An uncountable countable set
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From: mueckenh on 23 Jun 2006 12:28 Daryl McCullough schrieb: > >This is proven by three invalid proofs. > > We're only discussing one right now: Don't you recognize that K(f) takes exactly the same function as the diagonal number D in Cantor's second argument? f enumerates the list numbers. If f can arbitrarily be replaced by g, then this proof is invalid and shows only what it does show in fact, namely the countability of all list numbers including all diagonal numbers which can be constructed. > 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. So drop this assumption. There is no mapping, even if surjectivity is not at all in question! Regards, WM
From: Virgil on 23 Jun 2006 13:34 In article <1151079269.293509.23750(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > > 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. > > > > Consider the mapping f : |N --> M, where M contains a countable set of > *infinite* subsets of |N and K. > > What follows now? All sorts of things. If M is countable, f could be, but need not be, a bijection. What does NOT follow is that f ever has {x inN:x not in f(x)} as a value.
From: Virgil on 23 Jun 2006 13:42 In article <1151079363.438974.123090(a)u72g2000cwu.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > If an assumption leads to a contradiction, then that assumption must > > be false. > > The assumption includes that |N and P(|N) do exist as complete sets. Those are prior assumptions occurring in the axiom set, and are not a part of the assumption that a surjection exists. > > >The assumption that there is a surjection from N to P(N) leads > > to a contradiction. But no one needs to assume that a surjection exists from N to P(N) since there is a valid direct proof that none does. > Therefore, it is false that there is a surjection > > from N to P(N). Therefore, P(N) is uncountable. > > Therefore, |N and P(|N) do not exist as complete sets. Non Sequitur. Unless you also claim that any non-surjection between tow alleged sets proves those alleged sets non-existent. Does the fact that there is no surjection from {} to {{}} prove that that {}and {{}} do to exist as sets?
From: Dann Corbit on 23 Jun 2006 13:52 <mueckenh(a)rz.fh-augsburg.de> wrote in message news:1151079756.995887.177600(a)p79g2000cwp.googlegroups.com... [snip] > It is ridiculous to believe in Copernicus in the frame work of the > bible. Nevertheless I believe in Copernicus. What (exactly) is the place in the Bible that contradicts anything said by Copernicus? The only astronomically related verses I can think of (offhand) are: (Job 26:7) 7 He is stretching out the north over the empty place, Hanging the earth upon nothing; (Isaiah 40:22) 22 There is One who is dwelling above the circle of the earth, the dwellers in which are as grasshoppers, the One who is stretching out the heavens just as a fine gauze, who spreads them out like a tent in which to dwell, [1] (Isaiah 40:25-26) 25 "But to whom can YOU people liken me so that I should be made his equal?" says the Holy One. 26 "Raise YOUR eyes high up and see. Who has created these things? It is the One who is bringing forth the army of them even by number, all of whom he calls even by name. Due to the abundance of dynamic energy, he also being vigorous in power, not one [of them] is missing. [1] One could argue that the world is not a circle but a sphere, but the Hebrew word chugh can also mean 'ball'. How I read those verses (and I admit that it is open to interpretation) is that the earth is not sitting on some other body, it is round, and the physical matter of the universe is made up of energy. None of these ideas are at odds with astronomical texts -- not even modern ones. At any rate, there may be a few more verses that touch on astronomical notions, but I have never seen anything that contradicts what Copernicus said. The notion that the earth is the center of the universe is not to be found in the Bible, for instance. Sorry about the deviation from math. Perhaps follow-ups should be set to some other forum, though I have no idea what it might be.
From: Virgil on 23 Jun 2006 14:01
In article <1151079433.912821.53280(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > A *surjective* mapping does exist, but it is not f. There exists a > > > surjective mapping g -> M(f). > > > > If the function f and the set M(f) exist it is because there is no n in > > N for which f(n) = K(f), and in that case, there are bijections between > > N and M(f), but f is not one of them. > > > Don't you recognize that K(f) takes exactly the same function as the > diagonal number D in Cantor's second argument? I recognise that the set K(f) is precisely analogous to the constructed "anti-diagonal" number in Cantor's "diagonal" proof, if that is what you were intending to ask. If f can arbitrarily be replaced by g, then this proof is > invalid and shows only what it does show in fact, namely the > countability of all list numbers including all diagonals which can be > constructed. The question is whether for any given set S there exists ANY bijection (or, more generally, surjection) from N to S. SO one must consider all possible such functions, not merely one. Your version is analogous to saying that no primes exist because 6 is not a prime. Let H be the set of all finite subsets of N, and for arbitrary functions f:N --> P(N) consider only those whose restriction to codomain M(f) = H union K(f) still are functions from N to M(f). It transpires that in none of these will K(f) be a value of f. It also transpires that for every such f which is a function there are multitudes of functions h: N --> M(f) which are bijections. In fact for each bijection h: N <--> H, we can construct g: N <--> M(F) by g(0) = K(f) and g(n+1) = h(n) |