Definability in the field of reals with a predicate for some powers of two In "The field of reals with a predicate for the powers of two", Van den Dries has proved that the set of integers is not definable in  $(\mathbb{R}, +,\cdot, \leq, 0, 1, 2^{\mathbb{Z}})$, where
  $2^{\mathbb{Z}}=\{2^n: n \in \mathbb{Z} \}$.
Question. Is there a subset $S$ of $2^{\mathbb{Z}}$ such that $\mathbb{Z}$ is definable in $(\mathbb{R}, +,\cdot, \leq, 0, 1, S)?$
 A: The answer is no, due to Friedman and Miller, Expansions of o-minimal structures by sparse sets, Fundamenta Mathematicae, 1(167), 55-64. Thanks to Erik Walsberg for providing the reference.
The main result of the paper is the following remarkable theorem:

Let $\mathfrak{R}$ be an o-minimal expansion of $(\mathbb{R}, <, +)$. Let $E \subseteq \mathbb{R}$ be such that, for every $m\in \mathbb{N}$ and $f: \mathbb{R}^m \to \mathbb{R}$ definable in $\mathfrak{R}$, the closure
  of $f(E^m)$ is a finite union of discrete sets. Then every subset of $\mathbb{R}$ definable in $(\mathfrak{R},E)^\#$ either has interior or is a finite union of discrete sets.

Here $(\mathfrak{R},E)^\#$ is the structure obtained by adding to $\mathfrak{R}$ predicates picking out every subset of every cartesian power $E^k$ of $E$. 
Moreover, they show that the same is true if "a finite union of discrete sets" is replaced by "nowhere dense", "null", "countable", or "discrete". 
As an example, Friedman and Miller observe that the work of van den Dries implies that the theorem applies to $E = 2^{\mathbb{Z}}$. So you can expand the real field by all subsets of $2^\mathbb{Z}$ at once, without defining $\mathbb{Z}$ (since defining $\mathbb{Z}$ implies defining $\mathbb{Q}$ which has empty interior but is not a finite union of discrete sets).
Note, however, that there are subsets $X\subseteq 2^{\mathbb{Z}}$ such that $(\mathbb{R},+,\cdot,\leq,0,1,X)$ interprets $(\mathbb{Z},+,\cdot)$. For example, let $X = \{2^{n^2}\mid n\in \mathbb{N}\}$. Since every natural number is the sum of four squares, $X^4 = 2^\mathbb{N}$, and by taking quotients, we can define $2^\mathbb{Z}$. Then $(2^\mathbb{Z},\cdot,X)$ is isomorphic to $(\mathbb{Z},+,S)$, where $S$ is a predicate picking out the perfect squares, and multiplication on $\mathbb{Z}$ is definable in $(\mathbb{Z},+,S)$.
