An uncountable countable set
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From: mueckenh on 22 Jun 2006 07:36 Dik T. Winter schrieb: > > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed according to > > > > > Hessenberg *does* exist. > > > > but it is not possible to determine that K(f). > > Why not? Because, in case f should be surjective, a number k need be contained in a set in which it must not be contained. > > > > > > The proof above. > > > > It does not disprove a surjection but the completeness of |N and P(|N). > > You are completely wrong. How does it prove that? Because it is the only explanation of this and other paradoxes. At least you cannot deny that it would resolve he problem. Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of all rational numbers. If you say that it exists, then I can prove that it can be ordered by magnitude without destroying the well-order. Obviously that is impossible. Hence {q_1, q_2, q_3, ...} does not exist. > Let's see. Let S be the set of finite subsets of N. And let us have > a bijection f: N -> S (they do exist). Now obviously f is not a > bijection from N to M(f), that is clear by the construction. Hence by constuction, the set M(f) does not exist, if f is required to be surjective. > Note again: M depends on f, a fact that you leave out every time. M in > itself is not a set, but for every given f, M(f) is a set. And for every given f, this set is not in bijection with |N. Regards, WM
From: mueckenh on 22 Jun 2006 07:47 Virgil schrieb: > No more than it proves Muecken does not exist. Please note: My name is Mueckenheim or short WM. It is fairly long, but so much time must be for polite people. And I am not going to talk to impolite people. Regards, WM
From: Daryl McCullough on 22 Jun 2006 09:15 mueckenh(a)rz.fh-augsburg.de says... >Dik T. Winter schrieb: > > >> > > > > 1. Given a mapping f: N -> P(N), the set K(f) constructed according to Hessenberg *does* exist. >> > >> > but it is not possible to determine that K(f). >> >> Why not? > >Because, in case f should be surjective, a number k need be contained >in a set in which it must not be contained. That case is impossible. To prove a case impossible, you only need to show that it leads to a logical contradiction. >Here is another one: Let {q_1, q_2, q_3, ...} the well-ordered set of >all rational numbers. >If you say that it exists, then I can prove that it can be ordered by >magnitude without destroying the well-order. On the contrary, you cannot prove that. Here's a specific well-ordering: Let r1 and r2 be two rational numbers. Then to figure out whether r1 comes before r2 in the well-ordering: Write r1 in the form p1/q1 where p1 and q1 are integers and q1 > 0 and there are no common factors between p1 and q1. Similarly, write r2 in the form p2/q2. 1. If |p1| + q1 < |p2| + q2, then r1 comes before r2. 2. If |p1| + q1 > |p2| + q2, then r1 comes after r2. 3. If |p1| + q1 = |p2| + q2, and p1 < p2, then r1 comes before r2. 4. If |p1| + q1 = |p2| + q2, and p1 > p2, then r1 comes after r2. This well-ordering gives the following first few terms of the sequence: 0/1, -1/1, +1/1, -2/1, -1/2, +1/2, +2/1 -3/1, -1/3, +1/3, +3/1 -4/1, -3/2, -2/3, -1/4, +2/3, +3/2, +4/1 -5/1, -1/5, +1/5, +5/1 etc. (The first row has |p| + q = 1, the next row has |p| + q = 2, etc.) But notice that this is not ordered by magnitude, since 1/2 < 1/1, but 1/1 comes before 1/2 in the ordering. The ordering by magnitude in not a well-ordering. Obviously that is >impossible. Hence {q_1, q_2, q_3, ...} does not exist. > >> Let's see. Let S be the set of finite subsets of N. And let us have >> a bijection f: N -> S (they do exist). Now obviously f is not a >> bijection from N to M(f), that is clear by the construction. > >Hence by constuction, the set M(f) does not exist No, that doesn't follow. What follows is that M(f) is *not* a subset of S. Since S contains all finite subsets of N, it follows that M(f) contains some element that is *not* a finite subset of N. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 22 Jun 2006 09:23 In article <1150975362.277573.39860(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de says... > > >Virgil schrieb: > > >> So the resolution is that either f, K_f and M_f are self-contradictory >> and do not exist at all > >if f is defined to be a surjection. That is right. And why do you >believe that a self contradictory approach should prove anything (for >instance about the surjectivity of certain mappings)? If an assumption leads to a contradiction, then that assumption must be false. The assumption that there is a surjection from N to P(N) leads to a contradiction. Therefore, it is false that there is a surjection from N to P(N). Therefore, P(N) is uncountable. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 22 Jun 2006 09:21
mueckenh(a)rz.fh-augsburg.de says... >Maybe. It is, however, the same approach as yours. If it is impossible >to find a direct surjective mapping, then first define the set and then >try to find a surjection. I can proudly declare in your words: > > A *surjective* mapping does exist, but it is not f. There exists a >surjective mapping g -> M(f). And those words are perfectly correct: f is not a surjective mapping from N to M(f), but g *is* a surjective mapping from N to M(f). So M(f) is countable. In contrast, there is no surjection from N to P(N). So P(N) is *not* countable. -- Daryl McCullough Ithaca, NY |