An uncountable countable set
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From: Dik T. Winter on 18 Aug 2006 09:40 In article <1155899756.334481.23700(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1155817978.033024.136310(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > All digits can be covered, but there is no n that covers all digits. > > > > > > What is the result of your assertion? > > > There is a list number which covers only the first digits of 0.111..., > > > while another covers not the first digits but some more behind. > > > > That is nonsense. > > I know. I told you several times. The nonsense is that that is not the result of the assertion. > > Each finite initial segment can be covered by a natural > > number. The complete number can not be covered by a natural number. > > Hence it cannot be indexed. Again a conclusion without proof. Use my definition of K: K[p] = 1 for all p in N, and there are no other digits. 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. > > > > But *all* digit positions can only be indexed by natural numbers. > > > > There are no digit positions that can not be indexed by natural numbers. > > > > > > These positions are all in the list - by definition. > > > 0.111... is not in the list by mathematical proof. > > > > Yes. But why should it be in the list? It is not an index position as it > > is not a natural number. > > But you pretend it would consist of index positions which are all in > the list. Not pretend, define. > 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..."? -- 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 18 Aug 2006 09:50 In article <1155900019.373047.209670(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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... . The last sentence makes no sense, if x is real 1/x is never 0. But I now understand that the series: sum{n = 1 .. oo} (-1)^n/n does not converge according to your logic. Are you rewriting all of mathematics? > > > > 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 That makes no sense. As I read it, we have: |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings = 1 + 1 = 2. 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}| or as: lim{n -> oo} |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| -- 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 18 Aug 2006 10:01 In article <1155900165.554388.224080(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > 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? 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. > > > 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. 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? > > > > 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. 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. -- 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 18 Aug 2006 10:11 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". -- 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 18 Aug 2006 10:31
In article <1155900505.345504.196110(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. You keep stating that, but it is false when you assume the axiom of infinity. > 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? 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. > > 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. Makes no sense. The definition makes clear it is different for each line with index number n, where n is a natural number. So it is different for all lines that have a finite distance from the first line, and so for all lines (because there are no other lines). > > > > > 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? 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. But you asked for the *first* 10%, which is something different. > > 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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |