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From: albstorz on 19 Oct 2005 04:22 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
From: imaginatorium on 19 Oct 2005 05:16 albstorz(a)gmx.de wrote: > imaginatorium(a)despammed.com wrote: > > stephen(a)nomail.com wrote: > > > albstorz(a)gmx.de wrote: > > <snip: my goodness this stuff goes on and on...> > > > > 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. Remind me: "Cantor's first diagonal proof" shows what? Look, you're a crank, and you're muddled. We expect that. I'm trying to *help* you by pointing out that even if what you are trying to say is not quite right (the term "count in diagonal sequence" is not standard), at least we can see what you mean. > 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. Fool. OK, here's a thought of mine - I see you claim to have a list (well, actually a list of lists, but it comes to the same thing) of all subsets of the naturals. In this list, which comes first: the set of even numbers, or the set of prime numbers, and why? Brian Chandler http://imaginatorium.org
From: albstorz on 19 Oct 2005 07:27 William Hughes wrote: > 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. Since we talk about x e {1,2,3,4,5,...}, why not? > > > 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). But with 2^n you will reach infinity? Than you also will reach it with 1+1+1+1+... > > > > > 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". Of course. > > > > (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). > > -William Hughes No. Regards AS
From: albstorz on 19 Oct 2005 07:31 Virgil wrote: > In article <1129684564.158859.119300(a)o13g2000cwo.googlegroups.com>, > albstorz(a)gmx.de wrote: > > > Virgil wrote: > > > In article <1129622094.019699.233300(a)g44g2000cwa.googlegroups.com>, > > > albstorz(a)gmx.de wrote: > > > > > > > David R Tribble wrote: > > > > > > > At first, you should show, that bijection means something to > > > > notwellordered infinite sets. > > > > > > > > Bijection is a clear concept on finite sets, it also works on > > > > wellordered infinite sets of the same infinite concept. > > > > Aber: Show me a bijection between two infinite sets with the same > > > > cardinality, where one of the sets is still not wellorderable. > > > > > > If either of two sets is well-orderable and there is a bijection between > > > the sets, then that bijection induces well-order-ability on the other. > > > > > > So that you cannot have a bijection between two sets when one is > > > well-orderable and the other is not. > > > > > > That's right. > > > > So biject two non-well-orderable sets. > > > With or without an axiom of choice? I think it is more interesting for the most people without AC, since AC is not widely loved by the mathematics. But you may show both if you want. Regards AS
From: albstorz on 19 Oct 2005 07:40
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. Regards AS |