Bill Thurston wrote: << 1. A point of density of a set is a point such that for small enough neighborhoods, the density of the set in that neighborhood is arbitrarily close to 1. I.e. every neighborhood smaller than delta has 99.99% of points in the set, etc. The Lebesgue density theorem asserts that almost every element of any measurable set is a point of density. I.e. for any measurable subset of R (or any other space) almost every point has neighborhoods with density approaching 1 or density approaching 0. This is a fundamental principle of measure theory, I think it is (or at least it should be) in the basic texts. It really comes from the way measure is defined.
I think the following is right, though I haven't done measure theory in a while: If H is a measurable subset of [0,1] such that (*) meas(H \int [a,b]0i) = (b-a)/2 for all 0 <= a <= b <= 1 then of course meas(H) = meas(H \int [0,1]) = (1-0)/2 = 1/2. But this means that the infimum of |open cover| of H by intervals is 1/2, where |{o_i : i in Z+}| is defined as (sum over i in Z+ of meas(o_i)). So let O = {o_i} be an open cover of H with (**) 1/2 <= |O| < 1/2 + eps for any fixed eps > 0. But by (*) we also have meas(H \int o_i) = meas(o_i)/2. So for each o_i, the set (H \int o_i) has its own open cover C_i = {c_ik : k in Z+}, such that (***) meas(o_i)/2 <= |C_i| < meas(o_i)/2 + eps/(2^i) But the union the c_ik (over all i and k) contains H. Hence (R to L, then top to bottom) 1/2 <= sum over i in Z+ of (sum over k in Z+ of meas(c_ik)) = =sum over i of |C_i| < sum over i of (meas(o_i)/2 + eps/(2^i) (by ***) =(1/2) |O| + eps < (1/2)(1/2 + eps) + eps (by **) = 1/4 + (3/2)eps. By choosing eps = 1/6, this long inequality becomes 1/2 < 1/2 => <=. QED. << 2. You can take an irrational rotation of the circle R/Z like x -> x + sqrt(2). The orbits of this rotation are all isomorphic to Z --- i.e. points never return to themselves. Pick a point from each orbit, using the axiom of choice. Partition the orbit into the even images and the odd images of this point. The union of all odd points in all orbits is "half" the circle in the sense that its congruent by a translation to its complement. In fact you can take odd powers of the rotation arbitrarily close to the identity, so you can make the set match its complement by an arbitrarily small rotation. This example is covered by an example in R --- you could think of this as taking any action of Z^2 (or Z^k would also work) by translations on R with dense orbits, then partitioning the orbits into cosets of a subgroup of index 2 acting on a selected point It's known that there's no definition or rule or humanly comprehensible procedure to actually make such a selection. Bill
This is the same example in R/Z that I had in mind. While the two pieces of R/Z are dense, and isometric by a translation, I don't think it's clear they have a third quality I claimed: that each is homogeneous. (But I believe I've seen a construction that causes this, too, to be true.) Also: When I first learned that the Axiom of Choice is equivalent to the statement that the cartesian product of nonempty sets is nonmpty . . . I became a believer. --Dan -------------------------------------------------------------------------------- On Apr 18, 2005, at 9:24 PM, Michael Kleber wrote:
Dan Asimov wrote:
<< Can someone give me an example of a set of density 1/2 on every interval of the real numbers?
If such a set must be measurable, then no such set exists.
On the other hand, there do exist partitions of the reals into two dense homogeneous subsets that are related by a translation.
Probably I should know why both of these are true, but right now it seems that I don't. Can you explain?