Cantor Confusion
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From: Virgil on 30 Jan 2007 14:58 In article <1170162837.477985.97730(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 29 Jan., 03:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > If I consider the path length as the union of path length: = > > > {0,1,2,...,n-1}, I am right too. > > > > Yes, for this path. I never did have problems when you were uniting paths > > that you get infinite paths and all that that entails. > > But path-lengths are natural numbers! > > > But you are uniting > > *sets* of paths. In that case you do not unite paths at all. > > I unit finite trees. This implies the union of sets of paths. Paths are derived from the tree they exist in, and what is a path in one tree will generally not be a path in another. So that the union of sets of paths will, in general, contain things which are not all paths in any one tree. > This > implies the union of paths, because the union of two or more paths > need not but can be a path (notice: the indexes are of the form > {0,1,2,...,n-1}). The union of trees implies the union of outmost > left-hand side paths for example. This union is a path. While a union of a nested sequence of paths in a nested sequence of trees may be a path in the WM pseudo-union of those trees, the union of the sets of paths from the separate trees is not a set of paths in any single tree. > > You are correct. > The union of all natural numbers is not a natural number. Therefore > you think it (the cardinality) can be infinite. It certainly cannot be finite. > The union of all paths is a path in many cases. Only in unary trees. > There is no infinite union of finite numbers unless there is an > infinite number. There is, the union itself is such an infinite number but it is not a member of itself, at least not in any well founded set theory compatible with ZF. This persistent delusion of WM's that a set cannot be non-finite without containing a non-finite member precludes his ability to deal honestly with any set theory.
From: MoeBlee on 30 Jan 2007 15:02 On Jan 29, 10:33 pm, David Marcus <DavidMar...(a)alumdotmit.edu> wrote: > 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. My point still stands: If one gives a list of real numbers in a given interval and shows that the anti-diagonal of that list is not in that interval, then one has not thereby refuted the uncountability of the reals that is proven without reference to that particular list, interval and anti-diagonal. MoeBlee
From: Virgil on 30 Jan 2007 15:03 In article <1170163473.431673.242130(a)a34g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1170085646.542472.199010(a)k78g2000cwa.googlegroups.com>, > > > > Of course we can show in unary representation (IET) that the sum of > > > all natural numbers is countable > > > > You can't even show that it exists, WM. > > Of course I cannot show that this sum exists, because, in fact, N does > not exist. Right conclusion for the wrong reason. There are lots of infinite series whose partial sums are indexed by members of N whose sum exists (they converge). The real reason that WM's sum does not exist is that the series (sequence of partial sums) diverges, despite the exisence of the index set, N.
From: MoeBlee on 30 Jan 2007 15:05 On Jan 30, 2:48 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > David Marcus <DavidMar...(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. Who hadn't looked up what proof? You hadn't looked up a usual variant of the proof of the uncountability of the reals? Construct what injection into the reals? MoeBlee
From: MoeBlee on 30 Jan 2007 15:14
On Jan 30, 4:42 am, Dave Seaman <dsea...(a)no.such.host> wrote: > On Tue, 30 Jan 2007 01:33:03 -0500, David Marcus wrote: > > 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. > > Actually, the main problem with Andy's argument is that he is using a > defective variation of the diagonal argument. When correctly done, the > diagonal argument does not produce a number that has a dual > representation. > > >> Your lack of understanding of basic logic in mathematics has lead you > >> to this irrelevent red herring/strawman about [0 1). > > It's not irrelevant. A correct argument applied to that list shows that > there is a number in [0,1) that is not in the list. One way is to > represent the numbers in decimal and to carry out the diagonal argument > in such a way as to produce a number, all of whose digits are either 4 or > 5. The irrelevancy is in the non sequitur from giving a list of reals in an interval such that the anti-diagonal (in a given base) is not in that interval to the conclusion that that somehow works against a correct diagonal argument. We only need to show that for any list, there is a real not in the list. That we can do that for a particular interval if we wish to do so is gravy as far as the mere question of the uncountability of the reals is concerned; and that there might be a way to rig a list of reals in an interval such that the anti- diagonal (in a given base) is not in that interval does not refute that for any list there is a real not in the list. That is the sense I mean. MoeBlee |