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From: David R Tribble on 19 Oct 2005 19:08 William Hughes wrote: >> A natural number is a set, a cardinality is an equivalence class. > Albrecht Storz wrote: >> You make me hopefull. One sees a >> correspondence between natural numbers and von Neumann sets after all. >> 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. > William Hughes wrote: > You have made exactly this mistake before. Yes every number is a set. I agree that we can assign a correspondence between natural numbers and certain sets. But which sets? von Neumann sets are only one such kind of set, where each one has a successor: 0 = {} = { } 1 = {0} = { {} } 2 = {0,1} = { {}, {{}} } 3 = {0,1,2} = { {}, {{}, {{}} } } etc. But there are other, equally valid sets that can be associated with natural numbers. E.g.: 0 = {} 1 = {a} u {{0}} = { a, {{{}}} } 2 = {a} u {{1}} = { a, {{{a, {{{}}} }}} } 3 = {a} u {{2}} = { a, {{{a, {{{a, {{{}}} }}} }}} } etc. This is just one example of the many kinds of sets that can be associated with natural numbers. http://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory So when Albrecht says that natural numbers are sets, I ask the question, which sets? Any kind of set in particular?
From: David R Tribble on 19 Oct 2005 19:24 David Kastrup said: >> Uh, no. The proof depends merely on the fact that some value has to >> be either a member of a set, or not. And if some value is in one set, >> bur not another, then those two sets are different. > Tony Orlow wrote: > Oh! So, N is a different size from N/{1}? Finally we reach agreement. No, he's saying that N and N\{1} are different sets. That's obvious, because they have different members. But both sets are also the same size, because they have the same number of elements. Removing an element (or any finite number of elements, for that matter) from an infinite set results in another infinite set of the same size.
From: Virgil on 19 Oct 2005 22:07 In article <MPG.1dc07f10e985127698a4f7(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > > Here is a simple mapping from N to P(N). > > > > 1 -> {1} 2 -> {1,2} 3 -> {1,2,3} 4 -> {1,2,3,4} > > ... > > > > Every element is contained in the subset that it is mapped to. For > > this mapping, w={}, and no element is mapped to {}. > And not every subset is included, so this is not a bijection. Precisely! Nor is any other map from N to P(N) a bijection, since none of them are even surjections. > No need. I gave my general bijection, and this fact is true for it. I > see no other bijection between a set and its power set, but this one > holds, and that fact si true. Only in the twilight zone of TOmatics is there any bijection between any set and its power set. In the regular world outside of TOmatics it cannot happen. > If set S has S elements If S is not a number, then it cnnot be a number of elements. > > If y is not in S, then it is not part of the bijection. f is > > function from S to P(S). It makes no sense to plug in a value that > > is not in S into f. > That all depends. If you place a value range on the set Might have known TO would appeal to that (value ranges) which have no existence outside the twilight zone of TOmatics, at least for sets which do not both have and contain a LUB and a GLB. > There is nothing in this proof that > makes a bijection impossible. The TO is not reading the same proof as the rest of us are reading.
From: Virgil on 19 Oct 2005 22:18 In article <MPG.1dc078e83d0081bc98a4f6(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > I don't understand why an inductive proof of an equality relationship > does not extend to infinite numbers. Informally, the Peano axioms may be stated as follows: 1 There is a natural number 0. 1 Every natural number a has a successor, denoted by S(a). 3 There is no natural number whose successor is 0. 4 Distinct natural numbers have distinct successors: if not(a = b), then not(S(a) <> S(b)). 5 If a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers. (This axiom ensures that the proof technique of mathematical induction is valid.) More formally, we define a Dedekind-Peano structure to be an ordered triple (X, x, f), satisfying the following properties: 1 X is a set, x is an element of X, and f is a map from X to itself. 2 x is not in the range of f. 3 f is injective. 4 If A is a subset of X satisfying: 5 If [x is in A, and (If a is in A, then f(a) is in A)] then A = X. A member of the set of such objects so defined is defined as finite if the set of such objects of which it is ultimately a successor is a finite set by the Dedekind definition of finite sets. With that definition of finiteness of such objects, all such objects are provably finite.
From: Virgil on 19 Oct 2005 22:23
In article <MPG.1dc0779a8507946398a4f5(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > stevendaryl3016(a)yahoo.com said: > > albstorz(a)gmx.de says... > > > > >I don't know what you are talking about. The proof for finite sets > > >needs just a complete induction. This will not hold for infinity I > > >think. > > > > There is no induction involved in the finite case, and the > > infinite case is *exactly* the same proof as the finite case. > > > > Here it is once again: > > > > Let A be any set whatsoever, finite or infinite, it doesn't matter. > > Let f be any function from A to P(A). > > Let w = { x in A | x is not an element of f(x) }. > > Let x = any set in A. > > Let u = f(x). We prove that u is not equal to w. > > > > By definition of w, we have x in w <-> x is not an element of f(x). > > So x in w <-> x is not an element of u. That means that there are > > two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > > equal w. > > > > So what we have proved is that forall x, w is not equal to f(x). So > > w is not in the image of f. So f is not a bijection between A and P(A). > > > > There's no induction. There's no assumption that A is finite. > But there is an assumption that y is in S. Where? Nothing prohibits S from being the empty set, and for functions of form, say, f(x) = {x}, S is forced to be the empty set. > If you are assuming you have the > complete set of naturals TO seems unable to read English. This is not about sets of naturals but about arbitrary sets. |