Discrete Versions of the Beckman-Quarles Theorem

Let F ⊆ R denote the field of numbers which are constructible by means of ruler and compass. We prove that: (1) if x, y ∈ R (n > 1) and |x − y| is an algebraic number then there exists a finite set Sxy ⊆ R containing x and y such that each map from Sxy to R n preserving all unit distances preserves the distance between x and y; if x, y ∈ F then we can choose Sxy ⊆ F, (2) only algebraic distances |x − y| have the property from item (1), (3) if X1, X2, ..., Xm ∈ R (n > 1) lie on some affine hyperplane then there exists a finite set L(X1, X2, ..., Xm) ⊆ R containing X1, X2, ..., Xm such that each map from L(X1, X2, ..., Xm) to R n preserving all unit distances preserves the property that X1, X2, ..., Xm lie on some affine hyperplane, (4) if J,K, L,M ∈ R (n > 1) and |JK| = |LM | (|JK| < |LM |) then there exists a finite set CJKLM ⊆ R containing J,K, L,M such that any map f : CJKLM → R that preserves unit distance satisfies |f(J)f(K)| = |f(L)f(M)| (|f(J)f(K)| < |f(L)f(M)|). Let F ⊆ R denote the field of numbers which are constructible by means of ruler and compass. Theorem 1 may be viewed as a discrete form of the classical Beckman-Quarles theorem, which states that any map from R to R (2 ≤ n < ∞) preserving unit distances is an isometry, see [1], [2] and [5]. Theorem 1 was proved in [10] in the special case where n = 2 and the distance |x− y| is constructible by means of ruler and compass. Theorem 1. If x, y ∈ R (n > 1) and |x−y| is an algebraic number then there exists a finite set Sxy ⊆ R containing x and y such that each map from Sxy to R n preserving all unit distances preserves the distance between x and y. Proof. The proof falls naturally into two sections. 1. Elementary facts Let us denote by Dn the set of all non-negative numbers d with the following property: If x, y ∈ R and |x− y| = d then there exists a finite set Sxy ⊆ R such that x, y ∈ Sxy and any map f : Sxy → R that preserves unit distance also preserves the distance between x and y. Obviously 0, 1 ∈ Dn. We first prove that if d ∈ Dn then √ 2 + 2/n·d ∈ Dn. Let us assume that d > 0, x, y ∈ R, |x − y| = √ 2 + 2/n · d. Using the notation of Figure 1 we show that Sxy := {Sab : a, b ∈ {x, y, ỹ, p1, p2, ..., pn, p̃1, p̃2, ..., p̃n}, |a− b| = d} satisfies the condition of Theorem 1. Figure 1 shows the case n = 2, but equations below Figure 1 describe the general case n ≥ 2; z denotes the centre of the (n− 1)-dimensional regular simplex p1p2...pn. Figure 1 1 ≤ i < j ≤ n |y− ỹ| = d, |x−pi| = |y−pi| = |pi−pj | = d = |x− p̃i| = |ỹ− p̃i| = |p̃i− p̃j | |x− ỹ| = |x− y| = 2 · |x− z| = 2 · √ n+1 2n · d = √ 2 + 2/n · d