IMO 1991 SL 29

Origin: FIN

Problem

We call a set S on the real line R superinvariant if for any stretching A of the set by the transformation taking x to A(x) = x0 + a(x −x0) there exists a translation B, B(x) = x + b, such that the images of S under A and B agree; i.e., for any x \inS there is a y \inS such that A(x) = B(y) and for any t \inS there is a u \inS such that B(t) = A(u). Determine all superinvariant sets. Remark. It is assumed that a > 0.

Solution

One easily observes that the following sets are super-invariant: one-point set, its complement, closed and open half-lines or their complements, and the whole real line. To show that these are the only possibilities, we first observe that S is super-invariant if and only if for each a > 0 there is a b such that x \inS ⇔ax + b \inS. (i) Suppose that for some a there are two such b’s: b1 and b2. Then x \in S ⇔ax + b1 \inS and x \inS ⇔ax + b2 \inS, which implies that S is periodic: y \inS ⇔y + b1−b2 a \inS. Since S is identical to a translate of any stretching of S, all positive numbers are periods of S. Therefore S \equivR. (ii) Assume that, for each a, b = f(a) is unique. Then for any a1 and a2, x \inS ⇔a1x + f(a1) \inS ⇔a1a2x + a2f(a1) + f(a2) \inS ⇔a2x + f(a2) \inS ⇔a1a2x + a1f(a2) + f(a1) \inS. As above it follows that a1f(a2)+f(a1) = a2f(a1)+f(a2), or equiva- lently f(a1)(a2−1) = f(a2)(a1−1). Hence (for some c), f(a) = c(a−1) for all a. Now x \inS ⇔ax + c(a −1) \inS actually means that y −c \inS ⇔ay −c \inS for all a. Then it is easy to conclude that {y −c | y \inS} is either a half-line or the whole line, and so is S.