Cantor Confusion
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From: Eckard Blumschein on 27 Nov 2006 07:54 On 11/26/2006 3:25 AM, Dik T. Winter wrote: > Remove all irrational points and in addition those > points that have x = 0. You have a disconnected set. A set is thought of elements put into it. Can you really set irreal points into it? This cannot work because the set has to have an order p/q. Otherwise it is not a set but a continuum.
From: Bob Kolker on 27 Nov 2006 08:08 Eckard Blumschein wrote: > On 11/26/2006 3:25 AM, Dik T. Winter wrote: > >>Remove all irrational points and in addition those >>points that have x = 0. You have a disconnected set. > > > A set is thought of elements put into it. Can you really set irreal > points into it? > This cannot work because the set has to have an order p/q. Otherwise it > is not a set but a continuum. A continuum is a compact connected metric space. Learn some mathematics. Bob Kolker > >
From: mueckenh on 27 Nov 2006 08:30 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > If the set 1,2,3,...,n is finite, then the set 1,2,3,..., n+1 is > > finite too. > > Therefore, by induction we (that is: those who can think logically) > > prove that the generated set is always finite. > > Any set {0, 1, 2, ..., n} n e omega is finite. You have not proved that > the set of all such sets { {}, {0}, {0, 1}, {0, 1, 2}, ... } is finite. I am only interested in sets of he form {1,2,3,..., n}. Only such sets can be generated by induction. Their finiteness is sufficient for my purposes. Regards, WM
From: mueckenh on 27 Nov 2006 08:34 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Virgil schrieb: > > > > > > > > > By induction one can even prove that always a finite number of words > > > > has been mapped on the natural numbers. > > > > > > By induction one can also map an infinite set of words onto the natural > > > numbers. > > > > Then you think the natural numbers are covered by induction? You read > > my heart. I knew they were not actually infinite. > > > > > > > > > > Now, after you have "completed" this finite list of words, take it and > > > > construct a diagonal number. > > > > > > I prefer to take the infinite set of words. > > > > I know, but you don't get it by induction. > > > > > > > This is always a finite number, as the list is finite. > > > > > > Maybe yours is, mine isn't. > > > > > > > Fine. > > > > So we have proved that there is a finite word which was missing in the > > > > "complete" list. > > > > > > You haven't proved anything, > > > > If the set 1,2,3,...,n is finite, then the set 1,2,3,..., n+1 is finite > > too. > > Therefore, by induction we (that is: those who can think logically) > > prove that the generated set is always finite. > > The two statements: > > The elements of N satsify property X > and > The set N satisfies property X > > are independent.. Induction proves that the elements of N are finite. > Induction does not show that N is finite. Of course it does not. Induction is a sensible method. By induction only sensible assertions can be proved. The *set generated by induction* is always finite. Regards, WM
From: mueckenh on 27 Nov 2006 09:07
Virgil schrieb: > > It is asserted in that paper that the so called first Cantor proof of > the uncountability of the reals applies equally well to the set of > rationals. > > Cantor's proof relies on the fact that if one has a strictly increasing > sequence of reals with each term less than all terms of a strictly > decreasing sequence of reals, there is at least one real strictly > between the two sequences, not being a member of either but larger than > every member of the increasing sequence and smaller that every member of > the decreasing sequence. Ok. This is what I wrote when describing Cantor's proof. > > This is not the case for rationals. For example, there are strictly > increasing sequences of rationals whose squares converge to 2 and > strictly decreasing sequences of rationals whose squares converge to 2. > But there is no rational number between the sequences. > > For example a_1 = a, a_{n+1} = 4*a_n/(2 + a_n ^ 2) is an increasing > sequence of rationals and b_1 = 2, b_{n+1} = (2 + b_n ^ 2) / (2 * b_n) > is a decreasing sequence of rationals with only sqrt(2) between all the > a_n and all the b_n. Nevertheless the rational numbers and the irrational algebraic numbers are countable. So, what does the proof, as you described it, show? The uncountability of a countable set? Where is the error of mine? In addition, two sequences of transcendental numbers can converge to a rational number, such that we have the same situation as described above. Therefore both sets, Q ï and T (transcendental numbers), have the same status with respect to *this* uncountability proof. And we are not able, *based on this very proof*, to distinguish between them. On the other hand, the proof can show the uncountability of a countable set. If, for instance, the alternating harmonic sequence (-1)^n/ n --> 0 is taken as sequence (1), yielding the intervals (-1 , 1/2), (-1/3 , 1/4), ... we find that its limit 0 does not belong to the sequence, although the set of numbers involved is obviously denumerable. The alternating harmonic sequence does not, of course, contain all real numbers, but this simple example demonstrates that Cantor's first proof is not conclusive. *Based upon this proof alone*, the uncountability of this and every other alternating convergent sequence must be claimed. Only from some other information we know their countability (as well as that of Q), but how can we exclude that some other information, not yet available, in the future will show the countability of Q or T? > > Not so, as the example above proves. There is an increasing sequence of > rationals converging to sqrt(2) and a decreasing sequence of rationals > converging to sqrt(2) such that there is no rational at all caught > between the two sequences. Nevertheless the algebraic numbers are countable. > > No, it is you who misunderstand Cantor's proof. Where is an error of mine? I only see that you understood that the algebraic numbers are uncountable. > > > I claim that Cantor's proof is > > "symmetric". It can be applied to the algebraic numbers, showing that > > the limit is not algebraic. > > It can be but need not be. That is the same with the rational sequences. Some converge to 0. > > It can be applied to any proper subset of the reals showing that there > is a sequence of values in that subset which does not converge to a > value in that subset. Cantor's proof can be used to show the uncountability of the rational numbers. In Q there are sequences which converge to rational numbers. > > >This > > has nothing to do with rational numbers and sqrt(2) because that would > > not prove any uncountability at all (sqrt(2) is algebraic and as such > > belongs to a countable set). > > The point is that for every sequence of reals, there are reals not in > that sequence. And for every sequence of algebraic numbers converging to an algebraic number, there are algebraic numbers not in that sequence. Regards, WM |