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From: David R Tribble on 21 Oct 2005 12:18 William Hughes said: >> Well, as you have not even defined 2^N for infinite TO naturals, you >> have some work to do. > Tony Orlow wrote: > Defined 2^N? It's 2 to the Nth power. What do you need to know? In binary, > it's a 1 followed by N 0's. Please state your complaint in the form of a > question. It is entirely reasonable to ask you to define 2^N. In the past, you've defined 'N' as the sum of an infinite series: N = 1 + 1 + 1 + 1 + ... or perhaps: N = 1 + 2 + 3 + 4 +... or perhaps: N = 1 + 2 + 4 + 8 + ... Most of the time, you simply say that 'N' is an arbitrary infinite number. For 2^N, "a 1 followed by N zeros" doesn't cut it, if for no other reason than because your 'N' is arbitrary. Perhaps you should define N a bit more precisely before you go throwing it around so freely. Defining your "infinite naturals" in more concrete terms, perhaps sets and successors, would be a step forward in making sense of your number theories.
From: stephen on 21 Oct 2005 12:27 Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> > Well, it's really not possible, given the general bijection I have offered >> > between an infinite ordered set and its power set. No natural is a member of >> > the subset of naturals which it denotes, as I demonstrated. >> >> What is really not possible? It really is possible for y to >> be a member of f(y). What is really not possible is that >> f is a bijection from S to P(S). > Not in the bijection I offered. y is never in f(y). Then it is not a bijection. If y is never in f(y), then there does not exist a y such that f(y)=S. S is a member of P(S), and in order for f to be a bijection, there must be a y in S such that f(y) =S. You plainly state that such a y does not exist, so f is not a bijection. Stephen
From: Daryl McCullough on 21 Oct 2005 12:17 Randy Poe says... >You claim to have a bijection f from N to P(N). > >One element of P(N) is the set N itself. > >You admit that there is no y in N such that f(y) = N. > >Therefore f is not a bijection from N to P(N). Actually, Tony is not claiming that there is a bijection between the *finite* naturals and the power set of the finite naturals. He is claiming that there is a bijection between *N and P(*N), where *N is some kind of nonstandard model of the naturals that includes infinite naturals. Not that that makes any difference to Cantor's proof... -- Daryl McCullough Ithaca, NY
From: David R Tribble on 21 Oct 2005 12:35 Stephen said: >> Do you consider the following a bijection from S={a,b,c} to its >> power set? >> [...] >> >> The above function is a bijection from {a,b,c,d,e,g,h,i} >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > Tony Orlow wrote: > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. It is true > that no bijection is possible in the finite case. In the proof, a last > element of the set is assumed when we assume there is some string > representing the entire set. Again with the "last element" red herring. Just let it go, Tony. Consider the set of naturals N = {0, 1, 2, 3, ...} which obviously has no last element. I can map an infinite binary natural to it, so that f(x) = N: x = 2^0 + 2^1 + 2^2 + 2^3 + ... So I've managed to identify an x such that it mapped to the entire (infinite) set N. Granted, x must be an infinite natural, but it still works. And all without resorting to any "last" element in either the set or in the series. Tony Orlow wrote: > However, to whatever extent we have considered > the set, say to N elements, we can always consider it to N+1, or 2^N > elements. If this is the set of all naturals, for instance, then N in the > set implies 2^N in the set. There is no end to the bijection. The proof > assumes one. If there is no end to the bijection, then the bijection is not complete, and it therefore not a bijection. If we can't account for all of the mappings of numbers to subsets, then we can't say the mapping is a bijection. None of the standard proofs assume a last element. Just let it go.
From: Tony Orlow on 21 Oct 2005 12:46
Daryl McCullough said: > Tony Orlow says... > > >stevendaryl3016(a)yahoo.com said: > >> 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. > > You mean A. Yes, that's what it means to have a > surjection f between two sets A and P(A) (a bijection > is a kind of surjection). It means that for every w in P(A) > there exists a y in A such that w = f(y). If y is not in > A, then the fact that w = f(y) is irrelevant to whether > there is a bijection between A and P(A). But, is S the COMPLETE set, and if so, how many bits does it have? You are assuming you have some subset represented by some string of 1's. Does this string actually end? No. So how can you define a natural that corresponds to it? It's whatever that infinite string of 1's is which doesn't end. If you have two unending strings of 1's, can you not establish a bijection between them? > > >If you are assuming you have the complete set of naturals, > >that you have identified the last, and can therefore > >identify the element that maps to the entire set, then > >you indeed run into a contradiction. However, both the > >infinite set of naturals and the infinite power set go > >on forever, so you never run out of naturals to map to subsets, > >nor subsets to map to naturals. > > Yes, if you imagine generating subsets of the naturals > one at a time, then you are only going to generate countably > many subsets. I am not generating a "countable" number of them. I am saying that there is a bijection between them, a 1-1 unique correspondence between elements, which does not end. > > The point is that for any rule for generating > subsets, there exists a subset that will *never* be generated > by that rule. So no rule can generate all possible subsets > of the naturals. (Cantor's theorem is not limited to sets > and functions that are defined by any particular rule, > but for concreteness, let's restrict ourselves to those.) The process defined by the rule will never end, but as much as the Peano axioms define the infinite set of naturals, the bijection generates the infinite set of sets, the power set. > > Notice that this is not the case with *finite* subsets. > If S is any finite subset of naturals, then we can associate > it with the natural number > > n = sum of S_i 2^i > > where S_i = 1 if i is an element of S, and S_i = 0 > otherwise. This is a rule for associating every finite > set of naturals with a natural, and there are no > finite subsets that are left out. Therefore, there > is a bijection between the set N of naturals and the > set FS(N) the finite subsets of N. But, if there are N finite naturals, there are still 2^N finite subsets. This power set relation holds for all power sets. > > If you are limiting yourself to describable mathematical > objects, then Cantor's theorem has the following form: > > If you give me any rule for mapping sets of naturals > to naturals, then I will give you a rule for creating > a set of naturals that is not in the range of your > mapping. And what does that prove? That any attempt to place a range on an infinite set and declare it finished causes a contradiction. > > Even though you can never complete the process of generating > an infinite set, you *can* complete the process of giving > a rule for one. For example, the rule "x is in the set if and > only if x is even" defines the infinite set of even natural > numbers. The rule "x --> 2*x" is a rule for mapping the set > N of natural numbers to the set E of even numbers. Right, and the rule I offered gives a mapping to a unique subset for each natural in N. Please tell what rule I broke when forming my bijection? Qhat disqualifies it? > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony |