each member of N is finite. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: G. Frege on 23 Jan 2007 07:56 On Tue, 23 Jan 2007 12:38:04 GMT, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >> >> Btw, in the context of _this_ discussion you should omit the word >> "actually". Either a set (a sequence, a list) is infinite or not. >> We don't use the attribute "actual" (concerning infinite sets) in >> modern set theory. >> > OK. It is a reference to Aristotle et all, whose view of infinity was > that there was a "potential" infinity - e.g. the natural numbers, but no > "actual" infinity [...], no infinity could exist as a "completed thing". > Right. I know that. But see my comment from above. F. -- E-mail: info<at>simple-line<dot>de
From: William Hughes on 23 Jan 2007 08:32 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > 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. > > > > > > Which path is finite? Where does a path end? > > > > Every path in T1 ends at a node. > > Every path in a finite tree ends at a node. T1 is infinite and has no > finite paths. > > > The fact > > that you can continue a path does not mean that you must continue > > a path. > > > > Whenever there is an edge leading a path, this path is there. Yes, there are two paths. The path that stops and the path that goes on. - William Hughes
From: Dik T. Winter on 23 Jan 2007 08:25
In article <1169547646.733664.94830(a)m58g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > If the union of singletons {1} U {2} U {3} U... U {n} U ... is an > > > infinite number omega, then the union of domains of sequences {1} U {1, > > > 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an > > > infinite seqeunce {1, 2, 3, ...}. > > > > Pray explain what you understand under "domain of", and what you understand > > under "union of domains". That is complete non-standard terminology. > > The domain of a sequence is a natural number. The domain of an infinite > sequence is omega. That is not an explication. That is obfuscation. That is, it is still clear as mud. > > But > > whatever, whenever in a collection of sets A_k, none of the sets A_k > > contains a particular element E, it is also not in the union. > > Therefore there is no uncuntable set of path in the union tree. But > there are all paths in the union trees. Conclusion is false. > > > How could a path be in the union of finite trees if it were not a path > > > of a finite tree? > > > > Ah, that is the key question. Clearly 0.010101... is not a path in any > > of the finite trees. But it is because by one of your definitions of path > > and tree such a path does exist. You never stop to rigurously define what > > a tree and a path is. I have a few times attempted to define a tree, but > > each time you skipped my definition and resorted to intuitive, bad-defined > > concepts. Once you stated that a tree was a set of nodes. That is > > obviously false. > > No, it is correct, provided the edges are determined by the > construction of the tree. So it is not only a set of nodes. > I use only this special type of trees (i.e., > every node has two child nodes). So the tree is actually defined by a set of nodes plus a set of directed edges with some particular properties, and so is not a set of nodes. > > A set of nodes is nothing more than a set of nodes > > until you introduce edges. When you introduce edges you get a graph. > > But, equally obviously, not every graph is a tree. A tree needs a > > set of nodes and a set of directed edges (in a particular fashion). > > I did grant you all that and gave (I think) a rigorous definition of a > > tree, and even how to produce unions of trees. > > It is clear how to produce the union of the finite trees to all who > discuss here. Virgil, William, and you agree that this union contains > all nodes and all edges and all levels. Franzisca also knows and > undertands that but refuses to admit it, because she also knows the > consequences. The consequences are not what you think they are. > > > Two trees which are identical by all nodes and edges are identical by > > > all paths. > > > > Right. > > > > > T1 as the countable union of all finite sets of finite paths contains > > > only a countable set of finite paths. > > > > No. It *is* the countable set of finite paths. > > Right. It is the union of all nodes (and edges and levels and paths). Wrong. If you state that T1 is the countable union of all finite sets of finite paths, it is a set of paths, and so does not have nodes, edges or levels as element. > > > And T1 = T2. > > > > No. T1 is a set of paths, it is not a tree. T1 does not contains nodes > > or edges as elements. Only paths. > > The nodes can be enumerated in various ways. That does not matter. T1 as you defined above is a set of paths. *Not* a tree. And a set of paths does not have nodes or edges as elements, it has paths as elements. And if you consider a path as a set of node, then T1 *still* does not have nodes as elements but sets of nodes. > > > Here it is (for trees of the kind weeping willow): Taking the union of > > > two trees corresponds to taking the union of their sets of paths. > > > > Why? The union of the sets of paths is a set of paths, not a tree. > > The union of trees is a union of their nodes. The paths are merely > subsets which are defined by the nodes and the special kind of tree. But you have to be careful when uniting sets of paths, that is uniting sets of sets of nodes. That union contains all paths from both tree, you can only omit duplicates from that union, so only paths that are *identical*, i.e. consist of the same set of nodes. Suppose {a, b, c} is a path in tree T1, and {a, b, c, d, e} is a path in tree T2, than the union of the sets of paths of T1 and T2 contains *both* paths. > > > This holds if it holds for domains of finite seqeuences. The domain of > > > two finite sequences is the domain of the longer sequence. The domain > > > of all finite sequences is omega. > > > > But using this kind of defining and uniting trees, the complete tree will > > not have 0.010101... as a path. Do you see how it crucially depends on > > how you define things? If you define trees by nodes and edges, and paths > > as a consequence of that, you get a different result. (And I still wonder > > what you mean with "domain of".) > > The domain of the path 1,0,1 is 1,2,3. Still not clear. You can not define things by example. > A path is a sequence. A sequence > is defined as a mapping f from a natural number or omega into the > reals. The set of reals (here only 0 and 1) is called the range. > Abbreviations are usual: dom f and ran f. Finally now you define a the domain of a sequence. So it is now also clear that a path is an ordered set of nodes. And the domain is the initial subset of the natural numbers that terminates at the cardinality of the path (as set of nodes). Still if T1 contains the path {a, b, c} and T2 contains the path {a, b, c, d, e}, the union of the sets of paths contains both. > > > The tree is an (ordered) set of nodes. The nodes can be used to form > > > paths, i.e., subsets of the tree. > > > > As a tree is a set of nodes, any subset of it is a set of nodes. > > Odf course, therefore the subsets which are paths must exist in the > tree if the subsets exist there. Ok, if a path is defined as a set of nodes. But still see above about the union of sets of paths. > >Not > > every subset of it is a path. But whatever, the number of subsets of > > the final tree is not countable. The union of the sets of paths of > > the finite trees is not the set of paths of the complete tree. > > If all nodes and edges are in the union, then all subsets are in the > union too. You misread again. The set of subsets of the complete tree is *not* equal to the union of the set of subsets of the finite trees. The reason is clear: that union does not contain an infinite subset, while the complete tree *does* contain infinite subsets. I am talking about the union of *sets* of paths, which are *sets* of subsets. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |