An uncountable countable set
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From: David R Tribble on 8 Aug 2006 18:52 >> Yup, and aleph-0 is *not* in the inductive set of natural numbers. > Mueckenh wrote: >> It is not in the set, but the number of the elements of the set is >> aleph_0. Under this aspect aleph_0 should have some existence in the >> set, namely as the number of elements. Of course it doesn't. Therefore >> it cannot be at all. > That statement leads to a contradiction. If Aleph_0 is a member of the set of naturals, then it is a natural. But then Aleph_0+1 must then also be a natural. And if Aleph_0 is the cardinality of N, then Aleph_0+1 cannot be a member of N. But Aleph_0+1 must be a member of N because it is a natural. Contradiction: Aleph_0+1 cannot be both a member and not a member of N. Since your statement results in a logical contradiction, your statement must be false. So Aleph_0 cannot be a natural nor a member of N.
From: Dik T. Winter on 8 Aug 2006 21:14 In article <1155067631.761632.250240(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > But the *+ sum is defined for all digit positions which can be indexed > > > by a natural number. > > > > You still fail to see. There is a definition for (*+){k = 1 .. n} Ak, > > I still do not see a definition for (*+){k = 1 .. oo} Ak. Can you, > > please, once give a proper definition? > > There is no natural number oo. Therefore {k = 1 .. oo} is nonsense. > There is, as you say, a definition for (*+){k = 1 .. n} Ak. More is not > possible. More is possible, when you define what the notion means. But as you never define what the notion means, your arguments fall flat in their face. Balderdash. > > > n is a natural number. An index is a natural number. "Index" and > > > "natural number" are synonymous. > > > > You are assuming that only digits of natural numbers can be indexed. > > But 0.111... is not a natural number. And you are stating (in fact): > > (every digit of 0.111... can be indexed) ==> (0.111... is in the list). > > so, why does this follow? > > Because: index = natural number = "number that can be indexed" Again you are using a term in a single sentence to mean two different things. "number that can be indexed": if you mean a natural number can be assigned to it, I agree. If you mean a number for which each digit can be indexed, I disagree. So, what do you mean. > > > > Yes. This does *not* proof that the *+ sum of them all (once defined) > > > > is in the list, because there is no proof that it is an index. > > > > Rather, there is an easy proof that it is *not* an index. > > > > > > Then it cannot be indexed. Then it contains digit positions which are > > > not indexed by natural numbers. > > > > You state so, and remain stating so, without any proof. > > It is a definition. *What* definition? There is, as far as I know, no definition that states: A number can be indexed if all its digit positions can be indexed. so what definition are you alluding to. Pray, for once, be clear. > > > Indexing is a symmetric operation: > > > A digit position n can be indexed by a natural number 0.111...1 (with n > > > digits) if and only if the natural number (in unary representation) > > > can be indexed by the digit position of said number. > > > > Assuming again n natural I think, and unary representation? But this > > tells *nothing* about the indexing of digits of non-natural numbers. > > It is impossible, because index number = number which can be indexed. Again, you fail to provide a definition for "can be indexed". In normal terminology, a number can be indexed if we can assign a natural number with it in the indexing process. You appear to be meaning something completely different, namely: a number can be indexed if all its digit positions can be assigned a natural number to it. The two are not equivalent, unless you prove it. > Or he recognized that the first ordinal of a number class may well > serve as the cardinal of this class. But this topic is not very > interesting to me because neither of these numbers are numbers. So you know acknowledge that your statement: > On the contrary! Cantor distinguished very clearly in all his written > papers. was 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: Dik T. Winter on 8 Aug 2006 21:19 In article <1155067888.355942.225340(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > You are claiming that K is in the list of indices. That makes no > > > > sense at all. > > > > > > I am claiming that every index is in the list of indexes. This list is > > > infinite. There are infinitely many indexes. And, yes, 0.111... is not > > > among them and therefore cannot be indexed. > > > > No proof, yet. K = 0.111... can also be written as "for every n in N, > > K[n] = 1". > > Every number which can be written in this manner is contained in the > list, because there is every n and every digit which can be indexed. Makes no sense at all. Proof please. > EVERY INDEX IS THERE. Every index position is there. It is the set of > all indexes and o all index positions. But 0.111... is not there. > Because this number cannot be indexed. But when I state "for every n in N, K[n] = 1" can the number K not be indexed? To be clear, what do you mean with "can be indexed". You are using two different, conflicting, definitions in your articles. -- 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 8 Aug 2006 22:38 In article <1155069769.640195.127530(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1154968120.846531.209650(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > You agreed above that the triangle gets infinitely long and infinitely > > > broad. > > > > Yes, I ask for a proof that infinitely broad means infinite natural > > numbers. > > The triangle is defined as representing the natural numbers. The third > line contains he number 3 represented by three "x". There is no proof > because this is a model. So you acknowledge that there is no proof. On the other hand, you assert that it is in the model. I need proof of that. Pray show that infinitely broad means infinite natural numbers. > > If you complete the triangle (conforming to > > the axiom of infinity) the length of the list is infinite, > > of course, it has aleph_0 lines. > > > as is the width, > > Wrong. There are aleph_0 numbers, but all of tem are finite. Yes. But the width is infinite. > > but there is *no* element of the list that has infinite width, because > > there is no last element. (If there were a last element, indeed, that > > element would have infinite width, but as there is none...) > > So there are *not* aleph_0 lines now? Ah, you are assuming that a list of aleph-0 lines has an aleph-0'th line? You are wrong. There simply is no last line. Assuming that is in contradiction with the axiom of infinity. If you want to show that the axiom of infinity is self-contradictory you should show that contradictory facts can be proven based on that axioms. But in your reasoning you may *not* assume things that are in contradiction with it. > > The last one would be interesting. What does it mean? But I want to see > > a proof of "infinitely many natural numbers is equivalent to infinite > > natural numbers". That is what you asserted. > > You see it by your illogic dancing around the list above. Getting offensive? What does oo = oo mean? > > Yes, see above. > > Nothing tosee above except illogic dancing. Ah, yes, getting offensive. > > Still your assertion that "infinitely many natural numbers > > is equivalent to infinite natural numbers". Where is the *proof*? > > Note that the axiom of infinity asserts otherwise. And still assuming > > that you want to prove that that axiom is consistent, you should prove > > that assuming that axiom. > > I did so. I founf aleph_0 lines. But I found that they do not all exist > unless there is one line of infinite length. So there are less than > aleph_0 lines. You did *not*. You did find aleph-0 lines, but you did *not* find an aleph-0'th line. > > > Of course it does not. You cannot distinguish sqrt(2) and the same > > > number with the digit of position number 10^100^100 replaced by 5. > > > > So we are factoring numbers using things that do not exist. > > Not completely, not as a number. What do you mean with that? Numbers are factored using number fields (i.e. fields of numbers that you state do not exist). Have a look at the number field sieve, which does exactly that. The method is used to factor natural numbers. The largest RSA numbers that have been factored were factored using this method. > But you have never used more than > 10^100 digits of sqrt(2). Obviously not. I only used the concept. In things as above, there is no interest at all in particular digits. It is only about mathematical concepts. Using these concepts I can prove that (99831352 + 2146257.sqrt(-311))^(1/19) is a common factor of 7 and a root of a particular polynomial with integer coefficients. And I do not need *any* digit of the decimal expansion of any of those numbers (except where stated). Also, it is easily shown that 61508786157972776212869020804494515460494487 + 615303255419645369700183146701653918462324.sqrt(9993) is a unit in Q(sqrt(9993). I do not need *any* digit in the decimal expansion of sqrt(9993) to prove that. It is easy to show that that is indeed a unit. Take the conjugate, multiply, and see that you get 1. This is possibly not of much interest to you, but it is of interest to mathematicians, and other fields that use mathematics. > > > The set of edges is easily enumerated. > > > o > > > /1 \2 > > > o o > > > /3\4 /5 \6 > > > o o o o > > > /7... > > > > Yes, as long as you remain finite. > > The lines o Cantor's list can be enmerated. We do not remain in the > finite there. The edges of the infinite tree can also be enumerated. > See the scheme above. That is proof by example. I do not see how the infinite tree can also be enumerated. Pray, again, explain. > > So the edges leading to final nodes > > are countable. What is the natural number that can be mapped to the > > final edge in 0.111... ? > > There is no final edge. Nevertheless all edges can be enumerated. And > there are two edges per infinite path. Again, makes no sense to me. If there are final nodes, there are final edges. -- 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 10 Aug 2006 07:31
Virgil schrieb: > In article <1154967767.420316.33010(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > No. As 0.111... has more index positions than each and every natural > > number in unary notation, the natural indexes are not sufficient to > > index 0.111... . > > > > > Just as N has more elements > > > than each and every indivual natural number is irrelevant. > > > > This assertion is impossible. Compare the differences of 1 between the > > naturals which would sum up to a natural number infinity if there were > > infinitely many differences possible. > > That makes no more sense and not less > than to say that the sum of infinitely > naturals being infinite prevents existence of infinitely many naturals. Correct. Both conclusions are identical. > > Since the sum of more than two naturals must be a natural larger than > any of them, there are more than any finite number of naturals. Yes, the set is potentially infinite. But it is wrong to assert that all naturals existed somewhere, because "all naturals" makes not more sense than "the last natural". > > If "mueckenh" wishes to claim that "more than any finite number" is a > finite number, The idea that "more than any finite number" would be a number is wrong too. Regards, WM |