An uncountable countable set
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From: Dik T. Winter on 17 Jul 2006 21:16 In article <1153169497.380974.32190(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > 0 was invented less than 2000 years ago. Natural numbers need not be > > > invented. However, Bourbaki and Halmos tried this trick in order to > > > prevent set theory from too easily been demasked as inconsistent. > > > > Eh? I was not talking about invention but about definition. What problems > > do you have with definitions? > > I don't like definitions which define nonsense like the corner of a > circle. Check the manhattan-measure and you will find square circles. But whatever, if you do not like definitions that define nonsense in your opinion, you should have a hard time with mathematics. > > Yes, it was sloppy terminology. What happens when n grows without bound? > > Nothing happens with the *+ sum. > > > What is lim{n -> oo} SUM{i = 1 .. n} A_i ? How do you define that? > > Without such a definition I have no idea what the result is when I *+ > > all An. > > If there is at least one 1 in a column then the *+sum is 1. You need > not investigate how many 1's are there to follow if you want to > calculate the *+ sum. That is not an answer to my question. But I will take it in good faith, so lim{n -> oo} SUM{i = 1 .. n} A_i = 0.111... tell me where I have gone wrong. > > > Define: If any case includes at least one 1 the the *+ sum is 1. > > > > Again, in the finite case. You have not defined what you mean with the > > infinite sum. > > Hell and devil! Can't you read? The definition for *+ sum = 1 is: at > least one 1 must be encountered. That is enough in any case, finite or > not. Damnation leaving aside, you defined what SUM{i = 1 .. n} An is. You did *not* define what happens when n grows without bound. Now you state that in each column, whenever there is a 1, the final result should be 1. But what you are now meaning is that lim{n -> oo} SUM{i = 1 .. n} An = 0.111... . > > So the statement > > For all n, An[n] = K[n] > > is true? As is the statement > > For all p there is an n such that An[p] = K[p] > > also true? > > No. Then 0.111... would not differ from every n. No to which question? Is the first statement false? And if so, why? Is the second question false, and if so why? And how do you come at your conclusion? I think your logic is lacking. > > Why than do you write that it is false? > > Because it is not correct. According to the axiom of infinity [...] > infinite sets can not be exhausted in this sense. (Dik T. Winter) > Therefore, a unary representation of a natural number can never reach > the line next to the unary representation of aleph_0. You are arguing on two different lines at the same time, confusing one with the other. When we consider the An as unary representations of natural numbers the result is aleph_0, not a natural number. When we consider them as being decimal numbers, the result is 1/9. In both cases the resulting number is not in the list. What is the problem? > Therefore, a unary representation of a natural number can never reach > the line next to the unary representation of aleph_0. The latter does > exist according to set theory. Can you prove that? If the line next to the unary representation of aleph_0 does exist (and I think you mean preceding line) you have to show that aleph_0 does have a predecessor. But it has not. Unless you show how it can be proven through set theory. > The natural next to it does not. Hence, > the sum of > > 0.1 > 0.11 > 0.111 > ... > is *not* 0.111... With your definition (finally extracted) above (if in any column there is a 1, there is a 1 in the final result), it is. If you think that is false, please show me a column where there are only 0's. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 17 Jul 2006 23:09 In article <1153169497.380974.32190(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > 0 was invented less than 2000 years ago. Natural numbers need not be > > > invented. However, Bourbaki and Halmos tried this trick in order to > > > prevent set theory from too easily been demasked as inconsistent. > > > > Eh? I was not talking about invention but about definition. What problems > > do you have with definitions? > > I don't like definitions which define nonsense like the corner of a > circle. Then stop making them. > > > Yes, it was sloppy terminology. What happens when n grows without bound? > > Nothing happens with the *+ sum. As "*+" does not have the common values of a "sum", it is WM making square circles again. > the line next to the unary representation of aleph_0. The latter does > exist according to set theory. The natural next to it does not. Hence, > the sum of > > 0.1 > 0.11 > 0.111 > ... > is *not* 0.111... > There is a difference between appending one new value at a time to a finite list and collecting all members of an endless list in one step. The first is not possible to reach an end of, the second ends when it starts. > If exhaustion were possible, then I offered the natural oder by size > and simultaneoulsy the well-order of the rationals. WM presents us with another of those corners of a circle.
From: Virgil on 17 Jul 2006 23:16 In article <1153169803.310772.70530(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > > It reads: Either there is a column with only zeros, or there is at > > > least one 1 in each column spanned by the digit positions of 0.111... > > > , isn't it? > > > But you will "argue": Nice try but there is neither nor. > > > > What you call a proof is an unproven proposition. > > IF aleph_0 does exist, THEN 0.111... covers aleph_0 columns. > IF 0.111... covers aleph_0 columns, THEN aleph_0 columns do exist. > IF aleph_0 columns do exist THEN we can consider their contents. > IF we can consider the contents of each column, THEN we can ask how > many 1's are therein. > IF we can ask how many 1's are in each one, THEN the answer can be > "zero 1's" or "not zero 1's". > IF the answer is in each case is "not zero 1's", THEN in each column > at least one 1 must be present. In fact, in each "column" infinitely many 1's are present. > > However, there is no natural numbers with this property Precisely. > because > 0,111... has more 1's than each natural number. As does each column. > Hence 0.111... itself > must be present among its disciples. Non-sequitur. Except for the leading "0" and "." there will be "111..." in each column. But excluding the first column of all "0"s there re no others and except for the 2nd column of all "."s there are no others.
From: mueckenh on 18 Jul 2006 06:54 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > [...] > >> Yes, it is larger than all naturals, but I would not call it *the* > >> successor, but *a* successor, or, if you wish, *the smallest* > >> successor. > > > > However, not all of its 1's in unary representation can be indexed by > > natural numbers because they are smaller. > > My understanding of unary represenations is: strange! > > n e omega: unary(n) def= (a_i) > a_i = 1 if 0 <= i < n a_i =/= 1 if n = 0. Hence (what means the same as therefore etc.) you should write "a_i = 1 if 0 < i <= n". > a_i = 0 if n <= i < omega a_i = 1 if i = n. Example: 0.11 is the unary represenation of 2, not of 3. > > omega: unary(omgea) def= (a_i) a_i = 1 A i e omega > > > This is the deep dilemma of set theory: There is no actually infinite > > set of finite numbers. > > Non sequitur. Ever considered _your_ representation theory broken? If you insist that 0.111 represents 4, then something with your representation theory must be broken. How many letters x do you see here: x. Is zero the correct answer? But even this approach of yours does not prevent the dilemma, because even there the *+ sum of all naturals is omega, the *+ sum of all naturals and omega is omega too. This is a contradiction, and it is solved only by the observation that the *+ sum of all naturals is not omega, because a *+ sum of finite numbers alone cannot result in an infinite number. Regards, WM
From: mueckenh on 18 Jul 2006 07:01
Virgil schrieb: > > > > The successor of all naturals is not a natural and, therefore, must be > > > > larger (because it is not less). > > > > > > There is no such thing as the successor of all naturals any more that > > > there is a single successor common to both 3 a and 6. > > > > This is an expression coined by Cantor: "a number following after all > > natural numbers". > > But he does not call it a 'natural' number any more than he calls it an > even number or a prime number. I said above: The successor of all naturals is not a natural. > > There is no number following all naturals and, hence, there is no > > number omega at all. > > There is no such 'natural' but there is such a cardinal or ordinal. Both > are generically 'numbers' not all cardinals or ordinals are naturals. Why do you emphasize this? Of course it cannot be a natural if it follows after all naturals. > > > > > > One has the set of all naturals, and that set can have a successor under > > > the definition that the successor of any set, x, is (x union {x}). > > > > That is nonsense too. One cannot have a set of all naturals, because > > the step x U {x} defined by the axiom does never result in this set. > > No one said that the set of naturals was a successor, stupid, we only say > that it has one. Some say that omega is the set of naturals. And Cantor said that omega was the smallest successor of all naturals. > > Misrepresentations like that are the tools of trolls. Those trolls which troll set theory. > > > You recently asserted that such stepwise processes can never exhaust a > > set. Now you changed your mind? Just in time? > > One can have a set of all of an inexhaustable supply of objects even > though one cannot exhaust that supply by serial operations. such as a well-ordering undoubtedly is, because one number follows the other in a sequence or a chain or in steps. Regards, WM |