An uncountable countable set
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From: mueckenh on 18 Jul 2006 13:07 Dik T. Winter schrieb: > In article <1153148551.942037.97110(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > In article <1153052131.852951.273540(a)b28g2000cwb.googlegroups.com> muecke= > > nh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > ... > > > (1): 0.111... is not *the* successor of anything, we may sloppily say that > > > it is *a* successor of all naturals, just like 10 is *a* successor of > > > 2 > > > > It was Cantor who coined this expression saying " daß omega die erste > > ganze Zahl sein soll, welche auf alle Zahlen nu folgt." (Works, p. > > 195) So it must be larger anyhow. > > Yes. Not *the* successor, but *a* successor. The "first" successor. Tere is only one "first" successor! > > > > > The successor of all naturals is not a natural and, therefore, must be > > > > larger (because it is not less). > > > > > > 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. > > Are you again arguing that the statement > For all p, there is an n such that An[p] = K[p] That statement is false. (Without K including the index p = aleph_0, k cannt be different from all natural numbers.) > is false, or are you arguing that the statement > For all p, there is an n such that An[p] != K[p] > is false? That statement in the form An != K is correct. The seqence An[p] does not exist for all p which are in K. > > > There is no actually infinite set of finite numbers. > > Axiom of infinity. In contradiction with mathematics, obviously. Regards, WM
From: mueckenh on 18 Jul 2006 13:20 Dik T. Winter schrieb: > In article <1153169497.380974.32190(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > I don't like definitions which define nonsense like the corner of a > > circle. > > Check the manhattan-measure and you will find square circles. Then I'll better do without. > 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. You cannot exhaust the naturals. Therefore what you write as 0.111... is not the same as the unary representation which also is denoted by 0.111... but means that N is exhausted and there is at least one 1 which cannot be indexed by a natural number. The reason is that 0.1 0.11 0.111 .... gives allegedly the same sum as 0.111... (as representation of aleph_0) 0.1 0.11 0.111 .... > > > > > 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. Here is another example for the *+ sum: 0.1 0.11 0.111 The sum is 0.321, the *+ sum is 0.111. > But > what you are now meaning is that lim{n -> oo} SUM{i = 1 .. n} An = 0.111... . I said above: It is different from n really reaching infinity. But for 1/9 or aleph_0 infinity is reached. > > > > 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? The reason is that 0.1 0.11 0.111 .... gives allegedly the same sum as 0.111... (as representation of aleph_0) 0.1 0.11 0.111 .... This could not happen if 0.111... was a number with more 1's than any natural number (in unary representation). > > > > 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. How many digits has 1/9 in decimal representation? > 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. That column which cannot be covered by a natural number. (f all couldcovered then 0.111... was a natural.) Regards, WM
From: Virgil on 18 Jul 2006 14:38 In article <1153220508.661035.130520(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. No one else even says that such a thing exists. There is a set of all naturals but if is not the successor of any of them. > > > > There is no number following all naturals and, hence, there is no > > > number omega at all. Perhaps not in your philosophy, but you are not God, to command what is true or false. > > > > 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. I emphasize it because WM needs to learn it. > > > > > > > > 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. Does WM claim that the only sets that can exist must be of form (x union {x})? Then WM is wrong. > > > > 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. Not quite, omega is the cardinality of the set of naturals. Do try to get things right, WM. It makes you look so silly when you make such silly mistakes. > And Cantor said that omega > was the smallest successor of all naturals. Not quite. Cantor said that omega was the smallest cardinality greater than that of any natural. WM is being silly again. > > > > Misrepresentations like that are the tools of trolls. > > Those trolls which troll set theory. Which is what WM is doing. > > > > > 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. Well ordering is defining an order in which each non-empty subset has a first member. There is nothing that requires the defining process to be sequencial. One can, for example, well order the rationals by injecting them into an already well ordered set like the set of naturals, which will induce a well ordering on the rationals > > Regards, WM
From: Virgil on 18 Jul 2006 14:41 In article <1153222834.202538.101020(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > WM's version of logic is quite different from the sort used in ZF of NBG > > or any other part of mathematics. > > No the logic used in mathematics is the same as mine. That is WM's perpeetual delusion. > It is only in > modern set theory that such silly things as the rejection of the binary > tree occur. It is WM's sillyness which rejects the infinite binary tree.
From: Virgil on 18 Jul 2006 14:47
In article <1153223055.199107.315230(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > The difference is that our "traditions and folklore", which we chose to > > call axioms and definitions, are logically consistent, > > Only if you decide *arbitrarily* which infinite set can be exhausted > and which cannot. > Regards, WM On the contrary. Whether one can "exhaust"( deal with every member of ) such a set depends on one's methodology. If one attempts to exhaust ( deal with every member of ) any infinite set by operating on its members one at a time, it is not possible. If one attempts to exhaust ( deal with every member of ) any infinite set by simultaneously dealing with all its members, it is quite easy. At least for everyone except WM. |