An uncountable countable set
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From: mueckenh on 30 Jun 2006 07:47 Virgil schrieb: > > 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. Cantor's diagonal proof does not prove that all numbers of the list are included. > > 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. Cantor's diagonal proof does not prove that all numbers of the list are included. Either equal rights for all or for none. > > 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. Hence there must be one assumption which is wrong, isn't it? > > > > 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. Hence there must be one assumption which is wrong, isn't it? The only assumption I made is the existence of infinite sets. > 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. Do you have an idea why? (What is wrong with my assumptions?) Regards, WM
From: Dik T. Winter on 30 Jun 2006 07:47 In article <1151506559.824704.220480(a)i40g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > In article <1151441001.426436.208850(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > That is correct too, but the quote is irrelevant, because it is clear > > > that a countable number of transpositions cannot change a well-ordered > > > set into a not well-ordered set: There would have to be a first one to > > > destroy the well-ordering. Perhaps Cantor meant that. > > > > This makes no sense. The quote specifically states that an infinite number > > of transpositions does not change the "Anzahl" ("ordinal number"), but that > > is false, as I did show above. And the quote was only talking about > > changing "Anzahl". > > Your example makes use of k + n in the limit n --> oo. That may be > prohibited. I wrote: P(k) = lim{k -> oo} (k,k+1)(k,k+2)...(k,k+n) where I explicitly wrote that I used some intuitive form of limit here. This is the only way I can make sense of "infinitely many transpositions". If what I wrote is not valid, please provide me with a definition of "infinitely many transpositions". How do they operate on a set all together? Does the result depend on the order? So there are many questions. > > You state so (and I think it is true), but a quote from Cantor is not a > > proof, > > Cantor may have been wrong here. But my example is not. Eh? Your example just lists infinitely many transpositions without any definition of meaning. > > How do I interprete an infinite number of transpositions? You say it is > > unimportant, but it is very important. Unless I know that I can only come > > up with reasonable interpretations, but not with defined interpretations, > > whether the number is countably infinite or uncountably infinite. > > It is nothing else but the mapping between two infinite sets. And if > the elements of inifinite sets do exist, then the transpositions, > interpreted as elements of a countably infinite set, do exist. This is clear as mud. Indeed, each transposition can be seen as a mapping that changes order. Without some concept of limit, or whatever, there is no implicit meaning of "infinitely many transpositions". > > Moreover, there > > is no "for n = 1 to oo", I have no idea what you mean with that. Please > > be a bit more precise in the future. > > It is a mapping. Example: > The set {1,6,5,4,8} is mapped on {1,6,4,5,8} by the first procedure > which reorders the pairs with indices 2n-1 and 2n like indices 1 and 2 > and 3 and 4. The next mapping maps {1,6,4,5,8} on {1,4,6,5,8}. It > reorders the pairs with indices 2n and 2n+1 like 2 and 3 and 4 and 5. > Then again use the first mapping and so on. So you define a series of mappings. But the series is infinite. How do you define such an infinite series? > > Yeah. When are we done? This is a sequence of transpositions of order > > type w * w. I have shown how that would destroy the well-orderedness > > of the natural numbers. > > No, there is always one set indexed by the natural numbers. Again you are starting with a conclusion about the finite and extending it to the infinite. You have to prove that you can do that. > > > I am not sure whether you can define k+n for n --> oo. > > > > Why do you think I define that? I just give an intuitive meaning to an > > infinite number of transpositions. If you have a way to define such, pray > > give it. > > Above you used the expression k+n in the limit n -> oo. Yes. Similar as in lim{n -> oo} (1/k^2).(1/(k+1)^2)...(1/(k+n)^2). And you write {q_2n, q_2n+1) for n -> oo. Can you define *that*? > > Same with my sequence of transpositions of N given above. > > Then it is not an example for the the set w*w. Eh? Why not. -- 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 30 Jun 2006 07:53 Virgil schrieb: > > 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. > We consider only infinite paths without ends in a tree without leafs. We apply the simple rule which must every path obey: In order to split into two paths, two nodes are needed. Hence, the number of paths cannot double unless the number of nodes at doubles at least. Can you see that: | o /\ o o / \ / \ Where ever you look at the tree: you will see this pattern. > > > > 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). Therefore, this proof shows that set theory is inconsistent because another valid (and very lucid) proof leads to the opposite result. Regards, WM
From: Dik T. Winter on 30 Jun 2006 10:32 In article <1151667833.246403.204680(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Daryl McCullough schrieb: > > >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. 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? > > 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) We are checking whether 'f' (where f stands for *any* mapping). > > 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. What part of the definition fails? > > 5. Therefore, f is not a surjection. > > The existence of one or more non-surjective mappings doesn't tell us > anyting about cardinalities. > >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. That is a complete misstatement of the Hessenberg condition. To be clear, let's have given a set A and a set of sets B (I do not yet state what the elements of A and B are). Let's also have a mapping f from A to B. Now construct the set K(f) = {x in A and x !in f(x)}. This set is well-defined. The Hessenberg condition is as follows: When K(f) is an element of B, for f to be surjective, K(f) should be in the image of A. So K(f) should be the image of an element of A. (When K(f) is not an element of B nothing can be said about f.) And in the case of N -> P(N) it so happens that K(f) *always* is an element of B = P(N), so for surjectivity it should be the image of an element of A = 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: Virgil on 30 Jun 2006 14:19
In article <1151667833.246403.204680(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. But the non-existence of any surjective mappiings, combined with the existence of at least one injective map, does! And in the set of all mappings from N to P(N), there are injections but no surjections. > > > > That is a proof, valid for any function from N to P(N), that > > that function is not a surjection. > > Wrong. You are wrong to claim"Wong". Valid only for non-surjective mappings. So you proved that f is > a non-surjective mapping if f is a non-surjective mapping. It proves that EVERY mapping from N to P(N), or indeed from an set S to its power set P(S), is NOT an surjection. Theorem: There is no surjection from any set S to its power set P(S). Proof: For every set S and every finction f: S --> P(S), K(f) = {x in S: not x in F(S)} is a member of P(S). But for all s in S f(s) != K(f) . Thus for every S and every f:S --> P(S), f is not a surjection. and no surjection can exist. Q.E.D. > > >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. You are the one confused. In the above, K(f) is only required to be the image of a real. But K(f) will not, in general, even have to be a member of P(N), since there is nothing to prevent it from containing non-natural reals, like pi. > > 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(A)). 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. The set K(f) is perfectly well defined. "Mueckenh" is correct that it is not a subset of N but wrong about why. K(f) is not here a subset of N because it contains real numbers not in N. |