An uncountable countable set
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From: mueckenh on 19 Aug 2006 10:59 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. > > > > > |{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? > 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. Regards, WM
From: mueckenh on 19 Aug 2006 11:01 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. > > > > > 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 > > > > > > > That was not what I asked. You assigned a natural number to a subset of > > > the edges, not to all edges. > > > > 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. > 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? Is the number of 1's in my infinite list 1 11 111 .... uncountable? Regards, WM
From: mueckenh on 19 Aug 2006 11:02 Dik T. Winter schrieb: > In article <1155900279.092888.72790(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > > Let's have an easier example. The blocks are 1/2^n high and > > > > > 1-1/2^n wide. > > > > > > > > When you make a stair of them, when you complete, you have a > > > > > stair with height 1 and width 1. But there is neither a block > > > > > that is 1 wide, nor a block that is at height 1. On the other > > > > > hand, for every positive k, there are blocks that are beyond the > > > > > height of 1-k and beyond the width of 1-k. Both in height and > > > > > in width the blocks just do not reach the boundary line. And > > > > > note that in both "infinity is reached". > > > > > > > > What you argue is a potential infinity. Infinity, aleph_0, here reduced > > > > to 1, is present, actually existing, according to Cantor, in width. All > > > > steps do exist. But, according to him, infinity, here 1, is not > > > > present in height. > > > > > > It is, also according to him. Both in width and in height. > > > > Wrong. "Die unendliche Menge der endlichen Zahlen." > > The set is infinite. The numbers are all finite. > > 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). 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. 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. Regards, WM
From: mueckenh on 19 Aug 2006 11:08 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. Just the same is true for Cantor's list. > > I made my statement in accordance with my understanding of existence, > > which I supposed was the understanding of everybody: If something does > > exist, then every part of it does exist. To my great astonishment you > > denied this simple truth. Therefore I asked you what you think what > > existence is. How many percent of a set must exist in order to call it > > actually existing? > > 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. > 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 % ? > > > > But you succeeded again in diverting the discussion from your initial > > > assertion. The countable uncoutable set. > > > > That question has already been solved. The set of all non-generators > > does not exist. Therefore Hessenberg's proof of Cantor's theorem is > > false. > > > > 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)} Regards, WM
From: Franziska Neugebauer on 19 Aug 2006 12:08
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> You have agreed to that this limit does not exist ("There is no L in >> N"). So the sum is at least _not_ _finite_ (not in omega). >> > There is no L, nowhere. Hasty generalization. > It is wrong to say that L is in |N It is _false_. "L is in omega" is _not_ _true_. > and it is wrong to say that L is larger than any n e |N. _Nowhere_ did *I* write something like "L is larger than any n e N". This is probably your own wording. > Actual infinity does not exist! I don't discuss Cantorisms. > I told you that already several times. You persistently try to posit Cantorisms. > But refuse to understand what "to exist" means. An entity a exists if |= E a. > In no case infinitely many differeces of 1 can exist unless infinitely > many diferences of 1 do exist. But that means an infinite size. Balky reiteration. F. N. -- xyz |