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From: Tony Orlow on 19 Oct 2005 12:36 albstorz(a)gmx.de said: > 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? A > hidden megabooster for transcendental overflow? What is the very > special aspect of the powerfunction to be so magic? Why do you think there is magic in the notion that the set of all possible combinations of elements is larger than the set of single elements, especially when that power set essentially includes the set, in the form of the singleton subsets? > Why should all operations with transfinite numbers lead to results with > the same "level" of infinity, but only powerfunction beams up to the > next level? It shouldn't. I think the set of squares shpuld be considered to have a size which is the square root of the set of naturals. At least the standard theory recognizes that the power set is larger than the set. Unfortunately, that's about as deep as the standard distinctions go. > Is not true: a^2 = a*a? 2a = a+a? > Is the powerfunction something other than a very shortcut for multiple > additions? > Which amount you are able to reach with powerfunction which is > unreachable by succesor operation? As far as bijections go, none. Bijections alone are insufficient with infinite sets. Certainly, you don't disagree that any set's power set is larger than itself? Or, perhaps you do. > > I'm very sensible about this because this argument is found in very > much books although it's total meaningless. > > (Weak minds might be impressed by the big numbers which are easily > produced by powerfunction.) It's the rice on the chessboard and the wealth of the king. :) > > > What in finity holds may not (or do not) hold in infinity. Or, it may very well hold for all cases, finite and infinite. > > Regards > > AS > > -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 12:39 William Hughes said: > > albst...(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? A > > hidden megabooster for transcendental overflow? What is the very > > special aspect of the powerfunction to be so magic? > > Why should all operations with transfinite numbers lead to results with > > the same "level" of infinity, but only powerfunction beams up to the > > next level? > > Is not true: a^2 = a*a? 2a = a+a? > > Yes, but the powerfunction does not look like a^2 but 2^a. > > > Is the powerfunction something other than a very shortcut for multiple > > additions? > > Yes. You cannot represent 2^x as multiple additions. > > > Which amount you are able to reach with powerfunction which is > > unreachable by succesor operation? > > Infinity for one. You cannot get from a finite quantity to > an infinite quantity by using the successor operation (unless > like TO you are willing to wave a circular magic wand and apply > the successor operation an infinite number of times). > > > > > I'm very sensible about this because this argument is found in very > > much books although it's total meaningless. > > > > I suspect that you mean "sensitive" not "sensible". > > > > (Weak minds might be impressed by the big numbers which are easily > > produced by powerfunction.) > > Strong minds are impressed with the fact that there is no > bijection between X and P(X). > > > > > > > What in finity holds may not (or do not) hold in infinity. > > Words to live by. Start by noting that a finite set has a > "number of elements" that can be described by a natural number > while an infinite set (e.g. the set of natural numbers) does > not have a "number of elements" that can be described by a > natural number. However, some things are true for both > finite and infinite sets. e.g. the fact that there is no > bijection between X and P(X). Did you have an objection to the bijection between *N and P(*N)? What did I do wrong there, bijection-wise? > > -William Hughes > > -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 12:40 Virgil said: > In article <MPG.1dbf2de38cf6ec898a4da(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > David R Tribble said: > > > Tony Orlow wrote: > > > >> I already showed you the bijection between binary *N and P(*N). > > > >> What didn't you like about it? It is valid. > > > > > > > > > > David R Tribble said: > > > >> No, you showed a mapping between *N and R, which is equivalent > > > >> to a mapping between *N and P(N). That's easy. > > > > > > > > > > Tony Orlow wrote: > > > > No, it was specifically a bijection between two sets of infinite binary > > > > strings representing, on the one hand, the whole numbers in *N starting > > > > from 0, both finite and infinite, in normal binary format, and on the > > > > other > > > > hand, the specification of each subset of whole numbers in *N, where each > > > > bit which, in the binary number, represents 2^n denotes membership of n > > > > in > > > > the subset. This is a bijection between the whole numbers in *N and > > > > P(*N), > > > > using an intermediate bijection with a common set of infinite binary > > > > strings. > > > > > > But that's an incomplete mapping, because there are not enough infinite > > > binary strings in *N to enumerate all of the subsets of *N. Try it, > > > if you don't believe me. > > > > > > > > Not enough infinite binary strings? > > Precisely. > So, you need more than an infinite amount? How many more? Do you need more than an infinite amount to list all the integral multiples of 1/10? -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 12:45 Virgil said: > 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. > Except for the obvious bijection between *N and P(*N). But, hey, what's one measly counterexample? -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 12:53
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 |