Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Franziska Neugebauer on 27 Jan 2007 16:52 mueckenh(a)rz.fh-augsburg.de wrote: > I am interested in the fact that every set of natural numbers has a > finite maximum. Then you should perhaps not talk to contemporary set theorists who are accustomed to the fact that there *are* sets of natural numbers which do not have maxima at all. F. N. -- xyz
From: mueckenh on 27 Jan 2007 16:54 On 27 Jan., 16:52, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: Please define path: We need only path-lengths n. > > > > for instance the path p on the outmost left hand side of the tree.In the sequence-of-nodes picture this is the path > > p = { (0, 2^i) | i e omega } Ok, but it is not necessary to consider the whole tree and to count all its nodes and to sort out those belonging to the path. > The length of p_n is a number which has numbers (0 path-lengths) *as elements*. In the union of numbers, the numbers are elements. Every natural number is a set the elements of which are all smaller natural numbers. n = {0,1,2,...,n-1} The lengths of the paths in the union of path-lengths are represented by natural numbers. > We call p_n n e omega the finite left-path of length n + 1: Better say of length n: p_n := { (0, 2^0), ..., (n-1, 2^(n-1)) } , but this is not important. > > p_n := { (0, 2^0), ..., (n, 2^n) } > > Let P_l be the set of all finite left-paths: The length of p_n is a number which as elements has numbers. > > P_l := { p_n | n e omega } > > This set only exists if omega exists. Proof by definition? P_I includes all p_n which exist. The existence is guaranteded by the Peano axioms. We must not assume something which is under examination like omega. > (If we assumed omega not to exist > P_l and the set of all finite left-paths would not provably exist). It is sufficient that the path with length n implies the existence of the path with length n+1. > > Since we assume that omega exists (we do so in contemporary set theory) which is just under investigation. Therefore we do not do so here. > P_l exists. We are interested in the length of the path which is the union of the lengths of all finite paths. > According to the axiom of union even the union U P_l > exists. Maybe, but that is not at all an interesing question in this context. The interesting question is alone whether the union of finite lengths can be an infinite length. > As one can easily show it is true that > > U P_l = p We are not interesed in p but in the length n of p which length is a union of some finite lengths. > > > Therefore all the path-*lengths* in the union are natural numbers.1. The union of paths is itself is not "many paths" but (at least) is > single path. *In* the union there is no single paths as member. In the union of lengths there are all finite lengths. Is this infinite union of finite lenths a finite length? That is the question. > The > union is equal to the path p. The length of path p is (by definition) > not a natural number. How can a union of finite numbers yield an infinite number? This is by definition impossible. > Its length is card(p) = card(omega) +_card 1 > = card(omega). > > 2. P_l contains paths each of which has finite length (depth). Namely > depth(p_n) = card(p_n) + 1 = n + 1 n e omega. First you have to answer the question above before your definition can be accepted. > > True. U P_l = p has infinite cardinality. Maybe or not, but this is uninteresting. The union of pathlengths contains every single pathlength as a member. > > > then at least one of the paths in the union must be infinite.The union does not contain any single path as member (element!). > P_l does contain only finite paths p_n n e omega as members. So it cannot include an infinite path? > > > Is this so?You proved that silly questions do really exist. > > 1. Define "in the union": > a) there is no member (element) in U P_l = p which is a path Uninteresting. The *length* of P_I is in question. > b) there are subsets of U P_l which are paths. Namely > every p_n n e omega and p itself are subsets of U P_l. > (p is no a proper subset though). > > 2. You probably mean be a member of P_l (the set of all finite > left-paths). Here indeed every p_n n e omega is element of P_l. > But p is not an element of P_l, since P_l only contains finite > paths. The length of P_I is asked. > > Probably you confuse P_l and U P_l. No, I strictly mean the length of P_I, the union of all finite paths. I know that you think and define and mean to have proved that U P_I was infinite. That is boring. > > All in all you presented today next version of your 2003's theme > > X is not finite -> there must be an x in X which is infinite > > cast into paths. What comes next? Correct items will not change. The only improvement can concern better comprehensibility by you. Regards, WM
From: William Hughes on 27 Jan 2007 17:12 On Jan 27, 4:43 pm, mueck...(a)rz.fh-augsburg.de wrote: > On 27 Jan., 17:41, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Jan 27, 8:33 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 25 Jan., 23:33, Virgil <vir...(a)comcast.net> wrote: > > > > Then you best conclusion is that it holds for all natural numbers but > > > > not that it holds for N. > > >Precisely that is what I want and what I do! > > No proving that something holds for every natural > > number is not the same as showing that it holds > > for N. > > > Consider > > > Every set L_n = {0,1,2,3,....,n} has the property > > that the maximum of L_n can never refer to more that one natural > > number. > > > The maximum of the set N can refer to more than > > one natural number. > I do not want to prove something about a maximum of the set N. I am > interested in the fact that every set of natural numbers has a finite > maximum. > You claim that if something is true for every set L_n, then it is true for N. We know something about the maximum that is true for every set L_n. So you do want to prove something about the maximum of the set N. - William Hughes
From: Franziska Neugebauer on 27 Jan 2007 17:22 mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Jan., 16:52, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: [...] > How can a union of finite numbers yield an infinite number? This is by > definition impossible. The following applies: >> All in all you presented today next version of your 2003's theme >> >> X is not finite -> there must be an x in X which is infinite >> >> cast into paths. What comes next? > > Correct items will not change. Quantifier dyslexia? F. N. -- xyz
From: Andy Smith on 27 Jan 2007 17:26
Randy Poe <poespam-trap(a)yahoo.com> writes > > > (snip) >That's not an argument. It's a statement about why several of >your intuitions combine in a certain way to lead you to an >intuitive conclusion. I don't know what "uncountable" means >to you, or why "necessarily infinite binary expansion" implies >whatever that means to you. > >I do know that the set of infinite binary strings is uncountable, >but I know that because it is easily proven. > Well, at the high risk of ridicule and a rather lower probability of assassination by irate set theorists, I think that Cantor's diagonalisation argument has a terminal flaw. The flaw is that Cantor's argument, disregarding any qualms about the infinite sets involved, just demonstrates that one cannot construct an infinite list of all permutations of binary strings. this is not the same as showing that you cannot construct an infinite list of all reals, because, as has been pointed out, reals may have more than one representation in any base. Please consider the following, and then abuse me; I hope that this will cause intense discomfort, but I doubt that it will. Consider first Cantor's diagonalisation argument applied to the set of reals in [0,1), i.e. excluding the point 1.00.... Assume that we can form a hypothetically infinite list of the reals. The countability or otherwise is not a finction of arrangement, so consider the list set out in the following systematic order, defined by the mirror inverse of the indices to the hypothetical list: Index no. Index in binary Corresponding real number 0 0 .0000... 1 1 .1000.. 2 10 .0100.. 3 11 .1100.. 4 100 .0010.. etc. This arrangement systematically covers all the reals (subject to our assumption) - it is just the mirror inverse of counting from 0 upwards in binary. The maximum number of bits required to describe any index no. i is ceiling(log(i+1,base 2)). For all i>0, i+1> ceiling(log(i+1,base 2)). And for i = 0, the real number is 0.000 . So, all terms on the diagonal are 0 (and, more of this later, the distance in bits between the 0 on the diagonal and the first non zero bit goes as (n - log(n,base2)) for bit n. Following Cantor, we construct an antidiagonal number, different from the first row, the second row, etc. Because all the diagonal terms are 0, we construct the number .1111.... This string is certainly not in the hypothetically complete list, but it is a real number and = 1.000... , which was excluded anyway from our initial list. So we can make no inference that the list is incomplete/invalid - we have just generated a number. If we try to rectify that by considering the interval [0,1], on the systematic numbering scheme above we would insert 1.00.. as the second entry in our list.: Index no. Corresponding real number 0 0.0000.. 1 1.0000... 2 0.1000.. 2 0.0100.. 3 0.1100.. 4 0.0010.. The antidiagonal for this would be 1.11111... aka 10.000... which is another valid real which wasn't on our original list. It is worse than that. If we consider the open interval (0,1), our systematic ordering is: Index no. Index in binary Corresponding real number 1 1 .1000.. 2 10 .0100.. 3 11 .1100.. 4 100 .0010.. Our Cantor antidiagonal is then .00111111... aka .010..... which is the second element of our list. In general, with this arrangement of the reals, the diagonal must terminate in in a series of 00000000..., the antidiagonal as a series of 11111111... and the corresponding real either exist already in the list or outside of it. With this arrangement, Cantor's diagonalisation argument does not contradict a hypothesis that one can construct a list of the reals. that does not mean that it is possible to count the reals, it just invalidates the argument. Conceivably one might be able to define an arrangement of reals such that the antidiagonal was unambiguous in defining a real that should be in the list (but can't be becaue by construction it is different from those in the list). But if the argument above is valid, then I think Cantor's diagonalisation is dead and buried. That doesn't mean that the reals are countable - far from it. On the hypothesis that the rals are countable, we ahave a situation where, on our list, for all entries n in the hypothetical list n e N, the number of zeroes between the last non zero term and the diagonal bit increases monotonically as log(n)-loglog(n). So there can never be any non-zero bits crossing the diagonal. But what about e.g. sqrt(2) ? Any real in the set with an infinite binary expansion will intersect the diagonal. You can't construct a numbered list of the reals. Abuse/explanations/condescension please... -- Andy Smith |