Cantor Confusion
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From: William Hughes on 29 Jan 2007 13:50 On Jan 29, 10:40 am, mueck...(a)rz.fh-augsburg.de wrote: > On 28 Jan., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > On Jan 28, 7:21 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > You claim that if something is true for every set L_n, > > > > then it is true for N > > > I do not talk about N. This symbol has become the aim of heaviest > > > abuse.You talk about the union of all natural numbers. N is > > the union of all natural numbers.Something that is valid for each n in N need not be valid for N. I am > interested in properties which are valid for every n. For instance, > every n is finite. Every segment {1,2,3,...,n} is finite. Every path > being he union of paths (segments) is finite. > > > > > > > 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. > > > I want to see whether the union of all finite numbers can be an > > > infinite number. > > A union of numbers is a set of numbers. It is > > also a number. The union of all finite numbers is an > > infinite number.No. > > > > > > This question was raised in the framework of the > > > infinite tree. Set theorists asserted that a union of finite paths > > > cannot be / contain any infinite path. > > A union of finite paths is a set of finite paths. A set > > of finite paths is not a path.A union of two sets of nodes is a set of nodes. > A union of two paths need not be a path, but can be a path. > > > A set of finite paths > > does not contain an infinite path.That is the problem. > The union of all paths having only nodes with value 0, is a path. No it is a set of paths. In this case it is a set of paths that contains only one rather boring path. Now a set that contains just one element and that element are closesly related, so sometimes we treat them as the same thing. However, it is very important to note that they are not in fact that same thing. A union of paths can never be a path. >Is > it infinite? Potentially infinite, yes. Actually infinite, no. > > > > > Do not confuse paths and nodes, they are > > not the same thing. > > > Each path p has a set of nodes N(p). > > The union of finite paths p1 and p2 contains the set > > of nodes (union of N(p1) and N(p2)). > > (note that both N(p1) and N(p2) are finite). > > The union of all finite paths contains a set > > of nodes that is the union of finite sets. > > We know that the union of finite sets > > can be a (potentially) infinite set. The union > > of all finite paths contains a (potentially) > > infinite set of nodes. > Yes. But the same holds for path-lengths. All finite path-lengths are > natural numbers. Yes, and the number of nodes in a path is a path length. However, the fact that a set of nodes is in a set of paths does not mean that the path corresponding to the set of nodes is in the set of paths (Do not confuse paths and trees. The union of all finite paths, P1, is not the same as the union of all finite trees,T1) > This is your last sentence: The union of all natural numbers contains > an infinite natural number. No. The last sentence says that a set of paths, P1, contains an infinite number of nodes. However, P1 does not contain one path which contains all nodes. So the fact that P1 contains an (potentailly) infinite number of nodes does not imply that there is a path in P1 which contains an (potentially) infinite number of nodes. So it does not imply that there is a path in P1 which has an inifinite path length. The union of all finite paths (finite natural numbers) does not contain an infinite path (infinite natural number). - William Hughes
From: MoeBlee on 29 Jan 2007 13:51 On Jan 27, 2:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > 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. You're confused. And you don't understand the proof. If you understood the proof, you'd see that the fact that reals can be represented in more than one base does not vitiate the diagonal argument. We can formalize the diagonal argument in formal first order Z set theory. There is no step in the argument that is not justified by first order logic applied to the axioms. > 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) Except you haven't PROVEN that. You haven't proven that all reals in [0,1) are in the list. If you contend that [0 1) is a subset of the range of that list, then you need to define the function (not just give it ostensively), then you need to prove that every real in [0 1) is in the range of the function. > - it is just the mirror inverse of counting from 0 upwards > in binary. Whatever that means, it's not a proof. > 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. What you've shown is that applying the diagonal method to your list only shows a real number not in [0 1). So what? It doesn't contradict that for any list of real numbers, there is a real number not in the list. We don't have to prove: For any list of real numbers in a certain interval, there is a real number in that interval not listed. Rather, we only have to prove: For any list of real numbers, there is a real number not in the list. Your lack of understanding of basic logic in mathematics has lead you to this irrelevent red herring/strawman about [0 1). Your whole line of argument is ILLOGICAL since the fact that the anti- diagonal of a particular listing is of a real number not in the interval specfied does not contradict that there is no listing that has all real members listed. > Abuse/explanations/condescension please... Your arguments are illogical and based on your misunderstanding and virtually complete unfamiliarity with the basic logic and mathematics of this subject. MoeBlee
From: David Marcus on 30 Jan 2007 01:33 MoeBlee wrote: > On Jan 27, 2:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > > > 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. > > What you've shown is that applying the diagonal method to your list > only shows a real number not in [0 1). So what? It doesn't contradict > that for any list of real numbers, there is a real number not in the > list. We don't have to prove: For any list of real numbers in a > certain interval, there is a real number in that interval not listed. > Rather, we only have to prove: For any list of real numbers, there is > a real number not in the list. The usual diagonal proof proves that [0,1) or [0,1] (depending on how we do it) is uncountable. So, the fact that the anti-diagonal is in the specified interval is important in the usual proof. The main problem with Andy's argument is that he is using a specific list. > Your lack of understanding of basic logic in mathematics has lead you > to this irrelevent red herring/strawman about [0 1). -- David Marcus
From: Andy Smith on 30 Jan 2007 05:48 David Marcus <DavidMarcus(a)alumdotmit.edu> writes >MoeBlee wrote: (snip) >> What you've shown is that applying the diagonal method to your list >> only shows a real number not in [0 1). So what? It doesn't contradict >> that for any list of real numbers, there is a real number not in the >> list. We don't have to prove: For any list of real numbers in a >> certain interval, there is a real number in that interval not listed. >> Rather, we only have to prove: For any list of real numbers, there is >> a real number not in the list. > >The usual diagonal proof proves that [0,1) or [0,1] (depending on how we >do it) is uncountable. So, the fact that the anti-diagonal is in the >specified interval is important in the usual proof. The main problem >with Andy's argument is that he is using a specific list. > And that he hadn't looked up the proof, otherwise he would have realised that the set of binary strings IS uncountable, and can be used to construct an injection into the reals. >> Your lack of understanding of basic logic in mathematics has lead you >> to this irrelevent red herring/strawman about [0 1). > -- Andy Smith
From: mueckenh on 30 Jan 2007 06:36
On 28 Jan., 20:18, Virgil <vir...(a)comcast.net> wrote: > > Please do not mix up numbers of elements and natural numbers > > concerning path lengths. > > Translated to our problem they say: There is an infinite number of > > finite path length. > Since there are an infinite number of natural numbers to serve as path > lengths, and path lengths for all those naturals, why not? > > > Now I am in doubt whether these path length when > > put together (such that every path starts at 0 and ends at n) yield a > > finite length or not. > So now WM can't tell whether a path staring at 0 end ending at n is > finite or not? I can't tell whether the union of all lengths is finite or not. > > > > > > > So a union of different > > > > natural numbers must contain an infinite number should it exíst? > > > > > > The fact that p is a union of finite paths does not tell > > > > > us whether p is finite or (potentially) infinite. > > > > Fact is that p has no upper bound. Ok?So p does not have length n for any > > > > finite natural > > > number n. So p does not have a finite path length. > > > But the length of p *is* a natural number. > If p is the union of all paths with finite lengths it must have a length > at least as great as any natural, so which natural does WM claim is as > great as every natural? There is no natural number greater than every natural. The path lengths however must be such a number, if all natural paths are taken together. The consequence is that you cannot take all natural numbers together. There is not "every natural number". > > > > Either we say that p does not have a path length, > > > or we say that the path length of p is an infinite number. > > > That's it! Either there are not all natural numbers ( = the union p > > does not ave a length) > > or there is an infinite number ( = there is an ininiutr finite > > number). > There IS an infinite ordinal number, but it is not a natural number. Maybe. But the path-lengths are all finite numbers. > There IS an infinite path but it is not a path of finite length. So we can say: If there is an infinite ordinal or cardinal, then there must be an infinite path length? Regards, WM |