Cantor Confusion
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From: mueckenh on 21 Oct 2006 08:55 Dik T. Winter schrieb: > > If we could not get to omega, we would not need it. > > If you do not need it, so be it. Do not use current set-theory and base > your analysis on whatever is provided with a finitistic view. There is > *no* reason to attack people for using it, except that it is possibly > against your ethical views. I know that we cannot count to omega! But we use it in definitions of limits. > > > And one would not > > be able to count omega + 1 and so on. Cantor could have refrained from > > introducing it. > > In order to count all finite sets you need omega. That > > is the essence of Cantor's theory. > > Yes, that was Cantor's theory, but that was an error, and it was not the > essence. See my discussion with Dave Seaman about just that point. I > do not think he still had that position when he wrote his 'Contributions > to set theory' some 11 years after his 'About infinite linear point-sets', > page 213 in the 'Gesammelte Abhandlungen'. The essence of his theory was > that there were sets that could not be put in bijection with the set of > natural numbers. Yes. But this recognition is impossible or useless unless all natural numbers do exist. Therefore lim [n-->oo] {1,2,3,...,n} = N. Therefore lim [n-->oo] {1 + 1/2^2 + 1/3^2 + ... + 1/n^2} = (pi^2)/6. Therefore in the vase experiment: lim [t-->oo] X(t) = X(omega). > > > > Yes to the first, no to the second. f(oo) does not belong to set theory, > > > it belongs to analysis. Also note that Cantor at some time refrained from > > > using oo and started to use w, because there was too much confusion. > > > The oo from pre-Cantorian mathematics is different from the omega of > > > Cantorian mathematics. > > > > According to Cantor oo denotes potential infinity, omega denotes actual > > infinity. If all natural numbers exist, then the infinite oo of > > analysis in our vase problem is exactly the omega of set theory. > > No. The oo in standard analysis is still potential. That is, it can not > be attained. So the limit lim{n = 1 --> oo} 1/n = 0, does indicate a > potential infinity, not an actual infinity. For which natural number n is 1/n = 0? Potential means always finite. lim{n = 1 --> oo} 1/n = 0 proclaims "omega reached". > But you are dishonest in transforming the vase problem (where the answer > was asked at t = 0) to another problem (where the answer was asked at > t = oo). It is nothing but just a simplification in notation! > > > > > > Yes. You can construct a number from any list, but you can not > > > > > construct a constructable number from a list that is itself not > > > > > constructable. > > > > > > > > Every diagonal number is constructed and, therefore, is constructible. > > > > > > When constructed from a constructable list. > > > > What is an unconstructable list? Do you call any undefinite mess a > > list? > > Any list is construtced. Any diagonal number is constructed. > > Some confusion, I think (and I am myself also guilty). A constructible > number is a number that can be represented by a finite number of > additions, subtractions, multiplications and finite square root > extractions. So you can not prove that a list of constructable numbers > gives as the diagonal a constructable number. The very reason is that > taking the diagonal is not a construction according to the definition. So is it not. Good heavens, that is unimportant for the present argument. Every constructed number is an element of a countable set. The set of all constructed numbers is countable. Every diagonal number belongs to this set. > On the other hand, I think you are meaning computable No, I was not meaning that. I mean: constructed (by list or by formula). Regards, WM
From: mueckenh on 21 Oct 2006 08:59 Dik T. Winter schrieb: > In article <1161378187.155995.290420(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > In article <1161276574.792436.186750(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > Yes, but it is infinitely many times not continuous in each neighbourhood > > > of the limit point. > > > > Let t be the ordinal number of transactions. > > > > It *is* continuous like a staircase. > > X(t) = 9 for t = 1 until t = 2 where X(t) switches to 18. That is > > enough to excflude X(omega) = 0. > > How do you *define* X(omega). As far as I know X is only defined for real > numbers, and omega is not one of them. And I see no reason to exclude > X(omega) = 0, = 1, = -1 at all from this reasoning. lim {t --> oo} X(t) = X(omega) > > > > That is still not a mathematical formulation. More is required. You > > > need to actually state what you mean with 'number of balls in the vase > > > at noon' or 'natural numbers in the set at noon'. > > > > The number of transactions t is then t = omega at noon. > > Is *that* a mathematical definition? Pray provide a real mathematical > definition. lim {t --> omega} t = omega > > > > The limit can in mathematics *always* be determined by the terms (if there > > > is a limit). But the limit in no case defines the function value at the > > > limit point. > > > > Then the irrational numbers as limit points are undefined. > > You think so. The irrational numbers are defined to be the limits of some > particular sequences (or rather as equivalence classes of sequences). I Equivalence classes of sequences with same limit like lim {t --> oo} a_t. > think you have no idea how numbers (yes, I use that term while you think > it is disgusting) are defined. I will repeat: > (1) start with the natural numbers as defined by Peano (you may start with > 0, 1 or 2, but starting with 0 makes everything a bit easier). > (2) define arithemetic with those numbers, using the axioms. > (3) define negative numbers as pairs of two elements, the first is a single > bit (the sign), the second is a natural number. > (4) provide arithemetic with the negative numbers. Show that the natural > numbers from (2) can be embedded in this, Call this the integers. > (5) assume pairs of integers, and provide equivalence classes amongst > these pairs (i.e. (a, b) ~ (p, q) iff a.q = b.p). > (6) provide arithmetic with such pair. Show that the integers can be > embedded in this. Call these the rationals. > (7) assume sequences of rationals. Create equivalence classes amongst those > sequences (a_n ~ b_n if |a_n - b_n| goes to 0; but this is losely > speaking and quite a few other methods are known, all equivalent). > (8) provide arithmetic for those sequences. Show that the rationals can > be embedded in this. Call these the reals. > > So, at what stage in this process is the limit of a function used to > define the irrationals? At (7). The equivalence classes of sequences of rationals with same limit. (A sequence is a function f: N --> R, sometimes called "index function" in German) Regards, WM
From: William Hughes on 21 Oct 2006 09:27 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > No. If you restrict yourself to computable functions you have some > > > > counterintuitive results. Assume that > > > > the language you are working in has a finite alphabet. Then the set > > > > of all finite strings in the language is listable using a computable > > > > function > > > > (use dictionary order). And so the set of all finite strings is > > > > countable. > > > > Now, A, the set of all strings which define a computable number is a > > > > subset of the set of all finite strings. So A is countable, right? > > > > Wrong! > > > > It is not true that every subset of a countable subset is countable. > > > > > > It is true that every set can be well ordered and that any two sets can > > > be compared. Both are equivalent or one is equivalent to a sequence > > > (ordered subset) of the other. What you say is not counter intuitive > > > but wrong. > > No response? Here we work in our standard model, not in some externally > finite but internally uncountable model. > The response is a a different reply. > > > > > > > And that set can't be listed. > > > > > > > > > > > > And here is your problem. Uncountable means unlistable. > > > > > > > > > > Not my problem. Countably infinite means unlistable too. > > > > > > > > Yes, but what does unlistable mean? > > > > > > Listable means countable. > > > > A tad circular? > > An identity. > > > A set B is is listable (and therfore countable) > > if there exists a function, f, > > from the natural numbers to the set B. Whether we can immediately > > conclude that a subset of B is listable (and therefore countable) > > depends on what types of functions f we allow. > > Why should we refuse our admission for some functions? Because you have claimed that only computable functions exist (or more precisely, you need this claim in order to claim that there is no list of the computable numbers). - William Hughes
From: Virgil on 21 Oct 2006 15:23 In article <1161427471.109514.138610(a)m7g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > if your translation leads to my results, > you may be satisfied, and if it does not, then your translation or your > theory is a mess. Such arrogance is the pride of the foolish. > > =========================== > > You don't know what your're talking about. The convergence is proven. > > The identity of 1 and 0.999... in analysis is proven. The difference > 10^(-n) disappears for n --> oo (and not earlier!). The identity is not > proven in case of the diagonal number because it cannot be proven, > because of the lacking factor 10^(-n). Even for the digit with index n > --> oo the difference is as important as for the fist digit. Except for such cases as 1.000... = 0.999..., and others involving an endless string of 0's or an endless string of 9's, two reals differ if they digger in ANY digit, regardless of place value. So a number whose decimal representation does not contain any occurrence of the digits 0 or 9 will differ in value from any other number if it differs at digit from the other number. Since the Cantor diagonal rules all exclude use of 0 or 9, such a number can equal another only if they have the same unique decimal expansion, matching at every decimal place, with every digit equal. > > Here is the schisma, the point where mathematics is inconsistent: > Either the digit becomes negligible in both cases or in none of the > cases, i.e., either 1 =/= 0.999... in analysis or the later digits of > the diaonal number become more and more unimportant such that for an > infinite number there are negligible digits and Cantor's argument > breaks down. Not in any real mathematics. In real mathematics, there is no non-zero difference so small as to allow numbers differing by that amount to be equal.
From: Virgil on 21 Oct 2006 15:32
In article <1161434879.731750.127910(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1161378001.475899.279610(a)f16g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Every diagonal number is constructed and, therefore, is > > > > > constructible. > > > > > > > > When constructed from a constructable list. > > > > > > What is an unconstructable list? Do you call any undefinite mess a > > > list? > > > Any list is construtced. Any diagonal number is constructed. > > > > A list of reals for the Cantor construction need not be made up of > > constructable numbers, it is only required that the nth number be > > constructable to the nth decimal place. > > It is assumed that nearly all numbers of the list are contructible to > more than the n-th digit. > But the main point is: It is assumed that the diagonal number is a > constructed number. The set of constructed numbers, however, is > countable. Wrong again. The set of constructed numbers is finite. The set of constructible numbers is countable. The set of numbers within some interval of positive length, however small that length is, is uncountable. Thus for every list of reals with the nth real constructable to within epsilon/10^n, there is a real not listed. |