The right way to question the uncountability of the reals
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From: George Greene on 11 Jun 2010 17:43 On Jun 11, 2:39 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > And we don't need full ZFC to prove that any such carrier set is > uncountable. You certainly don't need the Axiom of Choice. What most people keep overlooking is that you don't need the Axiom of Infinity EITHER. The finite ordinals may Or May Not form a set, BUT IF THEY DO, then that set, LIKE ALL sets, IS NOT as big as its OWN powerset. This is simply a fact about sets IN GENERAL that has nothing whatSOEVER to do with the reals specifically! Or even with infinite sets specifically! This is a way in which infinite sets ARE EXACTLY LIKE finite sets. As OPPOSED to the more confusing usual ways in which infinite sets are UNlike finite sets (e.g. being bijectible with sparse subsets of themselves, or approaching a limit without actually containing it).
From: George Greene on 11 Jun 2010 17:44 "The reals" is not even the right entity to talk about. Any old infinite set will do, if infinity is where you want to take this. But it simply shouldn't be. It's called CANTOR's theorem and Cantor was talking about SET THEORY, so you HAVE to begin WITH A SET THEORY and NOT with "the reals".
From: Rupert on 11 Jun 2010 17:59 On Jun 12, 7:43 am, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 11, 2:39 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > And we don't need full ZFC to prove that any such carrier set is > > uncountable. > > You certainly don't need the Axiom of Choice. > What most people keep overlooking is that you don't need > the Axiom of Infinity EITHER. > The finite ordinals may Or May Not form a set, BUT IF THEY DO, > then that set, LIKE ALL sets, IS NOT as big as its OWN powerset. > This is simply a fact about sets IN GENERAL that has nothing > whatSOEVER > to do with the reals specifically! Or even with infinite sets > specifically! > This is a way in which infinite sets ARE EXACTLY LIKE finite sets. > As OPPOSED to the more confusing usual ways in which infinite sets > are UNlike finite sets (e.g. being bijectible with sparse subsets of > themselves, > or approaching a limit without actually containing it). If we don't have the axiom of infinity then it's a bit of a challenge to define the real numbers. Most definitions of the real numbers assume that infinite sets of naturals exist.
From: George Greene on 12 Jun 2010 02:04 On Jun 11, 5:59 pm, Rupert <rupertmccal...(a)yahoo.com> wrote: > If we don't have the axiom of infinity then it's a bit of a challenge > to define the real numbers. OF COURSE, yet it is NOT a challenge to prove that ANY set is not as big as its powerset. ANY set. "Of the reals" or "of all finite ordinals" or ANYthing -- IT DOES NOT EVEN *MATTER*!! People who are invoking "the reals" around this issue are MISSING THE POINT!
From: Daryl McCullough on 12 Jun 2010 09:23
George Greene says... > >"The reals" is not even the right entity to talk about. >Any old infinite set will do, if infinity is where you want to take >this. >But it simply shouldn't be. It's called CANTOR's theorem and Cantor >was talking about SET THEORY, so you HAVE to begin WITH A SET THEORY >and NOT with "the reals". Well, Cantor gave two different proofs of the uncountability of the reals. One of them generalizes to arbitrary sets, while the other one is very specific to the reals. The one that is specific to the reals goes like this: Let r_0, r_1, ... be an infinite list of reals. Then there is a real r that is not equal to any real on the list. Proof: We assume that the sequence r_n is dense, and has no largest element (otherwise, it's trivial to find a real that is not on the list). Define two sequences of reals a_n and b_n as follows: Let a_0 = r_0. Let b_0 = the first real r_k in the original sequence such that r_k > a_0. For n > 0, let a_{n+1} = the first real r_k such that a_n < r_k < b_n. Let b_{n+1} = the first real r_k such that a_{n+1} < r_k < b_n. Then we have the following inequalities: a_0 < a_1 < a_2 < ... < b_2 < b_1 < b_0 a_n is an increasing sequence of reals bounded from above, and b_n is a decreasing sequence of reals bounded from below, and there is no overlap between the sequences. So let a_lim be the limit of a_n. Let b_lim be the limit of b_n. Finally, let r = any number such that a_lim <= r <= b_lim. It can be demonstrated easily that r is unequal to any r_k. This proof uses essential properties of the reals, such as the fact that they are totally ordered, and that every bounded sequence of increasing (or decreasing) reals has a limit. -- Daryl McCullough Ithaca, NY |