Tamás Keleti, ^{∗}Department of Analysis, Eötvös Loránd University, Múzeum krt. 6-8, Budapest, H-1088, Hungary (e-mail: elek@cs.elte.hu)

A 1-dimensional subset of the reals that intersects each of its translates in at most a

single point

Abstract We construct a compact subset of R with Hausdorff dimension 1 that intersects each of its non-identical translates in at most one point. Moreover, one can make the set to be linearly independent over the rationals.

In 1984 P. Mattila [2] constructed compact subsets A and B of R with Hausdorff dimension 1 such that the intersection of A and any translate of B contains at most one point. In this note we show that - if we allow only non-identical translations - one can also have A = B.

We call a set of 3 or 4 real numbers x_{1} < x_{2} ≤ x_{3} < x_{4} a rectangle if x_{2} − x_{1} = x_{4} − x_{3}.

Note that a set intersects each of its translates in at most one point if and only if the set does not contain a rectangle. (Here and in the sequel by set we will always mean a subset of R and by translate a non-identical translate.)

Theorem 1 There exists a compact set with Hausdorff dimension 1 that intersects each of its translates in at most one point.

Proof. Let δ_{m} = 1/(6^{m−1}m!). We define inductively compact sets A_{m} as disjoint unions of the closed intervals [n_{i_{1}...i_{m}}δ_{m}, (n_{i_{1}...i_{m}} + 1)δ_{m}] for 1 ≤ i_{k} ≤ k, 1 ≤ k ≤ m. We will denote by I_{1}^{m}, I_{2}^{m}, …, I_{m!}^{m} the intervals of A_{m} and by (J_{1}, J_{2}, …) the sequence (I_{1}^{1}, I_{1}^{2}, I_{2}^{2}, I_{1}^{3}, …, I_{3!}^{3}, …).

Let n_{1} = 0. (Then A_{1} = I_{1}^{1} = J_{1} = [0, 1].) Assume that A_{1}, …, A_{m} have already been defined. If n_{i_{1}...i_{m}}δ_{m} ∉ J_{m} then let

n_{i_{1}...i_{m}i} = 6(m + 1)n_{i_{1}...i_{m}} + 6i − 6  (i = 1, …, m + 1),  (1)

Key Words: Hausdorff dimension, translation, linearly independent

Mathematical Reviews subject classification: 28A78

∗This research was done while the author was visiting the University College London having a Royal Society/NATO Postdoctoral Fellowship award.

and if n_{i_{1}...i_{m}}δ_{m} ∈ J_{m} then let

n_{i_{1}...i_{m}i} = 6(m + 1)n_{i_{1}...i_{m}} + 6i − 3  (i = 1, …, m + 1).  (2)

Thus

[n_{i_{1}...i_{m}i}δ_{m+1}, (n_{i_{1}...i_{m}i} + 1)δ_{m+1}] ⊂ [n_{i_{1}...i_{m}}δ_{m}, (n_{i_{1}...i_{m}} + 1)δ_{m}]

for i = 1, …, m + 1, which means that the intervals of A_{m+1} are contained in the intervals of A_{m}.

Let A = ∩_{l=1}^{∞}A_{m}. Then A has Hausdorff dimension 1, cf. [1] Example 4.6. Hence, by our previous remark, it is enough to show that A does not contain a rectangle.

Let x_{1} < x_{2} ≤ x_{3} < x_{4} be points of A. Take an m such that δ_{m} < x_{2} − x_{1}. Then if x_{1} ∈ I_{j}^{m} = J_{M} then none of x_{2}, x_{3} and x_{4} is in I_{j}^{m}. Thus, when we defined A_{M+1}, we used (2) for defining the interval that contains x_{1} and (1) for defining the intervals that contain x_{2}, x_{3} and x_{4}. This implies that x_{1} is of the form (6N_{1} +3)δ_{M} +ε_{1} but x_{2}, x_{3} and x_{4} are of the form 6N_{j}δ_{M} +ε_{j}, where N_{1}, …, N_{4} are integers and 0 ≤ ε_{i} ≤ δ_{M} for i = 1, …, 4. Thus x_{2} − x_{1} ≠ x_{4} − x_{3}, which means that (x_{1}, x_{2}, x_{3}, x_{4}) is not a rectangle.

Remark 2 Slightly modifying the above construction (by replacing 6 with a slowly increasing sequence of even numbers) one can also get a compact set with Hausdorff dimension 1 which is linearly independent over the rationals. (The existence of a linearly independent perfect set is well known, even in any non-discrete locally compact abelian group, see e. g. [3].)

References

[1] K. Falconer, Fractal Geometry, John Wiley & Sons, 1990.

[2] P. Mattila, Hausdorff dimension and capacities of intersections of sets in n-space, Acta Math., 152 (1984), 77-105.

[3] W. Rudin, Fourier Analysis on Groups, Interscience Publishers, New York - London, 1962.