Cantor Confusion
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From: Lester Zick on 27 Nov 2006 17:36 On Mon, 27 Nov 2006 14:06:09 -0500, Bob Kolker <nowhere(a)nowhere.com> wrote: >Lester Zick wrote: > >> On Mon, 27 Nov 2006 07:25:21 -0500, Bob Kolker <nowhere(a)nowhere.com> >> wrote: >> >> >>>Eckard Blumschein wrote: >>> >>>> >>>>Geometry is not outside mathematics. >>> >>>True. But Geometry is not all of mathematics either. >> >> >> Nor is arithmetic. >> >> >>>The two historical sources of mathematics has been land measurement and >>>counting objects. After which comes tracking the cycles in the heavens. >> >> >> So what? One historical source for chemistry was alchemy. Historical >> origins don't have any bearing on the nature of a science. > >They have some bearing. The earlier fields create a barrier to be >overcome. Just as modern math is a barrier to the study of mathematics to be overcome. Just as quantum physics is a barrier to the study of angular mechanics to be overcome. Just as relativity is a barrier to the study of mechanics in general to be overcome. > It is a challange. Detaching the notion of measure from length >and area was a big jump forward in the theory of real variables. Just as detaching the notion of a real number line from mathematics was a big jump forward in the theory of real variables. ~v~~
From: Virgil on 27 Nov 2006 19:10 In article <1164634499.426863.18200(a)l39g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > 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. I have no idea what WM misrepresents as being induction except that it is not mathematical induction. > By induction > only sensible assertions can be proved. WM's notion of sensible and mathematicians' notions do sensible disagree. >The *set generated by > induction* is always finite. The set required by mathematical induction requires an ordering on it under which there be a first member and FOR EVERY member, a next one different from it and all its predecessors. > > Regards, WM
From: Virgil on 27 Nov 2006 19:44 In article <1164636432.573982.57080(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? Which one? your arguments are rife with them. For one, any open connected subset of the reals can be proved uncountable by the same Cantor first proof argument, but disconnected subsets may not. > > In addition, two sequences of transcendental numbers can converge to a > rational number, such that we have the same situation as described > above. The set of transcendentals is not a connected set in R, so violates the hypotheses of the proof. > > Therefore both sets, Q ???and T (transcendental numbers), have the same > status with respect to *this* uncountability proof. Both are totally disconnected as subsets of the reals (between any two members there is a non-member) so that the hypotheses of the proof are violated. > And we are not > able, *based on this very proof*, to distinguish between them. Why should we, as the proof does not apply to either. > 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. That doesn't work it unless you show that NO sequence can be made including all of the values of S = {0} union { (-1)^n/n: n in N} But consider the sequence 0, -1, 1/2, -1/3, 1/4, -1/5, ... which does contain all of the members of S. > 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. Wrong! The theorem speaks of ANY sequence, not just hand picked sequences. > *Based upon this proof alone*, the uncountability of > this and every other alternating convergent sequence must be claimed. Not by anyone who has read the Cantor first proof with any understanding of what it says. > > > > 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 do not "understand" any such thing. It is your misunderstanding of the proof itself. It is the least upper bound and greatest lower bound properties of the reals ( together sometimes called the completeness property) which are essential here. In the reals, any subset which has a real upper bound has a real least upper bound and, similarly , any subset which has a real lower bound has a real greatest lower bound. The only subsets of the reals for which there is a similar property are real intervals. Thus it is only for the set of all reals, or for real intervals, that the proof appplies. > > Cantor's proof can be used to show the uncountability of the rational > numbers. In Q there are sequences which converge to rational numbers. But there are also sequences which do not, and it is the existence of those sequences that do not which scuttles WM's misrepresentation. Recall in the proof, that the analysis must hold for ANY sequence which is alleged to count every member! if it fails for any, it fails entirely. And it fails for every one of WM's so called counterexamples.
From: Dik T. Winter on 27 Nov 2006 19:48 In article <456AE00A.3050406(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > 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. I am talking mathematics, not what you are thinking is mathematics. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 27 Nov 2006 19:47
In article <456ACE78.2030902(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > On 11/25/2006 3:58 AM, Dik T. Winter wrote: .... > > > > Oh. What are "really real numbers"? > > > > > > Those, like the imagined numerical solution to the task pi, which are > > > just fictions. Those which are assumed as basis for DA2, > > > > In that case what are the "not-really real numbers"? > > Those putatively real numbers as defined by Dedekind's cut, nested > intervals, etc. without the step into a different quality. Oh. Well, I use them pretty regularly. > I learned the word "Grenzuebergang". Is "limit" the mathematically > correct translation? No. > In my understanding, a limit ia a border while Grenzuebergang means > border crossing. Indeed. > When Eudoxus coined the term, what did he mean? Why is that interesting? > > You completely misunderstood what I wrote. Cholesky decomposition > > While I did not understand you wrong, I nonetheless very appreciate your > hint to Cholesky decomposition. This way I found > > http://sepwww.stanford.edu/sep/prof/gem/hlx/paper_html/node11.html Yup, that is it. Note that in its basic form it requires the use of square roots. The form with separate diagonal is not used very much (although it will work with matrices that are not positive definite, but in that case in general no satisfactory error bounds can be found). What you still fail to see is to get information about numerical stability and information about error bounds, when doing it numerically on a computer, you *start* with a complete mathematical ideal decomposition. Similar things happen with similar problems in numerical mathematics. For instance the convergence of the QR method for finding eigenvalues of matrices (that are almost never rational numbers) is proven using ideal arithmetic, and also the order of convergence is calculated that way. It is only after you have established that that you look to see what is happening when you are doing it with numerical approximations (like floating point numbers). You will find that in most cases the properties do not change in that case, but in some cases there are very strong changes. It is very difficult to prove convergence in the numerical case, but adapting the general case with numerical properties is much simpler. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |