Sets meeting isometric copies of the lattice Z² in exactly one point

Abstract

The construction of a subset S of ℝ² such that each isometric copy of ℤ² (the lattice points in the plane) meets S in exactly one point is indicated. This provides a positive answer to a problem of H. Steinhaus [Sierpiński, W. (1958) Fund. Math. 46, 191–194].

In the 1950s, Steinhaus posed the following problem. Is there a set S in the plane such that every set congruent to ℤ² has exactly one point in common with S? The problem seems to have first appeared in a 1958 paper of Sierpiński (1). Steinhaus also asked several related questions that have been stated and studied in refs. 1–3. This specific problem has been widely noted (see, e.g., refs. 4 and 5) but has remained unsolved until now. Here by using a combination of techniques from analysis, set theory, number theory, and plane geometry we show the answer is in the affirmative.

Theorem 1.

There is a set S ⊆ ℝ² such that for every isometric copy L of the integer latticeℤ² we have |S ∩ L| = 1.

We call a set S as in Theorem 1 a Steinhaus set and note that whether there can be a Lebesgue measurable Steinhaus set remains unsolved. This problem has been the origin of many papers including those of Beck (6), Croft (7), Komjáth (3), and Kolountzakis (8). Kolountzakis and Wolff (9) showed that there is no measurable Steinhaus set for the higher-dimensional version of Steinhaus's problem for the standard lattice. Steinhaus's problem and variants were discussed in some detail by Croft (7) and have been updated in sections E10 and G9 of ref. 4.

A straightforward induction argument quickly runs into problems, as noted in refs. 7 and 10. We avoid this by using a hull construction that we describe shortly.

Let us say a lattice distance is a real number of the form , where n, m ∈ ℤ. Let ρ denote Euclidean distance in ℝ². Our methods allow us to prove a strengthening of Theorem 1.

Theorem 2.

There is a set S ⊆ ℝ² satisfying:

• 1.  For every isometric copy L ofℤ² we have S ∩ L ≠ ∅.

• 2.  For all distinct z₁, z₂ ∈ S, ρ(z₁, z₂)² ∉ ℤ.

The Steinhaus problem has a natural interpretation for smaller sets of lattices. Namely, given an arbitrary set ℒ of lattices (each of which is an isometric copy of ℤ²), we may ask whether there is a set S satisfying part 2 of Theorem 2 and such that S ∩ L ≠ ∅ for all L ∈ ℒ. We call a set S ⊆ ℝ² satisfying part 2 a partial Steinhaus set.

Our first step toward proving Theorem 2 is establishing this restricted version of the problem for the case where ℒ is the (countable) family of rational translations of ℤ². The proof involves only elementary number theory and combinatorics.

Lemma 1.

Let ℒ_ℚ denote the set of rational translations of ℤ², that is, lattices of the form ℤ² + (r, s) wherer, s ∈ ℚ. Then there is a set S ⊆ ℝ² satisfying the following:

• 1.  For every lattice L ∈ ℒ_ℚ, S ∩ L ≠ ∅.

• 2.  For all distinct z₁, z₂ ∈ S, ρ(z₁, z₂)² ∉ ℤ.

Proof:

We want to show there is a map f : (ℚ/ℤ) × (ℚ/ℤ) → ℚ × ℚ such that ρ²(f(x), f(y)) ∉ ℤ holds for x ≠ y and f is a selector: f([r], [s]) ∈ [r] × [s]. The set S is the image set of the selector.

For this we define a graph on the Abelian group ℚ/ℤ × ℚ/ℤ by joining x with y if and only if there are elements g(x) ∈ x + (ℤ × ℤ) and g(y) ∈ y + (ℤ × ℤ) such that the square of the distance between g(x) and g(y) is an integer. When constructing the selector we only have to worry about points that are joined. Let H be the subgroup of ℚ × ℚ consisting of elements with denominators only divisible by primes of the form 4k + 1. Note that if (r, s) ∈ ℚ × ℚ and r² + s² ∈ ℤ, then (r, s) ∈ H. For, suppose r = a/b, s = c/d (written in lowest terms), and r² + s² = e ∈ ℤ. Suppose p = 2 or p is a prime congruent to 3 mod 4, and p divides b or d. Then the exact power of p dividing b, say p^k, must be the exact power dividing d, as otherwise multiplying through by the square of the least common multiple m of b and d would give r²m² + s²m² = em² where the right-hand side and exactly one of the left-hand side terms are divisible by p, a contradiction. Consider the case p ≡ 3 mod 4. Since these exact powers are the same, neither term on the left is divisible by p, and both are nonzero mod p. This would give −1 is a square mod p, a contradiction, as p ≡ 3 mod 4. The case p = 2 is left to the reader. Thus, if [x], [y] are joined, then x − y ∈ H; that is, no edges go between distinct cosets of H, so it suffices to make the construction per cosets, and this again reduces to making the construction on H.

To motivate the following argument, let x = (i₁/d, j₁/d), y = (i₂/d, j₂/d), where d is only divisible by primes congruent to 1 mod 4, and suppose f([x]) = x + (k₁, l₁), f([y]) = y + (k₂, l₂) and ρ(f([x]), f([y]))² ∈ ℤ. Multiplying through by d² this becomes

Suppose p^a is one of the prime components of d, and p^e, p^f are the exact powers of p dividing i₁ − i₂, j₁ − j₂, respectively. If e ≥ a, then easily f ≥ a as well, and conversely. If e < a, then easily e = f. This gives (i₁ − i₂/p^e)² + (j₁ − j₂/p^e)² ≡ 0 mod p^a−e. Recall that for each prime power p^a, where p ≡ 1 mod 4, there are exactly two square roots, say λ_p^a, μ_p^a, of −1 mod p^a, and for any a′ < a, λ_p^a mod p^a′, μ_p^a mod p^a′ are the roots mod p^a′. So we must have (j₁ − j₂/p^e) ≡ λ_p^a(i₁ − i₂/p^e) mod p^a−e, where λ_p^a denotes one of the roots. So, j₁ − j₂ ≡ λ_p^a(i₁ − i₂) mod p^a holds in either case (e ≥ a or e < a). Since this holds for each prime power p^a of d, we conclude that for ρ(f([x]), f([y]))² ∈ ℤ to hold, we must have j₁ − j₂ ≡ λ(i₁ − i₂) mod d, where λ² ≡ −1 mod d. To further motivate the argument, suppose now that d = p^a is a prime power and j₁ − j₂ = λ(i₁ − i₂) where λ is chosen so that λ² ≡ −1 mod p^2a. Substituting in Eq. 1 and dividing through by 2d(i₁ − i₂) gives k₁ + λl₁ ≡ k₂ + λl₂ mod p^a−b, where p^b is the exact power dividing i₁ − i₂. Note, for example, that if a = 1 (i.e., d is a prime), then this equation would not hold if we had k₁ = k₂ = 0 and l₁ = i₁, l₂ = i₂. This suggests we somehow write each x = (i/d, j/d) in terms of the “basis elements” (1, λ_p^a), (1, μ_p^a) for each p^a dividing d and use this to define the k, l values [where f([x]) = x + (k, l)]. We now carry out the details of this argument.

Let P₁, P₂, … enumerate the positive powers of primes of the form 4k + 1 such that if P_i divides P_j, then i ≤ j. By recursion, for each i fix the distinct mod P_i natural numbers λ_i, μ_i > 0 such that P|λ + 1, μ + 1 and P|λ_j − λ_i, μ_j − μ_i, where P_j is the next member of the sequence that is a power of the same prime as P_i. Note that λ_i and μ_i > 0 are the distinct square roots of −1 mod P. Let B(P_i) be an integer divisible by every P₁, … , P_i−1 but not by P_i, and, if P_i = pⁿ, then let A(P_i) be an integer divisible by each of P₁, … , P_i−1, which are not powers of p, but (A(P_i), P_i) = 1. If (x, y) is a pair with x, y rational numbers, 0 ≤ x, y < 1, and the denominators of x, y are only divisible by primes of the form 4k + 1, then (x, y), we claim, can uniquely be written mod 1 as

with 0 ≤ α_i, β_i < p; here p is the prime whose power P_i is. To see this, first write (x, y) = [(e₁, f₁)/p] + ⋯ + [(e_k, f_k)/p] mod 1, where the e_i, f_i are integers and d = p ⋯ p. Each term in this sum is of the form (e, f)/P_j for some P_j, and it is enough to show that this term can be written as ∑[A(P_i)/P_i](α_i(1, λ_i) + β_i(1, μ_i)) where the P_i range over the divisors of P_j, and 0 ≤ α_i, β_i < p. Find α_j, β_j in the desired range with A(P_j) (α_j(1, λ_j) + β_j(1, μ_j)) ≡ (e, f) mod p, where P_j is a power of p. This is possible as (A(P_j), p) = 1, and the two-by-two system is nonsingular mod p. Then (e, f)/P_j − A(P_j)/P_j (α_j(1, λ_j) + β_j(1, μ_j)) is of the form (e′, f′)/P_k where P_j = P_kp. Continuing, we finish. The uniqueness part of the claim is easily checked. We now add (0, ∑t_iB(P_i)) to the point defined by Eq. 2, where t_i = (α_i + β_i). This will be f([x], [y]).

Assume that the square of the Euclidean norm of the difference of two such points is an integer. The difference of the two points is of the form

where S = ∑(u_i + v_i)B(P_i) with −(p − 1) ≤ u_i, v_i ≤ (p − 1) for P_i = pⁿ, and the sum here is taken over all i such that not both u_i and v_i are zero.

If the point given by Eq. 3 is (a/d, b/c + S), where (a, d) = (b, c) = 1, then d = c, as otherwise the square of the norm could not be an integer. From this we get that b ≡ λa (mod d), where λ² + 1 ≡ 0 (mod d²). Next, note that all the P_i that occur in the sum in Eq. 3 divide d. Thus, for every i, either λ ≡ λ_i (mod P) or λ ≡ μ_i (mod P), and the same case must hold for powers of the same prime. By renaming, if needed, we assume that λ ≡ λ_i (mod P) holds for every i.

Since b ≡ λa (mod d) and λ ≡ λ_i (mod P), we have

Notice that p does not divide λ_i − μ_i where P_i is a power of p, and from this we get by backward induction on i that v_i = 0 holds for every i.

What we have now is that

is an integer, where

and

Let i be the first index such that u_i ≠ 0, and let P_i = pⁿ. Then the first nonzero term of S, that is, u_iB(P_i), is divisible by pⁿ⁻¹ but not by pⁿ and all later terms are divisible by pⁿ [because of the factors B(P_j)]. So S is divisible by pⁿ⁻¹ but not by pⁿ. If we replace b/d by λa/d then we can easily conclude, using λ² ≡ −1 (mod d²),

which is certainly not an integer, as d is divisible by pⁿ while S is not.

We now argue that the same holds for (a/d, b/d) instead of (a/d, λa/d). To this end, it suffices to show that the integer λa/d − b/d is divisible by pⁿ. Indeed, if X is that last difference, then we can repeat the above argument with S − X in place of S.

To show the last claim, decompose it as

where ∑′ contains the terms with p|P_j and ∑" contains the other terms. In the first sum, using the fact that λ ≡ λ_j (mod P), every term is an integer divisible by pⁿ. The second sum is an integer of the form B/C where B is divisible by pⁿ [because of the factors A(P_j)] but C is not divisible by p.

Note that for any choice of the A(P_i), B(P_i) we have f([0], [0]) = (0, 0) in the above construction. However, for any (e, f) ∈ ℤ² we could add (e, f) to the value of f([x], [y]) for all x, y and still keep the squared distances between distinct points in the range of f noninteger. Thus, we are free to make f([0], [0]) any point in ℤ².

Actually, we require a slightly stronger form of Lemma 1. To state it we call sets of the form x + (ℤ × ℤ), (x ∈ ℚ × ℚ) ℤ × ℤ-subsets of ℚ × ℚ. Also, a subset E ⊆ ℚ × ℚ is small, if for every ℤ × ℤ-subset D, D ∩ E is contained in the union of finitely many lines. If L is isometric to ℤ² and Q is the set of points having rational coordinates with respect to L, then we define in an analogous manner the notion of E being small relative to L.

Let d be a positive integer. Let R_d be the subgroup of ℚ/ℤ × ℚ/ℤ of points ([x], [y]) where x, y can be written with denominator d. Let H_d = H ∩ R_d. If d = p⋯ pq⋯ q where each p_i ≡ 1 (mod 4) and each q_j = 2 or q_j ≡ 3 mod 4, then H_d are those ([x], [y]) where x, y can be written with denominator p ⋯ p. Note that R_d is isomorphic to the direct sum H_d ⊕ K_d, where K_d is the subgroup of ([x], [y]) where x, y can be written with denominator q ⋯ q. In particular, the distinct cosets of H_d in R_d are given by ([x], [y]) + H_d, where ([x], [y]) ∈ K_d.

Suppose now f is a selector on R_d. We say f is good if on each coset of H_d in R_d, f is defined as in the construction of Lemma 1. More precisely, we mean the following. Let d = p⋯ pq ⋯ q as above (if d = 1, we declare f to be good). Let d̄ = p ⋯ p. Then for each coset ([x], [y]) + H_d, where ([x], [y]) ∈ K_d, there is a sequence of prime powers P₁, P₂, … , P_i0 that “build up” d̄ (that is, d̄ is the product of the P_i that do not divide another term in the sequence) and integers A(P_i), B(P_i), i ≤ i₀, as in the proof of Lemma 1, and an initial translation (e, f) such that for all ([r], [s]) ∈ H_d, f(([x], [y]) + ([r], [s])) = (e, f) + f′([r], [s]), where f′ is defined on H_d as in Lemma 1 using the P_i, A(P_i), and B(P_i).

It was shown in the proof of Lemma 1 that if f is a good selector on R_d, then the range of f is a partial Steinhaus set.

Lemma 2.

Let d be a positive integer and f be a good selector on R_d. Suppose d|d′and E is a small set missing ran(f). Thenf may be extended to a good selector f′ onR_d′ whose range also misses E.

Proof:

Using the fact that the constructions on the different cosets of H are independent (compare the first paragraph in the proof of Lemma 1) and also that a rational translation of a small set is small, it is enough to show the following (changing somewhat the notation from the statement of the lemma). Let d be a product of positive powers of primes congruent to 1 mod 4. Let P₁, P₂, … , P_i−1 build up d (as defined above). Let A(P₁), B(P₁), … , A(P_i−1), B(P_i−1) satisfy the requirements given in Lemma 1. Let f be the selector on H_d defined from these quantities as in Lemma 1. Suppose finally that E is a small set missing ran(f), and P_i is given with P₁, … , P_i building up d′. Then we show that there are A(P_i), B(P_i) so that if f′ is defined using these extended sequences, then the range of f′ misses E. The point is we have enough freedom in choosing the values of A(P_i), B(P_i). We are assuming f([x], [y]) has been defined for all (x, y) having a representation as in Eq. 2 with the sum ranging over prime powers in the list P₁, … , P_i−1. Let A, B, satisfy the requirements for A(P_i), B(P_i) given in Lemma 1. Let U be the product of the P_j with j < i and P_j, P_i relatively prime. Then A′ = A + KP_iU, B′ = B + LP_iU also satisfy these requirements for any K, L ∈ ℤ. Let f′ be as constructed in Lemma 1 by using A′, B′. For any fixed x, y ∈ R_d′ − R_d, a computation shows that the corresponding values of f′([x], [y]) will have the form (x′ + KUa, y′ + KUb + LUP_ia) for some fixed x′, y′ ∈ ℚ, a, b ∈ ℤ with a, b ≠ 0. Since E is small, for each [x], [y] the requirement that f([x], [y]) ∉ E rules out only the K, L lying on finitely many lines in ℤ × ℤ. As there are only finitely many [x], [y] to consider at stage i, it is clear that we can choose K, L so that f′ misses E.

We thank one of the referees for pointing out the proofs of the preceding lemmas. These proofs are based on and motivated by the more complicated proofs that we give in ref. 10 for stronger results. To state one of these results we adopt the following terminology. For rationals r, s, let L_r,s = ℤ² + (r, s) be the rational translation of ℤ² by (r, s). Let R = ℚ² ∩ ([0, 1) × [0, 1)). For each positive integer d, let R_d ⊆ R be defined by R_d = {(i/d, j/d) : 0 ≤ i, j < d}. We prove the following in ref. 10.

Lemma 3.

Let d > 1 and suppose functions k, l mapping R_d into ℤ have been defined such that setting S_d = {(r + k(r, s), s + l(r, s)) : r, s ∈ R_d} we have:

Then for any d′ with d|d′, the functions k, l may be extended toR_d′ so as to satisfy (∗)_d′.

This last lemma assures us not only that there is a set S satisfying Lemma 2, but also that any partial Steinhaus set defined for the translates in R_d may be extended. We also prove similar extension theorems where we must avoid small obstruction sets. These extension lemmas are much stronger than is required for the proof of our main theorem, but we feel that they may be useful in studying analogous problems for other lattices, dimensions, and groups of isometries.

By a rational translation of a lattice L we mean a lattice of the form L + ru→ + sv→, where r, s ∈ ℚ, and u→, v→ are the unit basis vectors of L.

Definition:

Two lattices are equivalent L₁ ∼ L₂, if L₂ can be obtained from L₁ by rational rotations and translations.

In other words, L₁ ∼ L₂, if and only if all of the points of L₂ have rational coordinates with respect to the coordinate system determined by L₁ (and vice versa). This is easily an equivalence relation, and each equivalence class is countable. We also note that if two lattices are not equivalent, then there can be at most one point with rational coordinates with respect to both of them.

An important aspect of the construction is that if one has a Steinhaus set for all the rational translates of a given lattice L, then it is a Steinhaus set for the equivalence class of L:

Lemma 4.

Let L₁ be a lattice and suppose S ⊆ ℝ² satisfies the following:

• 1.  For every lattice L that is a rational translation of L₁, S ∩ L ≠ ∅.

• 2.  For all distinct z₁, z₂ ∈ S, ρ(z₁, z₂)² ∉ ℤ.

Then for every lattice L′ that is equivalent toL₁, we have S ∩ L′ ≠ ∅.

Naturally, this last observation suggests constructing a full Steinhaus set by induction on the equivalence classes of lattices. This leads us to the set-theoretic part of the proof. The transfinite induction considers collections of lattices that are sufficiently closed. The main closure property we need arises from Lemma 6, and the corresponding argument is given in the Claim below. Rather than specify at the outset exactly what closure properties we wish our sets at each stage to satisfy, it is more convenient to follow a standard set-theoretic practice. If CH held, we would at stage α consider those equivalence classes of lattices that lie in an elementary substructure (or hull) M_α of V_λ for some large λ (though actually λ = ω + ω will suffice). Here V_λ denotes the initial segment of the universe of sets of rank less than λ, and an elementary substructure denotes a subset closed under the functions f: (V_λ)ⁿ → V_λ, for some n, which are definable in V_λ (the so-called Skolem functions of V_λ). For the benefit of the reader unfamiliar with set-theoretic terminology, we note that this is merely a convenient shorthand for requiring that the sets we consider be sufficiently closed without having to specify in advance exactly which functions we want them to be closed under. Thus, in place of M_α the reader could substitute “a sufficiently closed collection of points and lattices” (the closure properties needed will be apparent from the following argument and could be specified in advance).

In the general case (not assuming CH) the construction is essentially an iteration of this substructure construction. Although the set-theoretic terminology could be largely eliminated (see also the comments at the end of this article), we believe doing so would lessen the generality of the method and hide our motivation (for example, it was by pursuing the “hull” strategy just outlined that we were led to consider Lemma 6).

We now describe the particular well-ordering of the equivalence classes that we will use.

First, some notation. If L ⊆ ℝ² is an isometric copy of ℤ², let [L] denote the equivalence class of L under the equivalence relation ∼ of the definition. Let denote the family of all equivalence classes. Let ℒ → L(ℒ) be a function that picks for each equivalence class ℒ a member L(ℒ) ∈ ℒ.

The construction of the enumeration begins by letting {M_α0 : α₀ < 2^ω} be an increasing continuous (i.e., at limit stages we take unions) chain of elementary substructures of a large V_λ with |M_α0| < 2^ω for all α₀ < 2^ω and such that each lattice and each equivalence class of lattices is in one of these substructures. Assume also that M₀ = ∅. (The starting set M₀ is an exception in that it is not an elementary substructure.) Let N_α0 = M_α0+1 − M_α0.

By simultaneous recursion we define a subtree T of ON^<ω and an ordinal κ(α→) and sets M_α→, N_α→ for α→ ∈ T. Set κ(∅) = 2^ω. In general, suppose that M_α→ is defined for α→ in a certain subtree of ON^<ω. If M_α0,...,αk is defined, we assume also that κ(α₀, … , α_k−1) has been defined and is an infinite cardinal. Furthermore, we assume in this case that M_{α0,...,αk−1,β} is defined if and only if β < κ(α₀, … , α_k−1). We let N_α0,...,αk denote M_{α0,...,αk+1}− M_α0,...,αk.

Suppose now that M_α0,...,αk is defined. If N_α0,...,αk contains only countably many equivalence classes of lattices, let {ℒ_α0, … , α_k;n}_n∈s, where s ≤ ω, enumerate these equivalence classes. In this case, (α₀, … , α_k) is a terminal node in the tree, T, of indices α→ for which M_α→ is defined. Otherwise, let κ(α₀, … , α_k) = |N_α0,...,αk| and express

as a continuous increasing union, where each M_{α0,...,αk,αk+1} is the intersection of N_α0,...,αk with an elementary substructure of V_λ, and each M_α0,…,α_k,αk+1 has size < κ(α₀, … , α_k). Assume also M_{α0,...,αk,0}= ∅. We note two facts. Easily, the tree of indices is well founded (since the cardinals κ_α→ are strictly decreasing along any branch). Also, if a₁, … , a_m ∈ M_{α0,...,αk,αk+1} and f is a Skolem function of V_λ and f(a₁, … , a_m) ∈ N_α0,...,αk, then f(a₁, … , a_m) ∈ M_{α0,...,αk+1}.

Notice if α→ is incompatible with β→, then N_α→ and N_β→ have no equivalence class in common. Furthermore, every equivalence class occurs as some ℒ_{α0,...,αk;n}. Thus, the ℒ_{α0,...,αk;n} precisely enumerate the equivalence classes of lattices. By an “ω-block” of lattices we mean the (equivalence classes of) lattices of the form ℒ_{α0,...,αk;n} for some fixed terminal index α₀, … , α_k. We consider the indices to be (well) ordered lexicographically.

The following easy lemma will be used.

Lemma 5.

Suppose α→ is an index for whichM_α→ is defined. Leta₁, … , a_m ∈ M_α→ and suppose b is definable from a₁, … , a_m inV_λ. Then b ∈ ∪_{β→≤α→} M_β→.

The idea for constructing a Steinhaus set is by transfinite induction along the terminal nodes of the tree T. However, it turns out we need another lemma in pure plane geometry to make an inductive extension.

Lemma 6.

Let c₁, c₂, c₃ be three distinct points in the plane, and let r₁, r₂, r₃ > 0 be real numbers. LetC₁ be the circle in the plane with center atc₁ and radius r₁, and likewise for C₂ and C₃.Let a, b, c be three distinct points in the plane. Then, except for the exceptional case described afterward, there are only finitely many triples of points (p₁, p₂, p₃) in the plane such that

• 1.  p₁ ∈ C₁, p₂ ∈ C₂, andp₃ ∈ C₃.

• 2.  The trianglep₁p₂p₃ is isometric with the triangle abc (we allow the degenerate case where the points a, b, c are collinear).

The exceptional case is when r₁ = r₂ = r₃ and the triangle abc is isometric with c₁c₂c₃.

This lemma seems to be known within engineering mathematics specifically regarding the geometry of mechanisms (11). Although not explicitly stated, Lemma 6 follows from the analysis of Gibson and Newstead in ref. 12. The Gibson–Newstead argument uses a significant amount of algebraic geometry. We also give two elementary proofs of Lemma 6 in ref. 10, an algebraic one using Gröbner bases and computer algebra, and the other a purely geometric proof.

Fix now a terminal index α→ = (α₀, … , α_k). Assume inductively we have defined for each terminal index β→ < α→ a set S_β→ ⊆ ℝ² that satisfy the following:

1. If β→₁ < β→₂ < α→, then S_β→1 ⊆ S_β→2.

2. For every terminal index β→ less than α→, S_β→ meets every lattice in every equivalence class ℒ_β→;n.

3. Every point of S_β→ − ∪_γ→<β→S_γ→ lies on some lattice of the form ℒ_β→;n.

4. For all distinct z₁, z₂ ∈ S_β→, ρ(z₁, z₂)² ∉ ℤ.

5. Suppose β→₁ < β→₂ < α→, x ∈ S_β→1, and y ∈ S_β→2 − ∪_γ→<β→2S_γ→. Then if ρ(x, y)² ∈ ℚ, then x, y both have rational coordinates with respect to some lattice of the form ℒ_β→2;n.

Let S_<α→ = ∪_β→<α→ S_β→. We show how to extend S_<α→ to a set S_α→ also satisfying 4, 5 and such that S_α→ meets every lattice in each equivalence class ℒ_α→,n. This suffices to prove Theorem 2.

To ease notation, let ℒ_n = ℒ_α→;n, and let L_n = L(ℒ_n). From Lemma 4, it suffices to maintain property 4, have property 5 when β→₂ = α→, and have S_α→ meet every rational translation of each L_n (recall a rational translation of L_n refers to a motion that is a translation in the coordinate system of L_n).

For integers n, d, i, j, let L denote the translation of L_n by the amount (i/d, j/d) (in the coordinate system of L_n). We also need the following easy lemma.

Lemma 7.

Let L be a lattice and z ∈ ℝ². Suppose z has coordinates (x, y) with respect to the lattice L, where at least one of x, y is irrational. Then there is a line l = l(z, L) such that if w has rational coordinates with respect to L and w ∉ l, then ρ(w, z)² ∉ ℚ.

These last two geometric lemmas give the following claim (see ref. 10 for details).

Claim:

For each n and rationals i/d, j/d, there is a finite set of lines G_n(i/d, j/d) with the following property: if c ∈ S_<α→ does not have rational coordinates with respect to L_n, if z ∈ L, and if ρ(c, z)² ∈ ℚ, then z ∈ ∪ G_n(i/d, j/d).

Fix a bijection (n, m) ↦ 〈n, m〉 ∈ ω between ω² and ω. We may assume this bijection is increasing in each coordinate. Let Q_n be those points having rational coordinates with respect to L_n. We construct partial Steinhaus sets T in stages, with T ⊆ T ⊆ ⋯ ⊆ Q_n. We will arrange it so that if Tⁿ = ∪_mT, then Tⁿ meets every lattice in ℒ_n. At stage i = 〈n, m〉 we extend T to T (or define T if m = 0). We will follow Lemma 2 in doing each extension.

Fix i = 〈n, m〉, and assume that for all 〈p, q〉 < i that T is defined. Assume inductively that the T so far defined satisfy the following (below, interpret T as ∅ if q = 0):

• (a)  T contains T, and each T is a partial Steinhaus set (that is, ρ(z₁, z₂)² ∉ ℤ for all distinct z₁, z₂ ∈ T).

• (b)  T is the range of a good selector f on R. Here R is the set of points having coordinates with respect to L_p that can be written with denominators d_q, and d₀, d₁, … is a sequence (depending on p) with d_j|d_j+1 for all j and such that every integer divides some d_j.

• (c)  Let E^p = ∪_i,j,d(G_p(i/d, j/d) ∩ L). Note that E^p is small relative to Q_p. Then (T − (S_<α→ ∪ ∪_{〈a,b〉<〈p,q〉} T)) ∩ E^p = ∅.

• (d)  Suppose z ∉ S_<α→, and i₂ = 〈p₂, q₂〉 is least so that z ∈ T. If i₁ = 〈p₁, q₁〉 < i₂ and p₁ ≠ p₂, then z does not have rational coordinates with respect to L_p1.

• (e)  Suppose z ∉ S_<α→, and i₂ = 〈p₂, q₂〉 is least so that z ∈ T. If i₁ = 〈p₁, q₁〉 < i₂ and y ∈ T does not have rational coordinates with respect to L_p2, then z ∉ l(y, L_p2), where l(y, L_p2) is as in Lemma 7.

Consider now i = 〈n, m〉, and we define T so as to also satisfy the above properties.

Consider first the case m = 0; that is, we are at the beginning stage in the construction of Tⁿ. We claim that there is at most one point in S_<α→∪ ∪_〈p,q〉<iT that is also in Q_n. For, suppose y, z were two such points. Note that ρ(y, z)² ∈ ℚ. Suppose first that y, z ∈ S_<α→. Say y ∈ S_β→1 − ∪_γ→<β→1S_γ→, z ∈ S_β→2 − ∪_γ→<β→2S_γ→ where β→₁ ≤ β→₂. If β→₁ = β→₂, then each of y, z lies on a lattice in N_β→2. Since L_n is definable from y and z, L_n is definable from two lattices in M_β→ for some β→ ≤ α→. From Lemma 5 it follows that L_n ∈ ∪_{γ→≤α→}M_γ→, a contradiction. If β→₁ < β→₂, then from inductive property 5, y, z both have rational coordinates with respect to some lattice L in N_β→2. This would again imply that L_n ∈ ∪_{γ→≤α→}M_γ→, a contradiction. Suppose next that y ∈ S_<α→ and z ∉ S_<α→. Let 〈p, q〉 be least so that z ∈ T (so p < n). Since by c, z ∉ ∪G_p(i/d, j/d), we must have that y is rational with respect to L_p [as otherwise ρ(y, z)² ∉ ℚ]. Thus, both y and z have rational coordinates with respect to both L_n and L_p, a contradiction of the fact that there can be at most one point with rational coordinates with respect to both lattices. Suppose now y, z ∉ S_<α→. Let y ∈ T, z ∈ T, with i₁ = 〈p₁, q₁〉, i₂ = 〈p₂, q₂〉 chosen minimally, so p₁, p₂ < n. Assume without loss of generality that i₁ ≤ i₂. If i₁ = i₂, then y, z are rational with respect to both L_p2 and L_n, a contradiction. If i₁ < i₂, then from e, y is rational with respect to L_p2. So y, z are both rational with respect to L_p2 and L_n, again a contradiction.

Let E be the union of ∪_i,j,d(G_n(i/d, j/d) ∩ L) together with ∪_z l(z, L_n) where z ranges over the points in ∪_〈p,q〉<i T not having rational coordinates with respect to L_n, together with the (finitely many) points of Q_n that are rational with respect to one of the lattices L_p with p ≠ n, and 〈p, q〉 < i for some q (for m = 0 this is equivalent to p < n). Clearly E is small with respect to Q_n. Let w_n be the unique point in (S_<α→ ∪ ∪_〈p,q〉<i T) ∩ Q_n if it exists, and otherwise let w_n be any point of L_n − E. Apply now Lemma 2 to get T avoiding E − {w_n} and with w_n ∈ T. From the definition of E it is clear that c–e are still satisfied.

Consider next the case m > 0. Define E exactly as above and apply Lemma 2 to get T, adding only points that avoid the set E. Again, c–e are satisfied.

Note for the arguments below that if z ∉ S_<α→ and 〈n, m〉 is least so that z ∈ T, then z ≠ w_n.

This completes the definition of the T, and we have verified a–e. Let T = ∪_n,m T. Let S_α→ = S_<α→ ∪ T. We must show that inductive hypotheses 1–5 are satisfied. Properties 1–3 are immediate from the construction. To see hypothesis 5, let β→ < α→ and y ∈ S_β→, z ∈ S_α→ − S_<α→, and suppose ρ(y, z)² ∈ ℚ. Let 〈n, m〉 be least so that z ∈ T. Since z ∉ ∪_i,j,d G_n(i/d, j/d) by construction, we must have that y is rational with respect to L_n.

Finally, we verify hypothesis 4; that is, we show S_α→ is a partial Steinhaus set. Let y, z be distinct points in S_α→, and assume ρ(y, z)² ∈ ℤ. We may assume z ∈ T − S_<α→. Suppose first that y ∈ S_<α→. Let i = 〈n, m〉 be least with z ∈ T. As in the previous paragraph we must have y rational with respect to L_n as otherwise ρ(y, z)² ∉ ℚ. Let i′ = 〈n, 0〉, so i′ ≤ i. In defining T, y would then have been the point w_n. So, y ∈ T, and since this is a partial Steinhaus set, ρ(y, z)² ∉ ℤ, a contradiction.

Assume now that y, z ∉ S_<α→. Let y ∈ T, z ∈ T, with i₁ = 〈n₁, m₁〉, i₂ = 〈n₂, m₂〉 chosen minimally. Clearly i₁ ≠ i₂; in fact n₁ ≠ n₂. Assume without loss of generality that i₁ < i₂. We must have y rational with respect to L_n2 as otherwise from the definition of E and l(y, L_n2) we would have ρ(y, z)² ∉ ℚ. Let i₃ = 〈n₂, 0〉 (note that i₃ ≠ i₁). If i₁ < i₃, then y is the point w_n2, which lies in T, and so ρ(y, z)² ∉ ℤ. If, however, i₁ > i₃, then from the definition of E we would have that y is not rational with respect to L_n2 [since into E we added the (at most one) point of Q_n1 ∩ Q_n2]. This again implies that ρ(y, z)² ∉ ℚ, a contradiction.

This completes the proof of Theorem 2.

Finally, we point out two simplifications that could be made to the proof. First, one could make the inductive construction more specific for this problem. Fix a transcendence basis for ℝ over ℚ, and let ≺ be a well-ordering of this transcendence basis. Then the finite ≺-decreasing sequences s from the basis are well-ordered lexicographically. For each such s, there are only countably many lattices that are algebraic over ℚ and s. Do the transfinite construction in the order of these sequences s, at each stage handling those countably many lattices that are algebraic over ℚ and s but not algebraic over ℚ and s′ for any s′ ≺ s. Then all of the arguments go through as before, if one replaces “definable” with “algebraic.” The key point is if three of the sequences β_m all precede α lexicographically and first differ from α at the same position k, then the union of these three sequences, arranged in ≺-decreasing order, still precedes α lexicographically.

Second, the full strength of Lemma 6 is not needed for the proof. Let again (using the notation of the Claim) E^p be those points z ∈ Q_p (i.e., having rational coordinates with respect to L_p) such that ρ²(c, z) ∈ ℚ for some c ∈ S_<α→ where c does not have rational coordinates with respect to L_p. Then it suffices to show that E^p is semismall with respect to Q_p, which means for each rational translation L = L of L_p there is a finite set of lines F such that for any line l ∉ F, l ∩ L ∩ E^p is finite. This is because the proof of Lemma 2 shows that we may actually avoid any Q_p semismall set in constructing the T. To see E^p is semismall, it suffices to show that there is a bound s ∈ ω such that if c₁, … , c_s are points in the plane with ρ(c_i, c_j)² ∉ ℤ for distinct c_i, c_j, and if z₁, … , z_s are collinear points with ρ(c_i, z_i)² ∈ ℚ and ρ(z_i, z_j)² ∈ ℤ, then the z_i are definable from {c₁, … , c_s}; in fact, for fixed c₁, … , c_s, distances ρ(c_i, z_i) and ρ(z_i, z_j), there are only finitely many such {z₁, … , z_s}. This fact follows from lemma 1.1 of ref. 3.