Addendum

(with O.Belegradek)

The aim of this note is to prove the following

Proposition Any L^{U}-definable subset of ℝ is a disjoint finite union of sets each of which is (i) an open interval, (ii) a point, (iii) a countable dense subsets of open interval or (iv) an open interval minus its countable dense subset.

0.0.1 Let S be a L^{U}-definable subset of ℝ given by a core formula in one variable x

⋁_{u∈U^{n}} ϕ(u, x).

0.0.2 Lemma S is finite union of sets S_{i} of the form

(a)

⋁_{u∈U^{n}} x = h(u),

where h is a continuous semi-algebraic function on some semialgebraic cell C ⊆ S^{n} or (b)

⋁_{u∈U^{n}} f(u) < x < g(u)

where f and g are continuous semi-algebraic function on some semialgebraic cell C ⊆ S^{n}, f(u) < g(u) everywhere on C.

Proof This is a statement about ϕ(u, x) in the language of the field of reals, so follows from o-minimality (or QE).□

0.0.3 Proposition. Assume C ∩ U^{n} is infinite. Let C′ ⊆ C be a cell of minimal real dimension such that C′ ∩ U^{n} is infinite. Then C′ ∩ U^{n} is dense in C′.

Proof We assume C′ = C. Recall that S ⊆ ℂ, so C ⊆ ℂ^{n}. Let C̃ be the Zariski closure of C in ℂ^{n}. Claim 1. dim_{ℂ} C̃ = dim_{ℝ} C. Indeed, let dim_{ℝ} C = d. Suppose C is defined over A. Consider a generic point ⟨x_{1}, y_{1}, …, x_{n}, y_{n}⟩ of C, z_{i} = (x_{i}, y_{i}) ∈ S. Then, up to enumeration, x_{1}, …, x_{d} are algebraically independent over A and x_{d+k} ∈ acl(x_{1}, …, x_{d}, A), k = 1, …, n − d. Then z_{d+k} ∈ acl(z_{1}, …, z_{d}, A), in the sense of ℂ. Let p_{k}(z_{1}, …, z_{d}, z_{d+k}) = 0, k = 1, …, n − d, be the polynomial equations over A which witness these dependencies. Let C̄ be the Zariski closed set defined by ⋀ p_{k} = 0. By construction dim_{ℂ} C̄ = d and ⟨(x_{1}, y_{1}), …, (x_{n}, y_{n})⟩ ∈ C̄. This proves that C ⊆ C̄ and so dim_{ℂ} C̃ = d. Claim proved.

We may assume that C̃ is irreducible.

Claim 2. C̃ ∩ U^{n} is a finite union of sets of the form

{u ∈ U^{n} : ⋀_{i≤k} u^{m_{i}} = g_{i}}.

This is a direct consequence of the Lang property.

By the irreducibility of C̃ and the minimality of C Claim 2 implies that C̃ is a coset of a torus, that is

{z ∈ ℂ^{n} : ⋀_{i≤k} z^{m_{i}} = g_{i}}.

So, C̃ ∩ U^{n} is dense in C̃ ∩ S^{n}. But

dim_{ℝ} C̃ ∩ S^{n} = d

(use the form of C̃), so C is an open subset of C̃ ∩ S^{n} and hence C ∩ U^{n} is dense in C.□

0.0.4 Lemma In 0.0.2(a) the set is a countable subset of an interval. In 0.0.2(b) the set is equal to the interval f(C).

Proof Under (a) by definition S is equal to h(C ∩ S^{n}) which is a dense countable subset of the interval h(C). Assume (b). Let a be the point in the closure C̄ where f reaches supremum, and b a point in C̄ where g reaches infimum. Obviously, S ⊆ [f(a), g(b)]. Claim. S ⊇]f(a), g(b)[. Proof of the claim. Let I ⊆ C̄ be a path, that is a Jordan curve homeomorphic to [0, 1], which connects a and b. We may also assume that f and g are monotone on I in the sense of the induced order. Let t ∈ I \ {a, b}. Choose t_{0}, t_{1} ∈ I so that a < t_{0} < t < t_{1} < b in the sense of the induced order on I. In the same sense we consider [t_{0}, t_{1}] ⊆ I and let

α = min_{c∈[t_{0},t_{1}]} g(c) − f(c).

By assumptions α > 0. Choose t′ ∈ I such that t_{0} < t′ < t and t′ close enough to t, so f(t′) < f(t) < f(t′) + α. By the density of P^{n} we can find and p ∈ P^{n} so that p and t′ are close enough to have f(p) < f(t) < f(p) + α. Hence [f(p), g(p)] contains f(t). So, we have proved that ]f(a), f(b)[⊆ S. Symmetrically we have ]g(a), g(b)[⊆ S. We are now done if f(b) > g(a). Otherwise for some q ∈ I ∩P^{n} close enough to b we have [f(q), g(q)] ⊇ [f(b), g(a)] and so we see that the whole ]f(a), g(b)[ is in S. □

0.0.5 Proposition Any L^{U}-definable subset of ℝ is a disjoint finite union of (i) open intervals, (ii) points, (iii) countable dense subsets of open intervals and (iv) open intervals minus a countable dense subset.

Proof Notice first that the family of disjoint finite unions of (i)-(iv) form a Boolean algebra, that is this family is closed under Boolean operations. Now, Lemma 0.0.4 proves that any core subset of ℝ belongs to this algebra. Theorem 2 finishes the proof. □