Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: G. Frege on 23 Jan 2007 06:43 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.) > > 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.) > > 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? :-? All natural numbers are "finite", but (still) there are infinitely many of them (in the set IN). 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, ...}. F. -- E-mail: info<at>simple-line<dot>de
From: Franziska Neugebauer on 23 Jan 2007 06:50 mueckenh(a)rz.fh-augsburg.de wrote: > 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. 1. You are in error about what is defined first in ZFC. The union of two sets a and b is defined a U b = { x | x e a v x e b } first. Thereafter for two distinct ordinals a and b "larger" is simply defined as a < b := a c b 2. Only in Virgil's (trees are not std-trees) and David's definition (union of tree is not std-union) the (tree-)union of all finite binary trees is defined due to the axiom of 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. If you assume that a max(N) exists you leave contemporary set theory. Any "contradiction" is then at the expense of yours. >> > 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. You have not defined UV* (the union of actually _all_ finite trees). > This construction does not come to an end. In ZFC there is no time and no processes. You must be writing on something entirely different. Have you scrapped your plan to show a contradiction in contemporary set theory? F. N. -- xyz
From: mueckenh on 23 Jan 2007 06:52 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). > > > 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. My cut tree (and also the weeping willow tree) is already completely defined by the number of levels, i.e., it is completely defined by a single natural number n. Regards, WM
From: mueckenh on 23 Jan 2007 07:01 Andy Smith schrieb: > 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). You are correct. Unless there is an infinite number the number of numbers, with difference 1, cannot be infinite. But there cannot be an infinite natural (= finite) number. It is very simple to see that the set {2,4,6,...2n} always, i.e., for every n, contains larger number than its cardinal number n. It is impossible that the "whole" set of even numbers has a cardinal number which is larger than every even number. There is no actual infinity. The set is potentially infinite, having no cardinal number. Regards, WM
From: G. Frege on 23 Jan 2007 07:10
On Tue, 23 Jan 2007 12:57:46 +0100, G. Frege <nomail(a)invalid> wrote: Typos... > > I guess it expresses some sort of disgust. > ~ ~ > F. -- E-mail: info<at>simple-line<dot>de |