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From: albstorz on 21 Oct 2005 05:37 Tony Orlow wrote: > albstorz(a)gmx.de said: > > > > imaginatorium(a)despammed.com wrote: > > > stephen(a)nomail.com wrote: > > > > albstorz(a)gmx.de wrote: > > > > > > <snip: my goodness this stuff goes on and on...> > > > > > > > > First of all you don't argue on my claim of the thread. > > > > > Second, your above argueing is not clear to me, since both sets are > > > > > well-ordered. But it's nice, so I give this: > > > > > > > > > N > > > > > {1},{2},{3}, ... > > > > > N/{1},N/{2},N/{3}, ... > > > > > {1,2},{1,3},{2,3},{1,4},... > > > > > N/{1,2}, ... > > > > > ... > > > > > > > > Is this supposed to be a list? My reading > > > > of this is that your list is: > > > > > > > > N, > > > > {1}, > > > > {2}, > > > > {3}, > > > > ... > > > > N/{1}, > > > > > > > > Right here we have a problem. What is the element > > > > before N/{1} in your "list"? ... > > > > > > > > > > > > Now count in diagonal sequence. You may think of Cantor's first > > > > > diagonal proof. > > > > > > > > What diagonal? > > > > > > Come on, come on! He means the zigzag diagonal, as in the standard > > > demonstration that the rationals _are_ countable. This isn't "Cantor's > > > first diagonal proof", > > > > I see, you are the real checker. You knows it all. You are famous. You > > are apodictic. All the authors who speak of the first diagonal proof of > > Cantor are wrong. > > > > > > > but if you're going to argue with cranks you > > > must expect them to be pretty muddled about things. > > > > > > Anyway, it's obvious that *if* the OP shows a "list of lists" that > > > include all the subsets that is enough. In practice, of course he's > > > given the standard crank non-list. > > > > > > Brian Chandler > > > http://imaginatorium.org > > > > You and many of the other checkers are not able to discuss my starting > > argument. You are only able to respond to the usual wrong arguments you > > know. And I think, your answers are memorized because you are unable to > > think your own thoughts. > > > > Regards > > AS > > > > > Albrecht, much as I appreciate some of your ideas, I am missing this one a > little. If you are trying to create an enumeration, then you must be doing what > Brian suggests, with the zigzag diagonal like for the enumeration of the > rationals. However, you only have one set on the top line and the bottom, which > you don't show, but which would have the null set. There is not really a zigzag > diagonal covering the set listed this way, as far as I can tell, because it's > not really a rectangular list. I highly recommend the binary natural ordering. > -- > Smiles, > > Tony I beg your pardon. But it was just a fake. Regards AS
From: David Kastrup on 21 Oct 2005 05:48 albstorz(a)gmx.de writes: > Tony Orlow wrote: >> albstorz(a)gmx.de said: >> > >> > Virgil wrote: >> > > In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, >> > > albstorz(a)gmx.de wrote: >> > > >> > > > David R Tribble wrote: >> > > > >> > > > > >> > > > > Because I can prove it (and it's a very old proof). A powerset of >> > > > > a nonempty set contains more elements that the set. Can you prove >> > > > > otherwise? >> > > > >> > > > This argument is stupid. Is there any magic in the powerfunction? >> > > >> > > "Proofs" are not stupid until they can be refuted. The proof that for an >> > > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. >> > >> > >> > Even if you think that the powersets of finite and infinite sets have >> > both a greater cardinality than their starting sets, you would not >> > really think it depends on the same cause in both cases. >> > >> > You must proof it independently for finite and for infinite sets. In >> > this sense the argument is stupid. >> Albrecht, do you accept the axiom of induction? If so, it is easily >> provable inductively that the power set of a set of size n has size >> 2^n, and since this is an equality property, it holds for the >> infinite case. The power set of an infinite set is infinite, but a >> larger infinity than the set. > > No Tony. > > If you think so, you should also argue, that since the cartesian > product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the > finite case is always of graeter cardinality than the starting sets, it > should be also in the infinite case. But it isn't. You are arguing with the guy who thinks that the cardinality of the set of even naturals is less than that of the naturals. The amusing thing is that although Tony is talking rubbish here, powersets _do_ have a larger cardinality than the original set, since any presumed mapping f:S->P(S) can't cover {x in S| x not in f(x)} since that set is different from all f(x) for all x in S. So while you win this particular argument against Tony (not too hard to do), you are still wrong in the main. The above proof does not depend on powers, calculation, finiteness or infiniteness. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 21 Oct 2005 06:16 albstorz(a)gmx.de wrote: > Tony Orlow wrote: > > albstorz(a)gmx.de said: > > > > > > imaginatorium(a)despammed.com wrote: > > > > stephen(a)nomail.com wrote: > > > > > albstorz(a)gmx.de wrote: > > > > > > > > <snip: my goodness this stuff goes on and on...> > > > > > > > > > > First of all you don't argue on my claim of the thread. > > > > > > Second, your above argueing is not clear to me, since both sets are > > > > > > well-ordered. But it's nice, so I give this: > > > > > > > > > > > N > > > > > > {1},{2},{3}, ... > > > > > > N/{1},N/{2},N/{3}, ... > > > > > > {1,2},{1,3},{2,3},{1,4},... > > > > > > N/{1,2}, ... > > > > > > ... > > > > > > > > > > Is this supposed to be a list? My reading > > > > > of this is that your list is: > > > > > > > > > > N, > > > > > {1}, > > > > > {2}, > > > > > {3}, > > > > > ... > > > > > N/{1}, > > > > > > > > > > Right here we have a problem. What is the element > > > > > before N/{1} in your "list"? ... > > > > > > > > > > > > > > > Now count in diagonal sequence. You may think of Cantor's first > > > > > > diagonal proof. > > > > > > > > > > What diagonal? > > > > > > > > Come on, come on! He means the zigzag diagonal, as in the standard > > > > demonstration that the rationals _are_ countable. This isn't "Cantor's > > > > first diagonal proof", > > > > > > I see, you are the real checker. You knows it all. You are famous. You > > > are apodictic. All the authors who speak of the first diagonal proof of > > > Cantor are wrong. > > > > > > > > > > but if you're going to argue with cranks you > > > > must expect them to be pretty muddled about things. > > > > > > > > Anyway, it's obvious that *if* the OP shows a "list of lists" that > > > > include all the subsets that is enough. In practice, of course he's > > > > given the standard crank non-list. > > > > > > > > Brian Chandler > > > > http://imaginatorium.org > > > > > > You and many of the other checkers are not able to discuss my starting > > > argument. You are only able to respond to the usual wrong arguments you > > > know. And I think, your answers are memorized because you are unable to > > > think your own thoughts. > > > > > > Regards > > > AS > > > > > > > > Albrecht, much as I appreciate some of your ideas, I am missing this one a > > little. If you are trying to create an enumeration, then you must be doing what > > Brian suggests, with the zigzag diagonal like for the enumeration of the > > rationals. However, you only have one set on the top line and the bottom, which > > you don't show, but which would have the null set. There is not really a zigzag > > diagonal covering the set listed this way, as far as I can tell, because it's > > not really a rectangular list. I highly recommend the binary natural ordering. > > -- > > Smiles, > > > > Tony > > > I beg your pardon. But it was just a fake. > > Regards > AS You can biject all constructable subsets of P(N) to N, but not the unconstructable. That's an argument against the diagonal proof, since, if a set will be constructable in form of an antidiagonal of a list, it's a constructable set and therefore it is element of the set of the sets which are bijectable with N. --> The diagonal argument can't proof if a list containes all subsets of N since there occures a contradiction. Regards AS
From: Tony Orlow on 21 Oct 2005 09:25 Randy Poe said: > > Tony Orlow wrote: > > stephen(a)nomail.com said: > > > What is element "2^S-1"? S is a set. Element "2^S-1" means > > > nothing to me. > > If set S has S elements, from number 0 through S-1, then the element which maps > > to the entire set is a string of 1's which is S long, which corresponds to a > > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - damned > > error of 1!). > > So you would say that no element x is in f(x), correct? Not given any particular x, no. > > Therefore for this mapping, the set > > w = {x in S: x not in f(x)} > > contains every element of S, right? That is, w = S. Yes. > > Now let's consider every element y of S. Obviously y is > in w, since w = S. But then y is not in f(y), since that's > what it means for y to be in w. Then f(y) can't be S. That's true, and this is an interesting argument. You certainly cannot identify any natural which maps to the entire set of naturals. That would require identifying some last element and and establishing a particular number of bits N for the subset representation, so as to say w maps to (2^N)-1. Of course, since w only includes N elements, it doesn't include y, and y is not in w, within the original range of N. This is why one needs infinite sets for their bijections. Since neither set ends, there is always a corresponding element for each element, even though the numbers in the subsets are always smaller than the numbers representing the subsets. That doesn't matter. There is always another one larger, so you never run out of corresponding elements. At least, this is the way your bijections normally work. What rule am I breaking with this bijection? Where does the correspondence break down? At element N+1? > > So this means that no matter what y you pick, f(y) can't > be S. So S is not mapped by your "bijection". Not within the range of S, no, but over the infinite range, yes. > > What's wrong with your "proof"? Simple. You have f(z) = S, > where z is NOT a member of your original set S. You don't > have f(y) = S for any y IN your set S. That is true. For any given upper bound you set on S, y is greater. You seem to think y is too larger to be in the boundless infinite S. So what is this upper bound which y exceeds, in order to be larger than all natural numbers, finite and infinite? It is larger than whatever value range you assign to S, but that is why you don't try to assign actual sizes to discrete infinite sets. There is no actual size, because there is no end to the process of generating elements. Without an endpoint, it cannot be measured in absolute terms, but must be measured by comparison with a discrete unit infinity, such as N, the identity function between element count and element value. In this bijection, for any given natural n, the subset has a largest element no greater than log2(n), but n is an element of subset 2^n, and where n exists, so does 2^n. None of this precludes bijection. > > - Randy > > -- Smiles, Tony
From: William Hughes on 21 Oct 2005 09:28
albstorz(a)gmx.de wrote: > William Hughes wrote: > > albstorz(a)gmx.de wrote: > > > William Hughes wrote: > > > > > > > > > > > > Coincidently natural numbers and cardinalities are undistinguishable in > > > > > finity. > > > > > > > > They are very similar, but they are not quite "undistinguishable". > > > > A natural number is a set, a cardinality is an equivalence class. > > > > > > You make me hopefull. Some experts make "äääh", "hömm" and > > > "üüüh" if I said "A natural number is a set." One sees a > > > correspondence between natural numbers and von Neumann sets after all. > > > You are free to say "A natural number is a set." without "äääh-", > > > "hömm-" and "üüüh-" comments. Be lucky. You are right. > > > And natural numbers don't behave in any other way than sets. So, if > > > there is an infinite set there is an infinite number. If there is no > > > infinite number there is no infinite set. And vic versa. > > > > > > > > You have made exactly this mistake before. > > What mistake? Assuming that since every number is a set that every set must be a number. > > > Yes every number > > is a set. No, not every set is a number. For example > > {peach, apple, plum, fiddle} is a set but not a number. > > Wrong. It's a number. The number is this aspect of the set which it > have in common with the set {diddle, daddle, doddle, duddle}. >To say, a > set is a number or a set has a number is a slight difference which has > no effect in this considerations. No, it is crucial. Your thesis is not that every set is a number, but that every set has a number of elements. If your thesis were every set is a number, then to state that there is no infinite number is to deny the existence of an infinite set. If your thesis is that every set has a number of elements, then to say there is no infinite number does not mean that an infinite set does not exists, just that an infinite set does not have an infinite number of elments (at least in the way you define number). >The important aspect is, that every > set has a number of elements, since a set, totally independent of the > kind of the members of the set, behaves like a set of numbers with same > cardinality in all math aspects. Yes all sets have a cardinality. Before you say that this is false, look up the definition of cardinality. This may save you from saying somethiing really silly. Notee. the cardinality of an infinite set does not behave in the same way as the cardinality of a finite set. -William Hughes |