An uncountable countable set
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From: Dik T. Winter on 3 Jul 2006 18:55 In article <1151962063.059935.176450(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Gc schrieb: > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > > not exist. > > > > Yes it does. For every function N to P(N) there exists K in P(N). > > Not for any surjective function. When such a function does not exist, of course the set K(f) also does not exist. > You will see it somewhat easier: > Map R on the set {1, 2, 3, K} with K(f) be { x in R | x is not an > element of f(x) }. That is again nonsense. Such a mapping does not exist, so K(f) does not exist, so the requirement (map R on the set {1, 2, 3, K(f)}) is nonsense as the right hand side does not have existence before we have defined a mapping f. -- 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 3 Jul 2006 18:51 In article <1151961136.639460.110410(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1151759090.073603.122700(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > Eh? K(f) is defined regardles the kind of mapping of f. How can you > > > > come to the conclusion that it is not defined for some particular f? > > > > What part of the definition fails for some particular f? > > > > > > Remember what I said: > > > Map |R (including |N) on P(|N) with the only condition that a natural > > > number has to be mapped on that set K e P(|N) which contains all > > > natural numbers which are not mapped on sets containing them. > > > > Yes, such a map is impossible, but that proves nothing about the impossibility > > to have a surjective mapping from R to P(N). If a mapping is surjective > > the requirement is that *each* element of the target should be the image > > of an element in the source. There is nothing that requires that element > > in the source to be a natural number, *unless* the source is the set of > > natural numbers (or a subset of it). So what does it prove that such a > > required mapping is impossible? > > Try to map R on the set {1, 2, 3, K} and you will see it. You again use K to stand for a particular kind of set that depends on the actual map. But let K(f) be the set {x | x in N and x !in f(x)}. As you see, that is alread sheer nonsense. Let's repair it to a mapping to {{1}, {2}, {3}, K}. Let's have f(x) = {1} when x in N. We immediately find that K(f) = N \ {1}. Now map pi to K(f) and e to {2} and the remainder of R to {3}. What is the problem? > > Next you come with something completely different: > > > Map |R (including |N) on P(|N) with the only condition that a natural > > > number has to be mapped on that set K e P(|N) which contains all > > > natural numbers which are not mapped on sets containing them. > > Yup, a mapping with such a condition can not be found. In the case of > > N -> P(N) this proves that there is no surjective mapping. In the case > > of R -> P(N) this proves nothing like that. > > And what is proved in case of a surjective mapping from N on the set of > even positive numbers & K? That such a maping is not possible. > And what is proved in case of a surjective mapping from R on the set of > rational numbers & K? Nothing. > > > Remember what I said: The set K e P(|N) which contains all > > > natural numbers which are not mapped on sets containing them. > > > > That makes no sense. It depends on the map. So you should talk about > > K(f) (with f the mapping assumed). But what is wrong with what I wrote? > > Wrong is the assumption that K(f) = { x in N | x is not an element of > f(x) } in case of a surjective f is an element of P(N). It is not. Why not? Either K(f) is empty or all elements of K(f) are integers, which shows that K(f) is an element of P(N), because P(N) contains *all* subsets of N (*). So it is not an assumption, it can be proven. There is only a single assumption in the reasoning: the map f is surjective. And so this assumption must be rejected. (*) Yes, I know you do not like that. But it follows in set theory with the axiom of infinity. If you do not like the mathematics based on that axiom, start your own mathematics without that axiom or with the negation of that axiom, and see how far you will get. With that axiom there is *no* surjective map from N to P(N). -- 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 3 Jul 2006 19:14 In article <1151962360.076787.291910(a)j8g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > But let's have f: {x | x in N or x = pi} -> P(N). Construct the following > > mapping: > > g: g(0) = pi, g(n) = n - 1; gf = h. > > this one is bijective, so the inverse exists. If f(x) is surjective, > > g(f(x)) is surjective, but that is a mapping from N to P(N) and can not > > be surjective because we can construct: > > L(g, f) = {x | x in N and x !in g.f(x)} > > we can translate it back to: > > L(g, f) = {g^-1(x) | (x in N or x = pi) and g^-1(x) !in f(x)} > > and see that that set is a member of P(N) and so should be in the image, > > but is not in the image. > > Try the same with Canor's diagonal. Works the same. Each time when it is shown that something (arbitrary) does not satisfy the condition of surjectivity (from N to P(N) or from N to R) you state, but what if we change either the source or the mapping? This makes no sense. You give us a map f: N -> P(N) or a map f: N -> R first. Next we show that the map you give us is not surjective. Period. This can be done for *any* map. So changing the mapping to a new mapping (a new list) makes no sense because the proof is valid for *any* mapping. Changing the source, like you did above, also does not help. There is a bijection (the 'g' constructed above) between N and N u {pi}. So assuming a surjective map f from this new set to P(N) or R would lead to the surjective map f.g (where I write f.g to mean: first apply g next apply f) from N to P(N) or R. But we know that such a surjective mapping is not possible, so the assumption that f is surjective is not valid. The basic fact is that it can be shown that *any* map from N to either P(N) or to R is not surjective. So there are *no* surjective maps. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Hartley on 3 Jul 2006 19:20 In message <1151961747.436643.271400(a)a14g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes > >David Hartley schrieb: > >> >-1, -2, -3, ... indexed by >> > 1, 2, 3, ... >> > >> >> and apply your process. The >> >> first few steps give >> > >> >the indices after the first step are >> >> >> >> 2 1 4 3 6 5 8 7 ... >> > >> >This is correct, but the next four steps would supply >> >2 4 1 6 3 8 5 10 7... >> >4 2 6 1 8 3 10 5 ... >> >4 6 2 8 1 10 3 ... >> >6 4 8 2 10 1... >> >because we must always alternate between pairs (2n-1, 2n) in one step >> >and (2n, 2n+1) in the next (for all n e |N). >> > >> I see we have different ideas of what these transpositions mean. I took >> (1 2) to mean swap the pair originally indexed by 1 and 2, wherever they >> now are. You seem to mean swap the current occupants of positions 1 and >> 2 in the list. This is certainly more likely to have the effect you >> desire. > >It is a very simple rule and it determines the result for any >well-order we are starting with. > >> Indeed, I agree that it is possible to apply a sequence of >> transpositions and change a well-ordering of the rationals to the usual >> ordering. (I posted my own example last night.) However, I take this to >> imply simply that such transformations do not preserve well-ordering, >> not that there is a contradiction in standard set theory. > >The bijection (one-to-one corespondence) between rationals and naturals >is preserved. This is if not a well-order so at least more then set >theory can supply for the natural order of Q. Your original claim was that your process would produce a bijection from the naturals to the rationals which still respected the ordering, which is now the usual one. This would of course be the contradiction you are looking for, but you have not offered any attempt to prove it. -- David Hartley
From: Dik T. Winter on 3 Jul 2006 19:25
In article <1151962633.174580.226060(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > You *explicitly* say: apply aleph-0 times. Consider the Cantor diagonal > > on a list l_n of reals in the range [0, 1): > > I define two subfunctions (only to shorten the definition line): > > remainder(x, a): x - entier(x / a) * a > > and > > digit(k): if k = 4 than 5 else 4 > > and now: > > D = sum{k = 1...oo} digit(remainder(entier(l_n * 10^k), 10)) / 10^k. > > there are no steps involved. > > Whether you believe it or not: the diagonal has aleph_0 digits. You consider aleph_0 to be a number. I think this is a problem, but I let it pass. > And to > make sure that every digit is covered by your proof, you must find out > whether "if" or "else" has to be applied. Why? We know that either "if" or "else" has been applied, that is sufficient. > You said the numbers of the > list are well known. For what sake should that be important unless you > compare the n-th digit of he n-th number with your prescription? The importance of that is that if the members of the list are not known there is no list. > > > Leave Cauchy out of the play. There is no justification to reach all > > > last digits of the diagonal number by his epsilons. His specification > > > would only show that all columns of the list up to nn have been > > > covered. > > > > Apparently you do not understand what Cauchy involves. > > Do you know it better than you know Cantor's writings? > > There is the positive epsilon arbitrarily small. But every epsilon > covers infinitely many digits of Cantors diagonal. To follow Cantor, on > the other hand, *every* digit has the same weight as the first one. > There is no limit-process at all. Apparently you do not know Cauchy well. According to Cauchy, any number that is written decimally has a limit, and that limit is a real number. And, no, I have not read much from Cantor, so I have no idea how I should interprete your representation here. > > The second sequence of transpositions defines the second "infinite > > permutation", but that one overlaps the first "infinite permutation". > > No problem. For a given well-order to start with, every result is > determined. When you use finitely many overlapping sequences of "infinite permutations". > > > This single definition leads to infinitely comparisons, unless the > > > numbers in the list are all known. > > > > By representing a list it is assumed that all members are known. If they > > are not all known, it is not a list. > > The given well-order to start with is also known, and every result is > determined. For finitely many "infinite permutations". Just like it is for a list (which is a map from N to something, where N contains finite elements only). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |