Cantor Confusion
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From: G. Frege on 23 Jan 2007 07:18 On Tue, 23 Jan 2007 12:43:38 +0100, G. Frege <nomail(a)invalid> wrote: >> >> ... with an actually infinite number of rows ... (Andy Smith) >> 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. F. -- E-mail: info<at>simple-line<dot>de
From: Andy Smith on 23 Jan 2007 07:33 G. Frege <nomail(a)invalid.?.invalid> writes >On Tue, 23 Jan 2007 10:45:22 GMT, Andy Smith ><Andy(a)phoenixsystems.co.uk> wrote: > >>> >>> 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. >> >See _what_? Do you understand the meaning of the word "not"? > >Are you a troll? (I hope not.) > No, just a curious nitwit. >> >> There is no greatest natural number. >> >Right. > >> >> So I can construct a finite ordered list of natural numbers, labeled >> 1 to n, and identify a number not in the list [labeled] n+1. >> >Right. > >> >> I can do that for all [any] n. >> >Right. > >In symbols: > > An e IN Em e IN m !e {1,...,n}. > >"For any (every) n in IN there is an element m (for example n+1) in IN >such that m is not in the "list" (here just the set {1,...,n}) from 1 >to n)." > >BUT we may n o t conclude from this fact that > >* Em e IN An e IN m !e {1,...,n}. > >"There is an m e IN such that for any (every) n in IN m is not in the >list from 1 to n." > >The inability to differentiate between this two statements is known as >"quantifier dyslexia". (It's a fact that most mathematical cranks >suffer from quantifier dyslexia, which makes it difficulty for them >(i.e. impossible) to comprehend correct mathematical arguments.) > Well I am not a Crank (I am not proposing that everyone is wrong and has failed to see the truth of some half-baked argument of mine). I wouldn't have proposed your second statement anyway. >> >> Cantor's hypothetical numbered list of the reals is also finite? >> >Huh? It's hard to see how a finite list might contain infinitely many >elements, to begin with. (Are you trolling?) > >> >> 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 >> [...] infinite list. >> >Right. Actually, there is a slightly more direct argument: We can show >(via Cantor's diagonal argument) that for _any_ (every) list of reals >there is at least one real not in the list. Hence there is no list >which contains _all_ real numbers. > >> >> But since you cannot have an actually infinite natural number, you >> cannot have an actually infinite list ... >> >Huh? Where did you get that nonsense from. (Did you read one of >M�ckenheim's papers? :-? > No, who he? >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 ? >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 set of bits {.. a_3 a_2 a_1} does not represent a natural number - you can consider an actually infinite set but no actually infinite member of it. A key distinction. So Cantor's argument doesn't rely on locating an actually infinite member of his list as some limit process n->oo, he just posits that an actually infinite list can exist and is complete, and then shows it can't be (or possibly alternatively that an actually infinite list can't exist?) OK, well I do see the argument better now, but if that was an argument that I had suggested for the first time, you would be on me like wolves .... Cheers -- Andy Smith
From: Franziska Neugebauer on 23 Jan 2007 07:35 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> Virgil wrote: >> >> So you obviously use a different notion of tree. > > He does not use any notion but the fact, as he believes, that the set > of all paths in a complete tree must be uncountable. The nodes of a > complete tree (complete concerning nodes and edges) are already > completely occupied by paths which are in the union of all finite > trees (or trees of type weeping willow). As Virgil has already pointed out: In the finite case every set of path (having certain properties) induces std-tree (M, E). Different sets of paths yield different std-trees. For infinite sets of paths this is no longer true. Let P* be the set of all finite paths then P* induces the whole tree I have named G elsewhere (the infinite binary tree). The set of all paths P^omega of that standard-tree G is not identical to P*. E.g. the path p(1/3) = (1, 2, 5, 10, (alternating-)...) = (2^0, 2^1 + 1, 2^2, 2^3 + 1, ...) is not a member of P* since P* contains only finite paths (or eventually left turning paths). So we note P* != P^omega. Or rephrased: ** The std-union of the set of sets of paths of every finite binary std-tree is not identical to the set of paths of the infinite binary std-tree G. ** ,----[ WM's wikipedia entry ] | Nach Aussage der transfiniten Mengenlehre besitzt der unendliche | bin�re Baum jedoch mehr Pfade als Knoten, also mehr Pfade als die | Vereinigung aller endlichen bin�ren B�ume. `---- My translation, broken into three claims: (WM-1) "According to transfinite set theory the infinite binary tree contains more paths than nodes.". When we interpret "more" as "cardinally greater than" this is true. (WM-2) "According to transfinite set theory the infinite binary tree contains more paths than the union of all finite binary trees." The German "also" means "therefore" hence: (WM-3) (WM-1) -> (WM-2) "union of trees" or "tree" cannot be interpreted without further choice of definition of "union of trees" or "tree". >> > and different sets of paths lead to the same sets of nodes and >> > edges. >> >> > Since it is sets of paths of a tree that WM has been going on >> > about, it seems more reasonable to consider those sets of paths >> > from the start. >> >> Then we need to use a revised definition of tree. > > We shouldn't. A tree is defined by its nodes and its type or by its > nodes and its edges, respectively. Then you must choose David's notion of tree-union. Using his definition your wiki-statement (WM-2) is wrong since the (non-std-)union of all finite binary std-trees *is* the infinite binary std-tree G. Since (WM-1) is true and (WM-2) is false your claim (WM-3) is also false. F. N. -- xyz
From: Andy Smith on 23 Jan 2007 07:38 G. Frege <nomail(a)invalid.?.invalid> writes >On Tue, 23 Jan 2007 12:43:38 +0100, G. Frege <nomail(a)invalid> wrote: > >>> >>> ... with an actually infinite number of rows ... (Andy Smith) >>> >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 e.g. a greatest infinite natural number, no infinity could exist as a "completed thing". I am trying to shed an Aristotolian (?) perspective ... -- Andy Smith
From: William Hughes on 23 Jan 2007 07:42
mueckenh(a)rz.fh-augsburg.de wrote: > 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. It does not matter what we call T1. It does matter what properties T1 has. Since T1, the union of all finite trees, is the union of all finite paths, and 0.11 is a finite path, T1 has the property that 0.11 is in T1. I guess you have a different definition for the union of all finite trees. Let R be a set of finite trees with the property that: there is a (fixed) tree t_D in R, such that: if s is in R, then s is a subtree of t_D Definition i: The union of all finite trees in R is the tree t_D. (Note that using this definition, the union of all finite trees in R is a finite tree.) Defintion i': The union of all finite trees in R is a set of paths, S, where S contains any path that is in a tree in R. (Note that i and i' are almost equivalent.) Now let W be the set of all finite trees. We know that there does not exist a (fixed) t_D in W such that every tree in W (that is every finite tree) is a subtree of t_D. So we cannot use definition i. Instead I use definition i'. By definition i', T1 contains every finite path. If you have a different definition for T1 would you care to share it? The problem is that you want a definiton under which -the set of paths in the union of all finite trees is the union of all finite paths -the set of paths in the union of all finite trees is the same as the set of paths in the infinite tree T2 However, you cannot have both. -William Hughes |