An uncountable countable set
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From: Dik T. Winter on 19 Aug 2006 20:12 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 yuo that. > > > 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? > > > 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". -- 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 19 Aug 2006 20:35 In article <1155999560.728582.29520(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > But I now understand that the series: > > sum{n = 1 .. oo} (-1)^n/n > > does not converge according to your logic. > > For every natural number n we have n/n = 1. But the notation lim n --> > oo is not clear. You do not know the convention that taking powers goes before division? What *do* you know about mathematics? But I will refrase: But I now understand that the series: sum{n = 1 .. oo} ((-1)^n)/n does not converge according to your logic. And why is the notation lim n --> oo not clear. You have defined it just a few days ago: > > > The same is meant as in sequences and series like: > > > lim [n --> oo] (1/n) = 0 > > > n becoming arbitrarily large, running through all natural numbers, but > > > always remaining a finite number < aleph_0. > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings > > > where the remainings are |{n+1,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 > > > > That makes no sense. As I read it, we have: > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings = 1 + 1 = 2. > > ? > 1 + remainings = 1 + 1 = 2 > Which error did you see? Read the fifth line above the question mark. > > But I will allow an error here. But I was thinking about: > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > which might be written as: > > lim{n -> oo} |{1, 2, 3, ..., n}|/|{2, 4, 6, ..., 2n}| > > = 1 by mathematics and by set theory. > > > or as: > > lim{n -> oo} |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| > > = 2 by mathematics but 1 by set theory. As set theory does not know about limits, I fail to see that, but even if set theory had knowledge about limits, I would state: No, also 2 by set theory. What you fail to see is that |{1, 2, 3, ...}| / |{2, 4, 6, ...}| is defined by neither mathematics, nor by set theory. Both do give *no* definition of aleph-0/aleph-0. The cardinal numbers do not form a field, only a ring (and not even a division ring). So arbitrary division is not defined, definition is not even possible. -- 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 19 Aug 2006 20:57 In article <1155999688.567981.312170(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > Why then did you want to allow to assign finite sequences of edges only? > > > > To simplify matters. Natural numbers have in any base notation a finite > > number of digits. In allowing finite sequences of digits I allow for > > the occurrence of (for instance) both 0 and 00 and 000. So I give you > > more freedom. > > But in fact that is less freedom because the sequence of edges per path > is infinite. But you want to assign natural numbers? I would think that assigning aribtrary finite sequences of digits would be easier? > > > > > 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. > > > Oh. So there is no final edge that leads to 1/3? And so 1/3 is not > > in the three? And all your edges (and hence paths) terminate? > > All edges terminate, no path terminates, because it is an infinte > sequence of edges I would think that the paths consist of edges, as each edge terminates so would each path. > > > Try to understand how the countability of the algebraic numbers is > > > proved. > > > I proved in the same way the countability of the set of edges. > > > > With the enumeration of algebraic numbers, we start with enumerating > > *finite* polynomials. And we subsequence within those polynomials. > > And each and every algebraic number can be found at a *finite* position > > from the start. > > Each and every edge of my tree can be found at a finite position from > the start. Yes. So the edges are countable. But none leads to 1/3. > > In your tree, 1/3 can *not* be found at a *finite* > > position from the start. So either 1/3 is not in your tree (and you > > have only terminating paths and a countable number of edges, and this is > > similar to the enumeration of the algebraic numbers), or you put all > > reals in your path, but in that case you have also non-terminating paths > > and an uncountable number of edges. > > Look at the real number which makes up Cantor's diagonal. By chance it > could even be 1/3, if 1/3 was not an entry of the list. It is a > non-terminating path. Is its number of edges uncountable in your > opinion? You are confusing issues again. Stay to a single subject. > 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? I think I see your problem. 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. 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. 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. -- 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 19 Aug 2006 21:04 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. > You said that a set exists if all its elements do exist. If so, then > all natural numbers do exist and make up an infinite number without one > of them being infinite. Right. -- 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 19 Aug 2006 21:22
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. > > 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? > > 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. And I do not even understand your notation at all. Hessenberg was talking about sets like M(f) as I defined above. And that the triple {f, N, M(f)} does not exist shows that there is no f that is surjective from N to P(N). And that only because surjectivity requires such a triple. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |