So, it is possible to destroy all circles by removing a small set; how about preserving "many" circles while removing a "relatively large" set? The following open problem by Phil Tracy and myself ("A Problem in the Open Disc", Annals of the New York Academy of Sciences, v. 704, 1993, p. 353-4) does address the issue:

Is it possible to write the unit open disc as a disjoint union of a dense circular set--that is, a set where every two points lie on a subset circle--and a preferably dense set that has positive planar Lebesgue measure? ["Positive planar Lebesgue measure" may be replaced by "second category and property of Baire".]

The quoted reference provides some information on the problem's origins (unpublished, undergraduate oriented work on circular sets' classification). We have good indication that the question is non-trivial, hence answering it could lead to a publication. If you post something about it here please remember to send me a copy as I will have no access to the net after Tuesday; and if you submit something elsewhere, please mail a preprint to me (address given below) and/or Phil Tracy, PO BOX 425, Liverpool, NY 13088, USA. Thanks!

George Baloglou--SUNY at Oswego, NY 13126, USA

>Is it possible to write the unit open disc as a disjoint union of a >dense circular set--that is, a set where every two points lie on a >subset circle--and a preferably dense set that has positive planar >Lebesgue measure? ["Positive planar Lebesgue measure" may be replaced >by "second category and property of Baire".]

The category version has an affirmative answer -- there is a dense circular subset of the open unit disc which is meager. Let D be the open unit disc. If p and q are distinct points in D, then one can parametrize the set of circles through p and q using the angle alpha between the (tangent line to the) circle at p and the segment pq (say, for definiteness, the one on the "right" as one looks from q toward p); this angle is in the open interval from 0 to pi. The set of alpha's for which the circle is within D is an open subinterval (a,b) of (0,pi). Using this parametrization, we can set up a measure on the set of circles through p and q within D; for instance, we can say that "more than half of the circles through p and q within D" are in a certain set if the corresponding set of angles alpha is more than half of the subinterval (a,b).

Note that one can easily compute the diameter of the circle from the length x of the segment pq and the angle alpha; this diameter is just x csc alpha.

Lemma: If C is a compact subset of D, and z is a point not in C, then, for any epsilon > 0, there is an open neighborhood U of z such that, for any distinct p and q in C, of the circles through p and q within D, at most the fraction epsilon meet U.

Proof: Suppose not. Then, for each n > 0, if we let U_n be the open disc with center z and radius 1/n, then we can find distinct points p_n and q_n in C such that more than a fraction epsilon of the circles through p_n and q_n within D will meet U_n. By compactness of C, we can find a subsequence of the sequence of pairs (p_n,q_n) which converges to (p,q) with p and q in C.

Case 1: p and q are distinct. We can find a neighborhood V of z so small that less than a fraction epsilon/2 of the circles through p and q within D will meet V. Then we can find a very large n such that p_n is extremely close to p, q_n is extremely close to q, and U_n is included in V; this will ensure that the fraction of circles through p_n and q_n within D which meet U_n is less than epsilon, contradicting the choice of (p_n,q_n).

Case 2: p = q. Let R be the minimum of the distance from p to z and the distance from p to the unit circle (the boundary of D). Choose r > 0 so small that 2r(1 + csc (pi epsilon / 2)) < R. (We may assume epsilon < 1.) Now find a very large n such that p_n and q_n are within distance r of p and 1/n < r. Then, by the Note preceding the Lemma, if pi epsilon / 2 < alpha < pi(1 - epsilon/2), then the circle through p_n and q_n given by angle alpha must lie within D and must not meet U_n. Therefore, again, the fraction of circles through p_n and q_n within D which meet U_n is less than epsilon, contradicting the choice of (p_n,q_n).

QED Lemma.

Now, let A and A' be disjoint countable dense subsets of D; say A = {a_0,a_1,a_2,...} and A' = {a'_0,a'_1,...}. We construct a decreasing sequence G_0,G_1,G_2,... of open subsets of D, each including A, as follows.

Let G_0 = D. Given G_n, let B be the closed disc centered at the origin with radius 1 - 1/(n+2), and let C be the set difference B \ G_n; then C is a compact subset of D which is disjoint from A. For each k, apply the Lemma to get an open neighborhood U_k of a_k such that, for any distinct p and q in C, of the circles through p and q within D, at most the fraction 1/2^{k+2} meet U_k. Let W be the union of the sets U_k; then, for any distinct p and q in C, there is a circle through p and q within D which does not meet W (in fact, at least half of the circles through p and q within D do not meet W). Let G_{n+1} = (W intersect G_n) \ {a'_n}.

Each G_n is an open dense subset of D, so the intersection of the sets G_n is a comeager subset of D; let X be the complement of this comeager set in D. Then X is a meager subset of D which is dense, since it includes A'. It remains to show that X is circular. Let p and q be distinct points in X. Find n so large that p and q are within distance 1 - 1/(n+2) of the origin and p and q are not in G_n. Then the preceding paragraph ensures that there is a circle through p and q within D which does not meet G_{n+1}, and hence is included in X.

Randall Dougherty Department of Mathematics, Ohio State University, Columbus, OH 43210 USA