An uncountable countable set
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From: mueckenh on 18 Jul 2006 08:18 Dik T. Winter schrieb: > In article <1153145345.281803.129520(a)35g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > ... > > > > Indeed. I recall. Therefore the linearity of the list numbers enforces > > > > a column with zeros. > > > > > > Nice try, but non sequitur. There is no such column. Otherwise show one > > > or prove its existence. > > > > The proof requires logic. Therefore I am afraid you will not accept it. > > 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? > > Yes, right. And now the remainder of the proof, please? If the *+ sum of the list is a unary representation of some 1's, then, by the construction of the unary numbers of the list, this sum must be a unary numer of the list. This is so whether you pretend it would not be so or not. A mathematic which does not reflect this fact does nt deserve its name. Regards, WM
From: Franziska Neugebauer on 18 Jul 2006 08:26 mueckenh(a)rz.fh-augsburg.de wrote: > 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. Of course. unary(0) = 000... The unary representation of 0 (the empty set) is the empty string (stiputalting only 1's are written). > Hence (what means the same as therefore etc.) you should write "a_i = > 1 if 0 < i <= n". If you are confused by the range you may drop "0 <=" entirely, since A i (i e omega -> 0 <= i). +--------------------------------------------+ | a_i = 1 if i < n. (sic!) | +--------------------------------------------+ >> 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. I presume omega = { 0, 1, 2, ... } and I presume i e omega. Therefor unary(2) = 11. The positions 0 and 1 are 1. If you are confused by the range you may drop "i < omega" entirely, since A i (i e omega -> i < omega). +--------------------------------------------+ | a_i = 0 if n <= i (sic!) | +--------------------------------------------+ summing up: a_i = 1 if i < n (1a) a_i = 0 if n <= i (1b) A tabulation may help you to understand what I mean: unary(n) (only 1's n 1-allocated i's 0-allocated i's written) -------------------------------------------------------------- 0 (none) 0 1 2 ... "" (empty string) 1 0 1 2 3 ... 1 2 0 1 2 3 4 ... 11 3 0 1 2 3 4 5 ... 111 ... .... ... .... If you want to drop index and number 0, you can simply use omega \ { 0 } instead of omega. >> omega: unary(omgea) def= (a_i) a_i = 1 A i e omega With (1a) and (1b) it is even not necessary to define unary(omega) specially. >> > 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, I never did and will not. > then something with your representation theory must be broken. Dead horse. > How many letters x do you see here: x. Is zero the correct answer? saucy. > 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. ,----[ http://en.wikipedia.org/wiki/Dilemma ] | A dilemma is a *problem* offering two solutions, neither of which is | acceptable `---- Please explain in detail what the *problem* is. I can't see any problem. > This is a contradiction, Once again: I can't see a "problem" or a "contradiction". Would you like to explain, what exactly _is_ the problem? > and it is solved only by the observation that the *+ sum of all > naturals is not omega, You are trying to "solve" a yet non-existing "problem". BTW: This can not be "ovserved" since every "observation" of the present *-sum of all unaries of naturals _is_ obviously representing omega. You confuse observation by hallucination. > because a *+ sum of finite numbers alone cannot result in an infinite > number. A *-sum of _infinitely_ _many_ representation of finite numbers does in the present case result in a representation of an infinite number. This is a fact. F. N. -- xyz
From: Franziska Neugebauer on 18 Jul 2006 08:49 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> > 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". >> >> We can. >> >> > IF the answer is in each case is "not zero 1's", THEN in each >> > column at least one 1 must be present. >> >> This is the case, since every a_jj = 1 j e N by definition. >> >> > However, there is no natural numbers with this property, >> >> Could you precicely _define_ which /property/ you are talking about? > > To contribute a 1 to each column. Should be a property of which object? >> For every column j e N a_mj has the 1 in position m(j) = j, since >> a_jj = 1 A j e N. Where exactly lies your problem? > > I told you recently: In a list like mine a *+ sum is defined. This * > sum is equal to the largest number of the list. The notion of "largest number" is misleading. There is no largest number. Neither in the list nor in omega. If it makes sense to talk about "a largest number", then please 1. name it (show us its name or position _in_ the list or _in_ omega), and/or 2. present a _proof_ that it exists. Until then we cannot (meaningfully) talk about "a largest number in the list" or "a largest number in omega". Wittgenstein applies here. > And if a larger number is included, then the sum grows. You pretend > that this is not valid for 0.111... because without 0.111... the list > has the *+ sum 0.111... and when you include it in the list (as > diagonal number or as the first line or anywhere else) the *+ sum does > not grow. Postponed until existential status of "largest number" is clarified by WM. F. N. -- xyz
From: Dik T. Winter on 18 Jul 2006 10:15 In article <1153225087.356970.296750(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153145345.281803.129520(a)35g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Franziska Neugebauer schrieb: > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > ... > > > > > Indeed. I recall. Therefore the linearity of the list numbers enforces > > > > > a column with zeros. > > > > > > > > Nice try, but non sequitur. There is no such column. Otherwise show one > > > > or prove its existence. > > > > > > The proof requires logic. Therefore I am afraid you will not accept it. > > > 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? > > > > Yes, right. And now the remainder of the proof, please? > > If the *+ sum of the list is a unary representation of some 1's, then, > by the construction of the unary numbers of the list, this sum must be > a unary numer of the list. As long as your list is finite. Suppose the list is infinite, and the *+ sum of all members of the list is a member of the list. That would mean that the *+ sum of the list is the last element of the list, contradicting that the list is finite. Hence the assumption is false. -- 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 18 Jul 2006 13:01
Dik T. Winter schrieb: > Sorry, we are now talking about 0.999..., please remain with the argument. > You said that the definition is silly. Why is the definition above silly? > And that there are undefined digits in irrational numbers does not bother > me in the least. I have no idea why you have a problem with that. In this case not the digits are undefined but the indices are undefined. See the discussion of the *+ sum of my list of unary numbers. > > But to go on: > > > And therefore we have *always* undefined digits in any > > irrational number and in the diagonal of Cantor's list. There are > > *always* most of its digits unknown, i.e., always there are more digits > > unknown than are known. No matter how small an epsilon you select. > > For Cantor's diagonal no epsilon is needed, neither is their for irrational > numbers. The only thing we need to know is that the digits are all in the > range [0, 9], The only thing we need to know for the binary tree is that no path can split without two additional edges. Why don't you use this kind of abstract mathematics in that case too? > > > Hence you cannot prove that the digits of the diagonal are all > > different from those of the line numbers. > > Why not? For the proof no exhaustion is needed. Just like (in another > thread) the proposition For the binary tree no exhaustion is requird either. It suffices to see that | o /\ is the ever repeating element which cannot be outwitted. > For all p there is an n such that An[p] = K[p] > does not need exhaustion. In the case of the list and the diagonal the > similar statement is > For all p there is an n such that An[p] != D[p] > proving that D is different from all An. BTW, in the first case also the > proposition > For all p there is an n such that An[p] != K[p] > proving that K is also different from all An. > In the first and second case take n = p, in the third case take n = p + 1. > This is *not* a proof by exhaustion. It is simply stating a fact, with > an easy proof. > > > You can prove it for each one > > but you cannot prove it for all. > > If something is true for each one, it is true for all. It is true for each natural, that it cannot index all 1's of 0.111... . > Let's reason > with the excluded middle (because that would complicate matters). Assume > it is not true for all, then there must be some for which it is not > true (say z). On the other hand, it is true for each one, so also > true for z. A contradiction, hence the assumption is false. And in > the cases above you can proof it for each one in one sweep, because > you prove it for an arbitrary element. It is true for each natural, that it cannot index all 1's of 0.111... . > > > That is the same problem as with the > > *+ sum of my list. You can prove the sum is 1 for each column but you > > cannot prove it for all columns. > > But you can. > > > Because stepwise exhaustion of > > infinite sets is impossible. Otherwise my (and Cantor's) reordering > > could be completed. > > If stepwise exhaustion is impossible, your reordering could be completed. > But you can with your *+ operation give the proof in one sweep, but you > failed to answer to my question: "what happens at infinity?". 1) There is o infinity. Ever Transposition has a natural number. 2) The same happens as with Cantor's diagonal. There is absolutely no difference. > > But, that was precisely what I intended when I asked you what you meant > with "*+ ..." in 0.1 *+ 0.11 *+ 0.111, because you gave no definition. > As you gave no meaning to that infinite "sum", only to finitely many > of them (and in that case going to infinity is impossible), there was > no way with your definitions. > > > > > But the assertion > > > > that the digits of these limits could be used to construct a diagonal > > > > number is simply nonsense. > > > > > > Only assertion and meaning. No content. > > > > Proven by epsilon. Set theory lives by contradiction. Some exhaustions > > of nfinite sets are accepted other are not. > > You can exhaust an infinite set if you take out all elements at once. You may think that would be possible, but it is not. You always need the belief expressen by the three "..." > That is why the function f(n) = n + 1 is a bijective mapping from > {1, 2, ...} to {2, 3, ...}, sic! > every definition works at once, you need > not to count to 100 to know what f(100) is. And the same is the case > with Cantor's diagonal number. You need not look at the first 100 > elements to find the 101-st element to know that the 101-st digit > of the diagonal is. That is what the definition of a list is. A > mapping from N to the members of the list. You need not look at the transposition number 100 to see what it does. > > Your set of transpositions are different. They do not work at once, They do work at once. Once they are defined the result cannot be changed. > you > need sequencing (transpositions are in general non-commutative). In the > case of a sequence you need limiting procedures to show what would be > the "final" result (like lim{n -> oo} 1/n = 0). But the problem with > limiting procedures is that the "final" result has not necessarily the > same properties as each and every finite result. Airy-fairy. My transpositions converge, i.e., a partial order by size can never become worse but can always only become better. I think there is no proof equired to see that. > > Let me give an example with transpositions. Let's say that (a, b) > means that we interchange the a-th and b-th element in an ordering. > Let us start with the ordered set of natural numbers {1, 2, 3, ...} > (yes, not Bourbaki this time). Let's define a sequence of transpositions: > (1, 2)(2, 3)(3, 4)(4, 5)... > with a proper measure and a proper definition of limit, I think that we > can show that that leads to the ordered set {2, 3, ..., 1}. Now interchange > the first two elements of the transpositions, in this case we get > {1, 3, ..., 2}. So sequencing plays a crucial role. But that is completely uninteresting, because my transpositions do never lead off an ordering by size. What |