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From: stephen on 18 Oct 2005 21:51 albstorz(a)gmx.de wrote: > David R Tribble wrote: >> David R Tribble wrote: >> >> Come on, now. It's up to you to prove your own claim, especially >> >> when it contradicts established mathematics. I know you cannot show >> >> a bijection between N and P(N). >> >> >> >> [Example snipped] >> > >> >> Albrecht S. Storz wrote: >> > 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}, ... >> > ... >> > >> > Now count in diagonal sequence. You may think of Cantor's first >> > diagonal proof. >> > >> > Which subset of N is not included? >> >> I don't know what you mean by "count in diagonal sequence". > It's a usual visualisation of Cantors first diagonal proof. You can go > through diagonally e.g. > N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ... Let us restate that as a function from N to your above list: 1 -> N 2 -> {1} 3 -> {2} 4 -> N/{1} 5 -> {1,2} 6 -> {3} 7 -> {4} 8 -> N/{3} 9 -> {1,3}, ... So if we consider w = { x : x is not in f(x) } we get w = { 2, 3, 5, 6, 7, 9, .... Where does that appear in your list? > It's a good way to understand, that the most subsets of the powerset > are this, No. Most of the elements of the power set are not this. This is a countable subset of an uncountable set. > which are infinit and and also their counter parts. And at > the end, are unconstructable. At the end? The end of what? The end of the unending list? There is no end to an unending list. That is what unending means. Unendlich!!! > But this is far away from the intend of this thread. > Exists the powerset? Exists the sum of all naturals? > The definition of infinity is wrong. I had shown an easy understandable > argument. You showed that you do not understand the definition of cardinality. Stephen
From: Virgil on 18 Oct 2005 23:07 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.
From: Virgil on 18 Oct 2005 23:08 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?
From: Virgil on 18 Oct 2005 23:15 In article <1129686187.158530.7270(a)g49g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > It's a usual visualisation of Cantors first diagonal proof. You can go > through diagonally e.g. > > N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ... This version of sequencing subsets of the set of naturals omits most subsets of the set of naturals. it includes only those which ase finite or whose relative compliments are finite, but most subsets are both infinite and with infinite compliments.
From: imaginatorium on 19 Oct 2005 00:34
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", 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 |