An uncountable countable set
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From: mueckenh on 21 Aug 2006 04:29 Dik T. Winter schrieb: > In article <1155999375.602088.28870(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > Again a conclusion without proof. Use my definition of K: K[p] = 1 for > > > all p in N, and there are no other digits. > > > > Why should I? > > I would think because we are talking about that number? You keep stating > that that number is not indexable, without proof. When you want to show > that you have to start with that definition. > > > Use my definition of the indexable digit positions and > > there are other digits than those of numbers of the list. > > This sentence does not parse. What is the meaning of the part after "and"? > Anyhow, the number, with my definition, does not have digit positions other > than those on the list. > > > > Show that that number can not be indexed or show that that number can > > > be covered by a natural number, give a *proof* of either. > > Again, I ask you that. Indexing is covering due to the form of the numbers of my list. If position n can be indexed, then 1 to n are covered. > > > > > But you pretend it would consist of index positions which are all in > > > > the list. > > > > > > Not pretend, define. > > > > I define the list such that your definition is wrong. if all natural > > numbers do exist, my list is complete. > > What is the wrongness of the definition? Let me refrase: > Let L be the list of all natural numbers > Let K be the sequence of digits such that for each p in L, K[p] = 1. > What is *wrong* with that definition? It cannot be satisfed. > > > > > Now find out how that can be: All index positions of 0.111... are in > > > > the list, but not in the form of 0.111... . In which form should they > > > > exist there? > > > > > > What do you mean with "in the form of 0.111..."? > > > > The required digit positions are not contained in the list in form of > > this infinite number but only disperged over many finite numbers (which > > is impossible, of course). > > Why is that impossible? There is no "of course" here, because that "of > course" is just stating "because the set of all naturals does not exist". Because all digit positions up to a given one are all contained in one single number. In my list holds the following law: If a digit position n is covered then there is at least one number not less than n in the list. If an infinite number of digit positions is covered than there is at least one number with infinitely many digit positions in the list. But in my list there is no number with infinitely many digit positions. If you would understand that, then all you question were answered. But am tired to discuss these simple things with you. Regards, WM
From: mueckenh on 21 Aug 2006 04:35 Dik T. Winter schrieb: > > > > > > 1/3 has the sequence of nodes 0.010101... in binary notation. > > > > > > > > > > > > 0. > > > > > > /1 \2 > > > > > > 0 1 > > > > > > /3 \4 /5\6 > > > > > > 0 1 0 1 > > > > > > /7\8/9\10........ > > > > > > > > > > > > The edges of 1/3 are enumerated 1, 4, 9, 20, ... . For any edge a > > > > > > natural number can be determined, somewhat cumbersome, but it > > > > > > can be done. > > > > Is the number of 1's in my infinite list > > > > 1 > > 11 > > 111 > > ... > > > > uncountable? > > No. Why should it be? I would state that at the n-th line the first > 1 is number 'n * (n - 1) / 2 + 1' and we count from there. Again, what > is the problem? Your wrong assertion that the set of edges in my infinite tree was uncountable. > > I think I see your problem. Better try to see your problem. Or ask an expert. The set of edges *is* countable. > In the tree we can assign natural numbers > that lead to a node that is a finite distance away from the root. No > problem, as I have stated over and over again, that kind of edges is > countable. The set of all edges in my infinite binary tree is countable, just like the digits of the infinite number pi or like the 1's in above infinite triangle. > Now, if we complete the tree (and the list above) the number > of edges is still countable, but *none* of the edges terminates at 1/3. > You want "a last line" for completion, but there is none. I don't want a last line, nevertheless the set of all edges is countable. > You simply > do not accept completion without there being a last one. And that is > precisely what the axiom of infinity asserts, the set of natural numbers > exist (that is, it is possible to talk about it), but there is no > largest natural number. Nevertheless they are all countable, just like the edes of my infinite tree. And by rational relation we find the set of paths being *not* larger than the set of edges. Have you got it now??? Regards, WM
From: mueckenh on 21 Aug 2006 04:38 Dik T. Winter schrieb: > In article <1155999765.712997.69780(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > Sorry, I do not understand. There is no block at height 1 and there is > > > no block with width 1. The smallest square that contains the complete > > > stair has height and width 1, so we can say that the complete stair has > > > height and width 1. But the top and right edge are never reached. You > > > could say they are "asymptotes". > > > > Correct. That is potential infinity. That is just what I claim. Width > > 1 is *not* reached (at least not without reaching 1 in height). > > So you claim the square I defined above has not width 1 and height 1? > And you claim that such a container does not exist? > > > But according to Cantor width 1 *is* reached while height 1 is not > > reached, infinity in number does actually exist infinity in size does > > not exist. > > I do not think so. What Cantor states is that a container exsits that > has both width and height 1. False. According to Cantor the height is never 1, width is 1. Remember: If the staircase has height H then at least one stair has height H. If the staircase has height 1 then at least one stair has height 1. That is a natural law for stairs and staircases. Regards, WM
From: mueckenh on 21 Aug 2006 04:42 Dik T. Winter schrieb: > In article <1156000079.633380.83620(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > Do you agree: If there are more than one can check then one cannot > > > > check all? > > > > Do you agree: If there are more than one can check then one cannot > > > > check every line? > > > > May one say so in set theory? > > > > > > But there is no need to check each and every individual line. When I > > > state: > > > sum{i = 1 .. n} i = n * (n + 1) / 2 > > > do I need to check for each and every n? > > > The *definition* of the diagonal makes clear that it is different for > > > each and every n. Like in the formula above I need not check for > > > each individual n, I need not check for a difference for each and > > > every n. > > > > Exactly. Above formula holds for every finite number, as can be shown > > by induction. But it doesn't hold for an infinite number of finite > > numbers. > > Eh? It holds for every finite number but does not hold for an infinite > number of finite numbers? What do you mean with that? If there are > infinitely many finite numbers, I would say that as it holds for > every finite number, it also holds for infinitely many finite numbers. What is the sum of infinitely many finite numbers? What is the sum of all finite numbers? > > > > 100 %? But with sets I would state that a set exists if all of its > > > elements do exist and all of its subsets do exist. > > > > Is it possible? If ZFC is consistent, then it has a countable model. If > > all of its subsets would exist, then it was uncountable. > > Is ZFC a set? The subsets of the set of the model are meant of course. > > > > But you asked > > > for the *first* 10%, which is something different. > > > > Then drop the "first". I will ask only for 10 % of the elements. Or > > would you prefer the last 10 % ? > > With ordering you lose, so not the last. {1, 11, 21, ...} are an example > of a set that contains 10% of the natural numbers (whatever that may mean). > > > > > Don't misunderstand me: All subsets of |N do exist. But the set of all > > > > non-generators does not exist. > > > > > > Oh. I must have missed something, because I have not seen a proof. > > > Given an injection f: N -> P(N), why does the set > > > M(f) = {n in N | n !in f(n)} > > > not exist? > > > > Excuse me, the non-existing set is the triple {f, n, M_f(n)} > > Excuse me, Hessenberg was not talking about such triples. But I am doing so in order to show that his arguing concerns impredicative definitions and is inconclusive.f is the mapping, n is a natural number, M_f(n) is a set which contains all nongenerators, including n if not including n which is mapped on M. Regards, WM
From: Dik T. Winter on 21 Aug 2006 07:26
In article <ecb84s$b1q$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Dik T. Winter wrote: .... > > > With the addition of infinitesimals and specific infinite quantities, > > > even division by 0 can be handled, probably even for complex numbers, > > > > Division by 0 can not be handled in a ring. > > According to MathWorld, indeed, there is an exception for 0 in the > optional Multiplicative Inverse condition. Only required due to the > aversion to infinitesimals. Alas! Why cannot we tie this all together? No. Required because if 0 has also to have an inverse in a ring, trivially there is only one ring. The ring consisting of the single element 0 with the standard operations. Because, suppose we have a ring. It can be proven that in a ring 0*a = 0 for every a in that ring. But if 0 has an inverse, say b, we also have 0*b = 1, and so 0 = 1. On the other hand, do not confuse 0 with the infinitesimals. And there is no aversion to the infinitesimals. Look at the surreals, look at non-standard analysis. Infinitesimals abound. Still, in order to be logically consistent, 0 has a special role. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |