An uncountable countable set
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From: mueckenh on 3 Jul 2006 17:25 David Hartley schrieb: > If this is defined for every element of A, we can say that the sequence > has a limit, and it is f. > > > Now, granting that your process works as you hope, you get a sequence of > well-orderings of the rationals whose limit is the usual ordering. But, > the corresponding sequence of bijections N -> Q does *not* have a limit. > (You need only consider the images of 1, these diverge to -oo.) As I only consider positive rationals, these go to the smallest positive rational, but that is the same problem. But to answer you argument of the missing limit: Cantor's diagonal has also this very same problem. In case of the Cauchy-limit of an irrational number, we see (a_k)*10^-k disappear, because it becomes as small as desired. In the Cantor-diagonal however, there is always a digit with the same weight and importance as the first one, even for k diverging to oo. > Your > example does not "preserve the bijective mapping between N and Q". Remember, here we have the same answer as in Cantor's diagonal: All natural numbers are finite. Though there are infinitely many, each one is finite. Hence even the smallest positive rational is in one-to-one correspondence with a finite natural number. There is no natural number oo (like in Cantor's list). Regards, WM
From: mueckenh on 3 Jul 2006 17:27 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. 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) }. Regards, WM
From: mueckenh on 3 Jul 2006 17:32 Dik T. Winter schrieb: > In article <1151844299.084352.59780(a)v61g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Daryl McCullough schrieb: > ... > > > But the assumption here is that f is a surjection from R to P(N), > > > not from N to P(N). So we have: > > > > > > 1. There is no k in N such that f(k) = K(f). > > > 2. There exists k in R such that f(k) = K(f). > > > > Why don't you try the same trick with the mapping (|N and pi) --> R. > > Let pi be mapped on the set of non-generators. > > Again, this makes no sense. Do you mean f: {x | x in N or x = pi} -> P(M)? to P(N). > Well, because K(f) contains pi, it is not an element of P(N), and so that > it is not in the image of f does not inhibit f from being a surjection. > > > Then the mapping can be > > surjective. > > I would not say "can be", but the proof does not show that it is not. It does not show non-surjectivity. > > > So if one mapping is possible but the other is not, then > > there must be something else prohibiting it, isn't it? > > The reason one might be possible and the other not is because one K(f) > is not a member of P(N) and is not necessarily in the image, while the > other K(f) is a member of P(N) and so (if f is surjective) *must* be > in the image. However this only does shows that the other is not surjective, > it does *not* show that one is surjective. ok. > > 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. Regards, WM
From: mueckenh on 3 Jul 2006 17:37 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. And to make sure that every digit is covered by your proof, you must find out whether "if" or "else" has to be applied. 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? > > > After that we can determine each and every digit of > > > the diagonal number without the need to know other digits. Moreover, > > > by Cauchy, such a sequence of digits defines a real number. > > > > 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. > > > > I do not know of any definition of a sequence of overlapping "infinite > > > permutations" that allows an inifinite sequence. Pray provide one. > > > > There are no "overlapping" permutations. > > There are. The first sequence of transpositions defines the first > "infinite permutation". Indeed, the transpositions do not overlap. Correct. > 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. > > > > Because there are not infinitely many comparisons, there is a single > > > definition. > > > > 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. Regards, WM
From: Virgil on 3 Jul 2006 17:55
In article <1151934331.994004.47450(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > imaginatorium(a)despammed.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > <snip> > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > > not exist. You can see this by the following argument: 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. > > > > Perhaps I missed it, but can someone explain what |N is and how (if at > > all) it differs from N (which I assume means the set of naturals)? > > It is the same. In Germany it is usual to denote N by |N, R by |R. > > Regards, WM On the other hand, let f map R on P(N) with the only condition on f be that f map a non-natural to K= {n in N: n !in f(n)}, then f can quite easily both exist and be a bijection. Which shows that "mueckenh" is again mucked up. |