An uncountable countable set
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From: Dik T. Winter on 6 Jul 2006 19:52 In article <1152208684.463235.35190(a)k73g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > Moreover, > > > > different orderings of the same subset of transpositions will yield > > > > different results. In the formula I gave for the diagonal number, > > > > calculation of one digit does *not* depend on the calculation of > > > > another digit. > > > > > > And what is the consequence of this? > > > > That the calculation of the digits can be done in parallel? > > It is done in zero time. It is determined from the beginning. Why > should something "be done"? It was you who remarked on the calculations... > > > > It is Cauchy's theorem that proves that the limit exists, using the > > > > epsilon argument. > > > > > > But Cantor's arguing is that without epsilon. > > > > I do not know Cantor's argument exactly. I think that he implicitly > > uses that result. In my formulation it was abundantly clear that I did > > use it. > > But he needs to consider every digit with equal weight. That is not a > limit process. What did he *mean* when he wrote "with equal weight"? For comparison purposes the digits have equal weight (i.e. it is not the case that some numbers on the list are less different from the diagonal than others). However, when you wish to know whether the number constructed is a real number you need limits. > > And again, it maintains the well order as long as you remain in the finite > > domain. So up to every finite n, you have ordered the first n elements > > of the well-ordered list, retaining a finite list. And back again to > > problem 1 with this. How do you define that when n grows without bound? > > I remain in omega! So I always stay with finite numbers, they may grow > as far as you like. Wha holds for small natural numbers does equally > well hold for big ones, as far as well-order is concerned. Yes, as I said, for every finite n your list of transpositions will have transformed the list (implying well-ordering) of rationals to another list of rationals. It will not have changed to the rationals under standard order. You have just sorted the first n rationals using a bubble sort. There are still infinitely many rationals that are not sorted according to their standard order. So you claim that with that process you get a well-ordering of the rationals with their standard order can only be made when, in some way, you define what happens when the process continues indefinitely. And for that you need some kind of limit. And next the question is whether well-ordering is preserved under taking the limit. And I think that is false. > > How do you define the "limit"? And if you define that, is that "limit" > > also well-ordered? Those are things you have to prove. > > I define limit by: *Using all finite natural numbers* just as like as > Cantor does. No. You do *not* and he does *not*. In the Cantor diagonal the number obtained is a real number using limits (in the Cauchy sense) with the definition of real number by many others (that can be proven to all be equivalent with each other). The limit used by Cantor is precisely the epsilon argument, together with majoration and minoration on the ordered set of rational numbers. By Cauchy any sequence of decimal digits has a limit, but it is not certain whether that limit is in the defined set of numbers. By Cantor, Dedekind, Weierstrass and a host of others (using various formulations), such a limit (when starting with rational numbers) is defined as a real number. That is the place where Cantor uses the limit (although he may not have expressed it as such), i.e. showing that the resulting number is a real. On the other hand, you have *not* defined what an infinite number of transpositions means. "Using all finite natural numbers" is not a definition. -- 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 6 Jul 2006 19:56 In article <1152208894.281047.49990(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1152193390.830398.326300(a)q16g2000cwq.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > > > Either the diagonal number 0.111... is not distinguished from all > > > > > finitely large numbers of the list > > > > > 0. > > > > > 0.1 > > > > > 0.11 > > > > > 0.111 > > > > > ... > > > > > then Cantor's proof fails. > > > > If we assume that that list contains natural numbers in some unary notation > > (as I think you do) than: > > > > > > > Or 0.111... is distinguished from all finitely large numbers of the > > > > 0.111... is not a natural number in unary notation. So it is inherently > > different from all elements of the list. > > OK. it is the unary representaton of omega. Interesting. What is the unary representation of w+1? But more clear, you admit that it is not a natural number, and so 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? > > I chose just this number because it is also the decimal representation > of 1/9. Interesting, although it makes no sense. > To be continued. I will not hold my breath. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 7 Jul 2006 10:22 Dik T. Winter schrieb: > > > > But Cantor's arguing is that without epsilon. > > I do not know Cantor's argument exactly. I think that he implicitly > uses that result. In my formulation it was abundantly clear that I did > use it. Cantor argues: If for all natural numbers something is defined (like the bijection n <--> 2n or the replacement of 5 by 4) then there is no further reason for any limit process. Cantor in his paper about the diagonal argument does not at all consider any limit process. Therefore: As my sequence of transpositions (1, 2) (2, 3) (1, 2) (3, 4) (2, 3) (1, 2) .... determines exactly which transposition will be executed as the n-th, everything is determined and everything is finite. > > And again, it maintains the well order as long as you remain in the finite > domain. So up to every finite n, THERE IS NO OTHER n ! ! ! > you have ordered the first n elements > of the well-ordered list, retaining a finite list. And back again to > problem 1 with this. How do you define that when n grows without bound? Also without bound n is always finite, or it would not be a natural number! > How do you define the "limit"? And if you define that, is that "limit" > also well-ordered? Those are things you have to prove. 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. Regards, WM
From: mueckenh on 7 Jul 2006 10:26 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. > > > 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. The case you mentioned above that position n may coincide but position n-1 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. This case would invalidate Cantor's diagonal proof. Regards, WM
From: mueckenh on 7 Jul 2006 10:30
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). >Now let a_n be the first element of the ordering > achieved by rearranging the first n elements of the original order. It > is clearly the minimum of the first n elements. The sequence <a_n> is > thus non-increasing and unbounded below (in the positive rationals). Let > f_n be the mapping from N which indexes the ordering achieved at the nth > stage. You claim this sequence "clearly" has a limit, say f. Given that > f_n(1) = a_n, what is f(1)? And don't say f(1) must be the non-existent > smallest positive rational and so set theory is inconsistent. You need > to prove, within standard set theory, that there *is* a well-defined > limit with the required properties before you can claim a contradiction. 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. Regards, WM |