An uncountable countable set
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From: mueckenh on 18 Aug 2006 07:20 Dik T. Winter schrieb: > > > And my question is unanswered. When I try to extract a definition from > > > that, I find: > > > lim{n -> oo} 1/n > 0. > > > because 1/n never becomes 0. > > > > That is correct. There is no natural number of infinite size, therefore > > for every natural we have 1/n > 0. > > Fine, I now understand. lim{n -> oo} > 0. An interesting assertion. In principle one must distinguish (n e |N) --> oo and x --> oo. In the latter case the result of 1/x may be 0, but that is undefined as we have seen from 0.111... . > > > > If a > b, then a has a sequence or subset equal to b. Now you can > > > > define a/b = 1 + remainings. > > Your definition. > > > > Let us try it with: > > > > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 > > > let's say a_n = {2, 4, 6, ..., 2n} and b_n = {1, 2, 3, ..., n}. What is > > > the subset equal to b_n in a_n? What are the remainings? > > > > Both sets have same cardinality. |b_n| = |a_n|. There are no > > remainings. Therefore the result of the division is 1. > > Your definition was about sets, not about cardinalities. What are the > remainings? According to the definition above I have no idea what the > remainings are. Can you define remainings in terms of cardinalities? |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings where the remainings are |{n+1,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 Regards, WM
From: mueckenh on 18 Aug 2006 07:22 Dik T. Winter schrieb: > > > > > Yes, of course. And what is your problem with the edges (how do they > > > > differ from the lines of Cantor's list)? > > > > > > The lines are countable (by assumption). But pray assign natural numbers > > > to each edge in your tree. I will even allow you to assign finite > > > sequences of digits 0 (go left) and 1 (go right) to your tree (the > > > mapping to natural numbers is quite standard in that case). What is the > > > finite sequence of digits assigned to the edge that leads to 1/3? You > > > claim that the edges are countable, so you should have an answer to my > > > question. > > > > The sequence of edges is countable but it is not finite! > > The sequence of naturals is countable but it is not finite! Why then did you want to allow to assign finite sequences of edges only? > > > 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. > > Edge 1 leads to 0.0 > Edge 4 leads to 0.01 > Edge 9 leads to 0.010 > Edge 20 leads to 0.0101 > > What edge leads to 1/3? I did not say that there is one single edge that leads to 1/3. > > 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. Then I proved by another means (rational releation) that the set of edges does not contain less elements than the set of paths. Regards, WM
From: mueckenh on 18 Aug 2006 07:24 Dik T. Winter schrieb: > In article <1155818644.699954.82400(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <J42BHx.IzE(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > > > In article <1155664930.866986.157410(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > ... > > > > > What I do not understand is how you can believe that someone (except, > > > > > perhaps, Virgil) could share your opinon that only in one dimension the > > > > > infinity was actually realized, but not in the other. > > > > > > > > Just because there is no last line. If there were a last line, it would > > > > have width aleph-0, but there is no last line. > > > > > > Let's have an easier example. The blocks are 1/2^n high and 1-1/2^n wide. > > > > I propose the following stair > > 1 - 1/2^n high and 1 - 1/2^n wide > > up to the n-th step, in order to have symmetry. > > I propose not (and I did not use those values). > > > > 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. Regards, WM
From: mueckenh on 18 Aug 2006 07:25 Virgil schrieb: > In article <1155817888.248664.91020(a)i42g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > Nobody ever has said that the diagonal is ever completed. There is a > > > > > definition of that number, and it is easy to show, with that > > > > > definition, > > > > > that that number (when completed) would be different from all numbers > > > > > on > > > > > the list. > > > > > > > > But if it is impossible to complete it, then the result is void. > > > > > > There are many real numbers that can not be completed. But by the > > > definition of "real number", each sequence of decimal digits defines > > > a real number, and it is easily shown, by the above, that that real > > > number is not in the list because it is different from each number > > > in the list. > > > > > > Oh no. It is not different for any infinite line but only for some > > finite lines. > > > Which numbers in the list is it NOT different from? > > We can show it is different from the first. > > We can show that if it is different from any one of them, it is also > different from the next one. > > By the inductive principle, which is valid in our axiom systems, it is > thus different from each and every single one of them. By the induction principle every line has a finite distance from the first one, hence there is no infinite set of lines. That is potential infinity. The induction principle does not cover the whole set of natural numbers. But if you admit that the induction principle is valid for the whole set of natural numbers, then you can prove that lim[n-->oo] |{2, 4, 6, ...., 2n}| / 2n > 1 i.e. lim[n-->oo] 1/2 > 1. Regards, WM
From: mueckenh on 18 Aug 2006 07:28
Dik T. Winter schrieb: > > > f the real cannot be > > completed, then it is unknown, whether it differs from every entry of > > the list. > > The number is different from each finite line. And as there are only > finite lines, it is different from all lines. No, if there were only finite lines, then the number of lines was finite too. The line n which you really can check is always in a finite distance from the first one. And the set up to this line is always a finite set of n lines. This means that you cannot check all lines, if indeed there should be infinitely many (which means: more than any finite number n = more than any you can check). 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? > > I stated this before, but you do not agree, there are sets without a > last element. This is what the axioms tell me. You may disagree with > that, and reject the axiom of infinity, but if you want to show an > inconsistency with the axiom of infinity you have to show how using > that axiom you can prove A and ~A. Your statement above "It is not > different for any infinite line" is a direct rejection of that axiom. No, what I meant is: it is not different for any line which has a finite distance from the first line. > > > > Here you see the inconsistence of set theory. The set shall exist > > > > actually = completely, but the first 10 % shall not exist. > > > > > > What is the inconsitency? Can you show what axiom of set theory or what > > > proposition from set theory it violates? > > > > If you tell me what is to be understood by "set" and by "existence" in > > set theory. > > Sorry. You made a statement. I asked you a question that you support your > statement. Now you tell me that you can not support that statement unless > I provide some definitions? To be precise, you state: > Here you see the inconsistence of set theory. > I ask: > What is the inconsistency, can you prove it? > So your initial statement is completely unfounded, because you stated it > without additional information and want information to give a foundation. 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? > > 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. Regards, WM |