An uncountable countable set
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From: Virgil on 22 Jun 2006 21:25 In article <1151010727.495487.142700(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > 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. > > This is proven by three invalid proofs. Do you know Canto's first proof > of 1874? Probably not, so we need not discuss it. If you are to claim that the first proof is an invalid proof you must either back up that claim or allow your claim to be called false and the first prof valid. And is that "first proof" for the theorem that there is no surjection from N to P(N) or for the theorem that there is no surjection from N to R, the set of all reals? They are not quite the same. > You certainly know > his second one. It is wrong, because all it shows is the construction > of a number of a set of countably many constructible numbers. A slight modification to the "diagoal" proof allows for any given list of reals, the constriction of one "anti-diagonal" real not listed for each real in the open interval (0,1) = {x in R: 0 < x < 1}. > It is ridiculous to believe this to be a proof of uncountability. It is even more ridiculous to believe otherwise. At least within ZFC or NBG.
From: Dik T. Winter on 22 Jun 2006 22:07 In article <1150975587.924677.62910(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1150957437.491685.293410(a)g10g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > > > > > A *surjective* mapping does not exist. The same does Hessenberg. > > > > > > > > A *surjective* mapping does exist, but it is not f. There exist a > > > > surjective mapping g -> M(f). > > > > > > Like this one?: > > > > > > 2) 0.12324389 > > > 3) 0.23123123 > > > 4) 0.85348714 > > > 5) 0.11133333 > > > 6) 0.31415161 > > > .. > > > > > > 1) 0.24446... > > > > > > This is the proof, that the proof that a surjective mapping f: |N --> > > > [0,1] does not exist, does not exist, isn't it? > > > > This is plain nonsense. > > 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: You are wrong again. Given the set S of finite subsets of N you can find a surjective mapping from N to S. Name that mapping f. Obviously that is not a surjective mapping from N to M(f), and it never claimed to be. But given M(f) we can find a surjective mapping from N to M(f) (as has been shown). Name that mapping g. Obviously it is not a surjective mapping from N to M(g), and never claimed such, it is not even a mapping from N to M(g). So I wonder where your problem is. > A *surjective* mapping does exist, but it is not f. There exists a > surjective mapping g -> M(f). Indeed. But your short list does not show anything at all, I wonder what I have to do with it. I think you intend to show something about the diagonal number of a list of reals. But we were talking about a mapping from N to P(N). So please keep to the subject. -- 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 22 Jun 2006 22:24 In article <1150976208.140566.164550(a)r2g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Yes, and that shows that whatever mapping f you give, it is simply not surjective. Given any mapping f, the proof shows that it is not surjective. Based on f you can determine K(f) and show that it is not in the image. See what I wrote. I did *not* say "given a surjective mapping f: N -> P(N)". > > > > 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. That is opinion, not proof. And I do not see any paradoxes. You simply do not understand that *in an abstract sense* infinite sets do exist, and are indeed guaranteed by the axioms. If you do not like that that is your problem. You might try to set up mathematics without the axiom of infinity. > 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. An interesting assertion. > > 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. Can you read plain English? Of course you can not construct a surjection to a set which you do not know when you start constructing. The condition "if f is required to be surjective" makes no sense at all in this context. The f I have *is* surjective, it is a surjective map from N to S, as I wrote. And so the set M(f) does exist. And f is not surjective to M(f), that is also clear. This does *not* mean that there is another map from N to M(f) that is 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. It is not in bijection with N by the function f, this does not preclude other mappings. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Daryl McCullough on 22 Jun 2006 22:25 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> 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. > >This is proven by three invalid proofs. We're only discussing one right now: If f is any function from N to P(N), then let K(f) = { x in N | x is not in f(x) }. Then 1. K(f) is a subset of N. 2. K(f) is an element of P(N). 3. K(f) is not in the image of f. 4. Therefore, f is not a surjection. That's all there is to it: If f is any function from N to P(N), then f is not a surjection from N to P(N). That is logically equivalent to "There is no surjection from N to P(N)", which is by definition what "P(N) is uncountable" means. >It is wrong, because all it shows is the construction >of a number of a set of countably many constructible numbers. What it shows is that the assumption that f is a surjection from N to P(N) leads to a contradiction. If something leads to a contradiction, then it is provably false. Therefore it is provably false that there exists a function f that is a surjection from N to P(N). -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 22 Jun 2006 22:46
mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> 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. >> >Consider a mapping |N --> P(|N) which need not be surjective but has to >satisfy only one condition, namely that the set K = {k e |N & k /e >f(k)} is in the image. This mapping is impossible. Right. There is no mapping f from N to P(N) such that K(f) is in the image of f. From this it follows that there is no surjection from N to P(N). >{f, k, K} is impossible to satisfy. But in the proof by Hessenberg, >you insist, it would prove non-surjectivity? Why do you always want to replace a clear statement by a confused one. I have no idea what you mean by "prove non-surjectivity". The more precise statement is this: For any function f from N to P(N), K(f) is not in the image of f. Therefore, f is not a surjection from N to P(N). >In a bijective mapping |N --> P(|N) there is every element of |N and >every subset of |N. Right. We have two facts: 1. If f is any function from N to P(N), then f is not a surjection from N to P(N). 2. If f is a bijection from N to P(N), then f *is* a surjection from N to P(N). 3. Therefore, it is a contradiction to assume that f is a bijection from N to P(N). If an assumption is a contradiction, then it is false. So it is false that there exists a bijection from N to P(N). -- Daryl McCullough Ithaca, NY |