Sparse subsets of the plane

Let ${\cal M}$ be an o-minimal expansion of a dense linear order $(M,\lt)$, and let $S \subseteq M^2$.

$S$ is sparse if $S$ has empty interior.

Lemma
Assume $S$ is definable. The following are equivalent:

1. $S$ is sparse;
2. the set $S’$ of all $x \in M$ such that $S_x$ is infinite is finite;
3. $S$ is nowhere dense in $M^2$.

Proof
1 $\Rightarrow$ 2: assume that $S’$ is infinite. Then there is an open interval $I \subseteq M$ such that $S_x$ contains an interval, for each $x \in I$. For each $x$, let $I_x$ be the first open interval contained in $S_x$ (with respect to $\lt$), and consider the definable functions $i:I \into M \cup \{-\infty\}$ and $s:I \into M \cup \{+\infty\}$ defined by $$i(x):= \inf I_x\quad \text{ and } \quad s(x):= \sup I_x.$$ By the Monotonicity Theorem, there exists an open interval $J \subseteq I$ such that $i\rest{J}$ and $s\rest{J}$ are continuous. Since $i(x) \lt s(x)$ for all $x$, the set $\set{(x,y):\ x \in J,\ i(x) \lt y \lt s(x)}$ is open and contained in $S$.

2 $\Rightarrow$ 3: assume that $S’$ is finite, and let $U \subseteq M^2$ be open and definable. Then $(S \cap U)’$ is finite, so $(U \setminus S)’$ is infinite. From the previous implication, it follows that $U \setminus S$ has nonempty interior, that is, $S$ is not dense in $U$.

3 $\Rightarrow$ 2 is obvious. $\qed$

Corollary
If $S$ is definable and sparse, then so is $\cl(S)$, and there is a finite set $F_1 = F_1(S) \subseteq M$ such that $S_x$ is finite for every $x \in M \setminus F_1$. $\qed$

In other words, a definable and sparse $S$ has “almost everywhere” finite fibers.

Exercise
Assume $S$ is definable. Show that the definable set $$\bd_1(S):= \set{(x,y) \in M^2:\ y \in \bd(S_x)}$$ is sparse.