An uncountable countable set
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From: Dik T. Winter on 7 Jul 2006 11:10 In article <1152282364.618937.306970(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > 0.111... is not a natural number in unary notation. So it is inherently > > different from all elements of the list. > > > > > "To be different" means for all unary representations of n > > > An : 0.111... - n =/= 0 > > > > How do you propose to define that subtraction when 0.111... is not a > > natural number? > > It is not by pure accident that I chose 0.111... . This number has also > a meaning as decimal representation of 1/9. So we can do the > calculations. I have no idea what you are meaning here. > > > not forall n: 0.111... - n =/= 0 > > > > But that is a wrong conclusion. Let's call 0.111... (as a sequence of > > digits) K. And let's define K[i] is the i'th digit of K and An[i] the > > i'th digit of An. The following statement is correct: > > There is no i such that for all n K[i] != An[i] (1) > > > > > En : 0.111... - n = 0. > > > > And that is wrong, because that means: > > There is an n such that for all i K[i] = An[i]. (2) > > pray tell us under what logical reasoning you transform (1) to (2). > > > > And to simplify it, take A0 = 0.10, A1 = 0.01, K = 0.11. (1) is > > satisfied, (2) is not satisfied. > > Your error is the invention of numbers like 0.01. They are not unary > representations of natural numbers. > By construction of the list we have to compare K = 0.111... only with > numbers > > 1 0.1 > 2 0.11 > 3 0.111 > ... > n 0.111...1 > ... > > Those are unary representations of 1, 2, 3, ..., n, ... which we may > abbreviate conveniently by n. > > There are (by tertium non datur) two possible alternatives: > 1) K has a 1 at a position p where no n has a 1. > E p A n : p =/= n > 2) K has not a position p where no n has a 1. > ~(E p A n : p =/= n) > > If (1) is true, then the position p in question cannot be enumerated by > a natural number, hence it is undefined. > If (2) is true, then we observe that for any position where the 1 > coincides in p and n also all positions with m < n do coincide. Indeed. > The > case you mentioned above that position n may coincide but position n-1 > may not, simply cannot occur. Where do I mention such a case (except in my example to show the fallacy of your reasoning). > may not, simply cannot occur. We get from (2) > A p E n : p = n > and by construction of n = 0.111...1 we obtain that all 1's in K > coincide with the 1's of n. Right. But this does *not* prove that there is an n that is equal to K. To reformulate clearly For all p there is an n such that An[p] = K[p] this does *not* imply that There is an n such that for all p An[p] = K[p], which is what you are arguing. Pray explain how you come (in logical steps) from the first statement to the second. -- 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 7 Jul 2006 13:52 In article <1152282125.757193.28320(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > There is no limit other than in Canrtor's diagonal. EVERY set initially > indexed by natural numbers will remain indexed by natural numbers. And > it will unavoidably become ordered by size too. At best you have proved that a countable subset of the rationals have ben restored to their usual ordering, but you have nowhere shown that your process ever re-orders ALL of them. After each of your countably many transpositions, there is still a countably infinite subset of the rationals left which are not in natural order. That is a necesssary consequence of the entire set being originally well ordered.
From: Virgil on 7 Jul 2006 14:00 In article <1152282364.618937.306970(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > This case would invalidate Cantor's diagonal proof. "Mueckenh" has merely manages to prove that there is a string not representing a natural number. This has nothing whatever to do with whether or not there is a surjection from N to a non-empty interval in R, nor whether Cantor"s "diagonal proof", or any of the many other proofs, establishes that no such surjection exists.
From: Virgil on 7 Jul 2006 14:17 In article <1152282618.774946.24830(a)s16g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Hartley schrieb: > > > In message <1152191768.157605.15960(a)b68g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de writes > > > > > >David Hartley schrieb: > > >> > > >> Infinite processes can be brought to an end if they have a clearly > > >> defined limit. You haven't defined what you mean by the limits of the > > >> orderings or mappings produced by your process. To make "this result > > >> occur unavoidably" you need to give definitions of both limits and show > > >> that the limit of the mappings is an order-isomorphism from N to the > > >> limit of the orderings. Using the definitions I offered, the sequence of > > >> orderings does have a limit, and it is the usual ordering of the > > >> positive rationals. The limit of the associated sequence of mappings > > >> from N would be an order-isomorphism if it existed but - no surprise - > > >> it doesn't. > > > > > >The limit is clear. The question is why it is not reached. > > >> > > >> Unless you at least try to define these limits, and to prove the > > >> required relationship between them, I see little point in continuing > > >> this discussion. > > > > > >Here is a set of transpositions of order type omega. Therefore we stay > > >once and for all in the finite natural numbers. > > >Start with transpositions of the indices: > > > (1, 2) > > > (2, 3) > > > (1, 2) > > >This orders the set of the first three elements by size. Then order the > > >set of the first 4 elements by size. > > > (1, 2), (3, 4) > > > (2, 3) > > > (1, 2) > > > > > >And so on. After the set of the first n elements has been ordered by > > >size order the set of the first n+1 element by size. At most n-1 > > >transpositions are required for a set of n elements. Now, the complete > > >set of transpositions is of order type omega. And it maintains the > > >well-order by index while it achieves well-order by size. > > > > Your quoted transpositions don't alter the ordering as you want, > > ? > > > but the > > intention is clear and that does work. (It is the sequence I suggested a > > couple of days ago.) > > Are you sure? As far as I remember, your sequence lead to oscillations > such that two elements changed there places on and on. My sequence > leads unavoidably to an order by size for any set indexed with natural > numbers (i.e. of order-type omega). No one has shown that any sequence of transpositions, nor any sequence of "finite" permutations (which leave all but finitely many elements fixed), can convert a well-ordering into a non-well-ordering. > > I need to prove that there is not a contradiction before I claim a > contradiction? > > If the whole well-ordered set 1/1, 2/1, 1/2, 3/1, 1/3, 4/1,... does > exist, then this whole set can be ordered by size using the countable > and well-ordered set of my transpositions. This is the limit set: Alle > positive rationals ordered by size. If this cannot be done, the > question is simply: why? As the transpositions work perfectly for any > finite set, the only reason can be the non-existence of an infinite > set. Since "mueckenh" cannot work with infinite sets, how can he be so certain of what can or cannot be done with them? While a transposition can increase by one the *finite* initial segment of the list of rationals which are also ordered by size, it can never decrease the infiniteness of the terminal segment which is not so ordered. So that each new transposition faces exactly the same problem as its predecessor, an infinite terminal list of rationals not in natural order. If there could be any permutation in the list of permutations which did not face the initial problem all over again, there would have to be a first one, but that is impossible.
From: Virgil on 7 Jul 2006 14:34
In article <1152282981.126840.166280(a)s53g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152182749.926627.310460(a)a14g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Dik T. Winter schrieb: > > > > > > > > > > > All sequences, even all single transpositions which I apply can be > > > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > > > > > > Yup, so there are infinitely many. > > > > > > Are yo sure? How could an enumeration by natural numbers supply > > > infinity? > > > > If an enumeration rule exhausts the naturals, the set being enumerated > > is Dedekind infinite. > > > > > > > > There are only finite natural numbers. > > > > But infinitely many (an endless supply) of those finite natural numbers. > > That does not make their sizes infinite. It is completely irrelevant > here. What IS irrelevant is the notion that in order to have infinitely many naturals (more than any finite number of them) one must somehow have one of them being infinitely large. > > > > > I have shown that the transpositions can be enumerated by natural > > > > > numbers. There is no number oo. There is neither such a number in > > > > > Cantor's list. All we do is enumerated. Othewise it would not be > > > > > defined at all. Neither in Cantor's list. Therefore it is irrelevant > > > > > whether or not something has to be "executed in order". We are in > > > > > the > > > > > countable domain and do not leave it. > > > > But transpostitions do not always commute: As right operators on a list > > (a b) (b c) = (a c b) but (b c) (a b) = (a b c) > > > > So changing their order of execution can change their effect. > > Why should we change their order? You have posited an infinite sequence of infinite sequences of transpositions to perform your alleged miracle. As you can never even finish the first subsequence in finite time, you must reorder your transpositions into a single sequence to even consider its effect. > > > > > > > > That makes no sense, again. The transpositions have to be execute in > > > > the > > > > order given. > > > > > > Of course. But that does not exclude that these transposition can be > > > executed and finished (if Cantor's list can be finished). Cantor does not have a list. Only those who challenge him need to have lists. What Cantor does is to refute each list presented to him.. > > > > But it prohibits them from being executed out of their prescribed order. > > Since the order of execution induces a well ordering of the set of > > transpositions, there would have to be a first transpostion producing > > any given effect. What is the first transposition on a well ordering of > > the rationals producing an ordering that is dense? > > What is the first real number which cannot be named by a finite set of > letlers? There are such numbers. Which is the first, e.g., by size, or > in an assumed well-ordering of the reals? But in ZF, there is no such assumption. Anyway, one cannot name a number which by its very definition cannot have a name. Even if such a number were known to exist, one cold not name it. Besides, properties of the reals are not the issue here. "Meckenh" claims to be able to have the rationals simulteneously well-ordered and naturally ordered, but has not been able to justify his claim. |