An uncountable countable set
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From: Daryl McCullough on 29 Jun 2006 12:40 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> >> The difference is that the claim that there is a surjection >> from N to P(N) is provably false, > >"provably" by a false proof is not provably. Right. As I said, there is a valid proof that there is no surjection from N to P(N), and there is another valid proof that there *is* a surjection from R to P(N). >> while the claim that there >> is a surjection from R to P(N) is provably true. > >Not that surjection which requires the set {k, K, f}. You are confused. The case N to P(N) and R to P(N) are not analogous. Look at the argument: 1. Let f be any function from N to P(N). 2. Let K(f) be { x in N | x is not an element of f(x) }. 3. Then K(f) is not in the image of f. 4. But K(f) is an element of P(N). 5. Therefore, f is not a surjection. That is a proof, valid for any function from N to P(N), that that function is not a surjection. Now try the same thing with a surjection from R to P(N): 1. Let f be any function from R to P(N). 2. Let K(f) be { x in R | x is not an element of f(x) } 3. Then K(f) is not in the image of f. But the next step doesn't work. K(f) is a set of *reals*, not a set of naturals. So there is no reason to believe that K(f) is an element of P(N). So Steps 4 and 5 don't go through. So we have a proof that works to prove there is no surjection from N to P(N), (or in general, to prove that there is no surjection from A to P(N)). But the proof *doesn't* prove that there is no surjection from R to P(N). -- Daryl McCullough Ithaca, NY
From: Virgil on 29 Jun 2006 15:37 In article <1151572793.102028.13570(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > The non-empty set of positive rationals has no first element, > > > > since for every positive rational, 1/2 of it is strictly between > > > > it and zero. > > > > > > This shows that set theory is self contradictory, because the following > > > mapping does not destroy well-order but establishes order by magnitude. > > > > > > > > > For n = 1 to oo: > > > (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, > > > q_2n) --> (q_2n, q_2n-1) > > > For n = 1 to oo: > > > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, > > > q_2n+1) --> (q_2n+1, q_2n) > > > And again: > > > For n = 1 to oo: > > > (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, > > > q_2n) --> (q_2n, q_2n-1) > > > For n = 1 to oo: > > > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, > > > q_2n+1) --> (q_2n+1, q_2n) > > > ... > > > repeat aleph_0 times. > > > > Where is your proof that it takes the rationals in some given > > sequential order and ever produces an ordering by magnitude? > > If actual infinity exists, then my mapping comes to an actual end. But that "end" does not prove that the well ordering of the rationals end up as the natural ordering. To establish your claim, you must prove that you have an infinite set that is well ordered and densely ordered under the same ordering. You have not done that. > > As there is no such thing as a smallest (most negative) rational, how > > does all of your transposing produce what does not exist? > Let's use only the positive rationals, but the problem is the same, > because there is no smallest one > > > How does sqrt(2) = m/n show that sqrt(2) is rational? It does not show > that but it shows that sqrt(2) does not exist as a rational. My proof > shows that infinity can never be completed. Your "proof" hasn't proved it to me, as there are essentials omitted. > > > > As far as I can see what you have is sort of like an infinite bubble > > sort, with large values being bubbled upwards past smaller ones, one at > > a time, until they bump into an even larger value. > > > > But at no stage of this operation does there ever get to be infinitely > > many values between two other values. > > There are not infinitely many values between two values, because the > set is and remains well-ordered, i.e. each element is indexed by a > natural number. Then the set of rationals never becomes naturally ordered, which would require exactly what you say never occurs. > And the difference between two naturals is never infinite. That is irrelevant to whether the set of rationals between two given rationals is finite or infinite. In the natural order of the rationals, all such subsets are infinite. I a sequentially ordered set, all such subsets are finite. Since no transposition can convert a finite set into an infinite one, no well ordered set of transpositions can do it either. > > > > For example, in the original sequence there are at most finitely many > > values between 0 and 1. You are claiming that there will be some single > > transposition which makes the up until then finite number of values > > between 0 and 1 in one step into an infinite number of values between 0 > > and 1. > > See above. Between two naturals in natural order, there are never > infinitely many naturals. So how is that relevant to anything? You are claiming a sequence (or at least a well ordered set) of transpositions applied to a well ordered set of all rationals will produce the densely ordered set of rationals in their normal ordering. If that were true, the set S = {x in Q: 0 < x < 1} would have to be originally finite but terminally infinite under your process. Then, by well ordering, there would have to be a first transposition for which S is infinite. This cannot happen.
From: Virgil on 29 Jun 2006 15:48 In article <1151577364.227200.197730(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151442361.464301.177450(a)u72g2000cwu.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > To show you that the condition K is impossible, independent of any > > > surjectivity. > > > 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. You see: > > > It is impossible a condition. > > > > > > There are all sorts of impossible conditions, most of which are > > irrelevant to anything. > > > > What is the alleged relevance of this one? > > The fact that whole set theory rests upon an impossible condition like > "no rule without an exception". Set theory does not have "no rule without an exception" nor any similar statement as an axiom, and, in fact, logic requires one assume the negation of that statement.
From: Virgil on 29 Jun 2006 16:12 In article <1151589330.892858.100380(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151505096.329945.87930(a)i40g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Daryl McCullough schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > > > >Daryl McCullough schrieb: > > > > > > > > >> So the answer is no? There is no function f enumerating > > > > >> all the real numbers? So the reals are uncountable. > > > > > > > > > >There is a mathematical form containing all real numbers of [0, 1]. In > > > > >binary representation it is given here. Why not call it a function in > > > > >an extended sense? > > > > > > > > The question is this: does it map the naturals onto the real numbers > > > > in [0,1] such that every real number is in its image? > > > > > > > > In particular, with your representation, which natural number maps > > > > to the real 1/3? > > > > > > It is not necessary to know this natural number (which remains unknown > > > or does not exist) in oder to prove that the infinite branches of the > > > tree are countable. > > > > > > If no natural number maps onto 1/3,, then you do not have a surjection > > from N ONTO R at all, and your mapping does not show the set of reals to > > be countable after all. > > If the paths are not more than the nodes It is only for finite trees, in which every path has a leaf (or terminal) node, that the number of paths is less that the number of nodes. For infinite binary trees, it is false. > and if the nodes are > countable, then the set of paths cannot have cardinality larger than > aleph_0. The set of paths in an ->infinite<- binary tree, in which no path ends, is uncountable, provably of the same cardinality as P(N). > > That is easy to explain. Proofs with the "wrong" result are declared > invalid. Which immediately invalidates all of "mueckenh"'s.
From: mueckenh on 30 Jun 2006 07:43
Daryl McCullough schrieb: > >> while the claim that there > >> is a surjection from R to P(N) is provably true. > > > >Not that surjection which requires the set {k, K, f}. > > You are confused. The case N to P(N) and R to P(N) > are not analogous. Look at the argument: > > 1. Let f be any function from N to P(N). > 2. Let K(f) be { x in N | x is not an element of f(x) }. K(f) is *only* defined in case f is a non-surjective mapping. K(f) is not defined and, hence, not a subset of |N, in case f is a surjective mapping f. > 3. Then K(f) is not in the image of f. Small wonder in case of a non-surjective mapping f. But by switching to a surjective mapping g, the problem is resolved as like as in the case of my original proposal f : |N --> (finite subsets of |N & K) > 4. But K(f) is an element of P(N). K(f) is *not defined* in case of a surjective mapping f. Hence, it cannot be a subset of |N. > 5. Therefore, f is not a surjection. The existence of one or more non-surjective mappings doesn't tell us anyting about cardinalities. > > That is a proof, valid for any function from N to P(N), that > that function is not a surjection. Wrong. Valid only for non-surjective mappings. So you proved that f is a non-surjective mapping if f is a non-surjective mapping. >Now try the same thing > with a surjection from R to P(N): > > 1. Let f be any function from R to P(N). > 2. Let K(f) be { x in R | x is not an element of f(x) } > 3. Then K(f) is not in the image of f. > > But the next step doesn't work. K(f) is a set of *reals*, > not a set of naturals. You are confused. K(f) is required to be the image of a natural number. > So there is no reason to believe > that K(f) is an element of P(N). So Steps 4 and 5 don't > go through. > > So we have a proof that works to prove there is no surjection > from N to P(N), (or in general, to prove that there is no > surjection from A to P(N)). But the proof *doesn't* prove > that there is no surjection from R to P(N). Of course, because only a mapping requiring K to be the image of a natural number cannot be realized. But the reason has nothing at all to do with cardinalities but with the fact that for such a surjective mapping f the set K(f) is not defined and hence is not a subset of |N. Regards, WM |