Second theorem of the complement

Let $\Delta_n$ be a collection of subsets of $\RR^n$, for $n \in \NN$, and set $\Delta = (\Delta_n)_{n \in \NN}$. As usual, we refer to the elements of the various $\Delta_n$ as $\Delta$-sets, and we call $\Delta$-sets that are manifolds $\Delta$-manifolds.

We assume the following axioms for $\Delta$-sets:

$(\Delta 1)$

every semialgebraic set is a $\Delta$-set;

$(\Delta 2)$

$\Delta$ is closed under taking finite unions, finite intersections, projections and cartesian products;

$(\Delta 3)$

if $A \subseteq \RR^n$ is a $\Delta$-set and $m \le n$, there exist $k \in \NN$, $n_i \ge n$ and connected $\Delta$-manifolds $N_i \subseteq \RR^{n_i}$, for $i=1, \dots, k$, such that $$\Pi_m(A) = \Pi_m(N_1) \cup \cdots \cup \Pi_m(N_k)$$ and, for each $i$, we have:

  • $\fr(N_i)$ is contained in a closed $\Delta$-set $B_i \subseteq \RR^{n_i}$ that has dimension and satisfies $\dim B_i < \dim N_i$;
  • $d_i:= \dim N_i \le m$, and there is a strictly increasing $\iota:\{1, \dots, d_i\} \into \{1, \dots, m\}$ such that $\Pi_\iota\rest{N_i}$ is an immersion.

By Step 2 and these properties, the collection of all global sub-$\C$-sets satisfies Axioms $(\Delta 1)$–$(\Delta 3)$.

Let $\tau_n:\RR^n \into \RR^n$ be defined as here.

  1. Let $A \in \Delta_n$. If follows from $(\Delta 3)$ that
    • $A$ has dimension,
    • $A$ has finitely many connected components, each of which is a $\Delta$-set, and
    • there exists a closed $B \in \Delta_n$ such that $\fr(A) \subseteq B$ and $\dim B < \dim A$.
  2. The system $\Delta$ is closed under permutation of coordinates.
  3. Let $A \subseteq \RR^n$. Then $A \in \Delta_n$ if and only if $\tau_n(A) \in \Delta_n$.

Second theorem of the complement
The complement of a $\Delta$-set is a $\Delta$-set.

Proof. Let $A \subseteq \RR^n$ be a $\Delta$-set. By the remarks above, we may assume $A \subseteq I^n$ and show that $I^n \setminus A$ is a $\Delta$-set, where $I = [-1,1]$. We proceed by induction on $n$; the case $n=0$ is clear. So let $n>0$ and assume that the theorem holds for $\Delta$-sets in $I^d$, for $d < n$. Let $A \subseteq I^n$ be a $\Delta$-set. By $(\Delta 3)$, we may assume that $A = \Pi_n(N)$ for some connected $\Delta$-manifold $N \subseteq \RR^q$, where $q \geq n$ and $N$ has the following properties:

  1. $\fr N$ is contained in a closed $\Delta$-set $B \subseteq I^q$ such that $\dim(B) < \dim(N)$;
  2. $d:= \dim(N) \leq n$, and there is a strictly increasing $\iota:\{1, \dots, d\} \into \{1,\dots,n\}$ such that $\Pi_{\iota}|_N$ is an immersion.

Since $\Delta$ is closed under permutations of coordinates, we may assume that $\iota$ is the identity map, that is, $\Pi_\iota = \Pi_d$ is the projection on the first $d$ coordinates. Since $\Pi_n\rest{N}$ and $\Pi_d\rest{N}$ have constant rank $d$, we have $\dim N = \dim A = \dim \Pi_d(A) = d$.

Case 1: $d < n$. In this case we first establish Claim
There exists $K \in \NN$ such that $\left|A_x\right| \leq \left|N_x\right| \leq K$ for $x \in I^d$.

The left inequality of the claim is obvious; for the right inequality, note that $\Pi_{\iota}\rest{N} : N \longrightarrow \RR^d$ is a local homeomorphism, and set $G := \Pi_d(B)$. Then $G$ is a closed $\Delta$-set of dimension less than $d$; in particular, every neighbourhood of every point in $G$ contains points of $G^c:= I^d \setminus G$. Hence, if $K \in \NN$ is such that $\left|N_x\right| \leq K$ for $x \in G^c$, then $\left|N_x\right| \leq K$ for $x \in G$ as well. So it suffices to show there is such a constant $K$ for $x \in G^c$.

However, for each connected component $C$ of $G^c$, the set $S:= N \cap \Pi_d^{-1}(C)$ satisfies conditions 1–4 of Exercise 1 (recall that every expansion of the real field is definably complete). By the inductive hypothesis, $G^c$ is a $\Delta$-set and therefore has finitely many connected components, so the claim follows.

Case 2: $d=n$. Then $\Pi_n\rest{N}$ is a local homeomorphism, hence $A$ is open. Note that $\Pi_n(B)$ is a closed $\Delta$-set of dimension less than $n$, so $I^n \setminus \Pi_n(B)$ is a $\Delta$-set by Case 1. Since $$I^n \setminus \Pi_n(N \cup B) = \left(I^n \setminus A\right) \cap \left(I^n \setminus \Pi_n(B)\right),$$ and since $A$ is open and $N \cup B$ is compact, it follows that $I^n \setminus \Pi_n(N \cup B)$ is open and closed in $I^n \setminus \Pi_n(B)$, so that $I^n \setminus \Pi_n(N \cup B)$ is a $\Delta$-set by Remark 1 above. Next note that $$I^n \setminus A = I^n \setminus \Pi_n(N) = \left(I^n \setminus \Pi_n(N \cup B)\right) \cup \left(\Pi_n(B) \setminus (A \cap \Pi_n(B))\right).$$ Since $A \cap \Pi_n(B)$ is a $\Delta$-set of dimension less than $n$, it follows from Case 1 that $\Pi_n(B) \setminus
\left(A \cap \Pi_n(B)\right)$ is a $\Delta$-set. Hence $I^n \setminus A$ is a $\Delta$-set. $\qed$

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.