We continue working with our rings of germs $\C_n$, for $n \in \NN$, as introduced here.
As with convergent power series, for $f \in \C_n$ such that the polyradius $(1, \dots 1)$ is $f$-admissible, we define a function $\bar f:\RR^n \into \RR$ by setting $$\bar f(x):= \begin{cases} f(x) &\text{if } x \in \bar B(1), \\ 0 &\text{otherwise,} \end{cases}$$ called a restricted $\C$-function. We denote by $\RR_\C$ the expansion of the real field by all restricted $\C$-functions.
Based on our normalization algorithm, we will establish the following version of our earlier goal:
Theorem
The structure $\RR_\C$ is model complete and o-minimal.
As pointed out in this post, we plan to prove this theorem by establishing a version of Gabrielov’s Theorem of the complement for all existentially definable sets in $\RR_\C$ (called global sub-$\C$-sets later). This will be done in three steps:
- using the normalization algorithm, obtain a precise description of the bounded semi-$\C$-sets;
- combine the description obtained in Step 1 with a “fiber cutting” argument to obtain a description of projections of bounded semi-$\C$-sets;
- define the collection of global sub-$\C$-sets and prove a theorem of the complement for them.
To state the precise description of Step 1, we need some notations.
Definitions
We call $M \subseteq \RR^{n}$ a $\C$-manifold if there is a polyradius $r \in (0,\infty)^n$ such that
- $M$ is a basic $\C$-set contained in $B(r)$, and
- there are $f_1, \dots, f_k \in \C_{r}$ such that $M$ is a submanifold of $B(r)$ of dimension $n-k$ on which $f_1, \dots, f_k$ vanish identically and the gradients $\nabla f_1(z), \dots, \nabla f_k(z)$ are linearly independent at each $z \in M$.
Example
A set $M \subseteq B(r)$ is $\C$-trivial if one of the following holds:
- $M = B\big(r,(x_{1}, \dots, x_{n}), \sigma\big)$ for some sign condition $\sigma \in \{-1,0,1\}^n$, or
- there are a permutation $\lambda$ of $\{1, \dots, n\}$, a $\C$-trivial $N \subseteq I_s$ and a $g \in \C_{s}$, where $s = (r_{\lambda(1)}, \dots, r_{\lambda(n-1)})$, such that $g(I_s) \subseteq \left(-r_{\lambda(n)}, r_{\lambda(n)}\right)$ and $\Pi_\lambda(M) = \gr\left(g\rest{N}\right)$.
Exercise
Let $M \subseteq \RR^n$ be $\C$-trivial. Show that $M$ is a connected $\C$-manifold and that $\fr(M)$ is a $\C$-set.
The following notion of dimension is useful for us: we say that a set $S \subseteq \RR^n$ has dimension if $S$ is a countable union of $C^1$ manifolds; in this case, we set
$$
\dim(S) := \begin{cases} \max\{\dim(M): M \subseteq S \text{ is a
$C^1$ manifold} \} &\text{if } S \neq
\emptyset \\ -\infty &\text{otherwise}.
\end{cases}
$$
Exercises
- If $S = \bigcup_{i \in \NN} S_i$ and each $S_i$ has dimension, show that $S$ has dimension and $\dim(S) = \max\{\dim(S_i): i \in \NN\}$. (Hint: use Baire category theory.)
- If $M$ is a manifold, then $\dim(M)$ in the sense of the previous definition agrees with the usual manifold dimension.
- If $M \subseteq \RR^n$ is $\C$-trivial, show that $\fr(M)$ has dimension and $\dim \fr(M) \lt \dim M$.
Step 1
Let $f = (f_1, \dots, f_\mu) \in (\C_n)^\mu$ be such that $f_j \ne 0$ for each $j$. Then there is an $f$-admissible neighbourhood $W$ of the origin with the following property:
- $(\ast)$
- for every sign condition $\sigma \in \{-1,0,1\}^{\mu}$, there is an $l \in \NN$ and, for $k = 1, \dots, l$, there are $n_k \ge n$, $r_k \in (0,\infty)^{n_k}$ and $\C$-trivial manifolds $N_k \subseteq B(r_k)$ such that $$B(W,f,\sigma) = \Pi_{n}(N_1) \cup \dots \cup \Pi_{n}(N_l)$$ and, for each $k$, the set $\Pi_{n}(N_k)$ is a manifold and $\Pi_{n}\rest{N_k} : N_k \longrightarrow \Pi_{n}(N_k)$ is a diffeomorphism.
The idea of the proof is now straightforward: if each $Tf_j$ is normal, $(\ast)$ follows from Exercise 14 (with each $n_k = n$). So we proceed as in the proof of this theorem, by induction on $h_n(Tf)$, replacing each $N_k$ obtained inductively by the graph of the considered substitution above $N_k$. This requires a bit more care of detail in order to obtain the “diffeomorphism” part of Step 1: we need to subdivide space into pieces on each of which the corresponding substitution is a diffeomorphism; this is where this lemma comes in.
(Show proof)Case 1: there exist $i \in \{2, \dots,n\}$ and $c \in \RR^{i-1}$ such that $h_n\left(l_{i,c}Tf\right)\lt h_n(Tf)$. Let $l_{i,c}:\RR^n \into \RR^n$ be the homeomorphism $x \mapsto (x_1 + c_1 x_i, \dots, x_{i-1} + c_{i-1} x_i, x_i, \dots, x_n)$. By assumption (D2) and the properties of the map $T$, we have $f \circ l_{i,c} \in \C_n^k$ with $T(f \circ l_{i,c}) = l_{i,c} Tf$. Therefore, by the inductive hypothesis, there is an $\left(f \circ l_{i,c}\right)$-admissible neighbourhood $V$ of the origin such that, for every sign condition $\sigma \in \{-1,0,1\}^{\mu}$, there is an $l \in \NN$ and, for $k = 1, \dots, l$, there are $m_k \ge n$, $s_k \in (0,\infty)^{m_k}$ and $\C$-trivial manifolds $M_k \subseteq B(s_k)$ such that $$B\left(V,\left(f \circ l_{i,c}\right),\sigma\right) = \Pi_{n}(M_1) \cup \dots \cup \Pi_{n}(M_l)$$ and, for each $k$, the set $\Pi_{n}(M_k)$ is a manifold and $\Pi_{n}\rest{M_k} : M_k \longrightarrow \Pi_{n}(M_k)$ is a diffeomorphism. Since $l_{i,c}$ is a diffeomorphism, the set $W:= l_{i,c}(V)$ is an $f$-admissible neighbourhood of the origin and $$B(W,f,\sigma) = l_{i,c}\left(B(V,f \circ l_{i,c},\sigma)\right).$$ Also, each $$N_k:= \set{(l_{i,c}(x),y,x):\ (x,y) \in M_k}$$ is a $\C$-trivial set, and we have $$B(W,f,\sigma) = \Pi_{n}(N_1) \cup \dots \cup \Pi_{n}(N_l)$$ and, for each $k$, the set $\Pi_{n}(N_k)$ is a manifold and $\Pi_{n}\rest{N_k} : N_k \longrightarrow \Pi_{n}(N_k)$ is a diffeomorphism.
Case 2: there exist $i \in \{2, \dots, n\}$ and $\alpha \in \D_{i-1}$ such that $\alpha(0) = 0$ and $\,h_n(t_\alpha Tf) \lt h_n(Tf)$. Since the map $t_\alpha:\RR^n \into \RR^n$ defined by $x \mapsto (x_1, \dots, x_{i-1}, x_i + \alpha(x_1, \dots, x_{i-1}), x_{i+1}, \dots, x_n)$ is a homeomorphism, the argument is similar to Case 1.
Case 3: there exist $i \in \{1, \dots, n\}$ and $q \in \NN$ such that $h_n\left(p^{\ast}_{i,q} Tf\right) \lt h_n(Tf)$ for $\ast \in \{+,-\}$. Let $p^*_{i,q}:\RR^n \into \RR^n$ be the map $x \mapsto (x_1, \dots, x_{i-1}, x_i^q, x_{i+1}, \dots, x_n)$, which is not a diffeomorphism if $q\gt 1$. However, it suffices to prove $(\ast)$ with $f\hat{}x_i$ in place of $f$, where $$f\hat{}x_i:= (f_1, \dots, f_{\mu}, x_i).$$ By part 1 of this lemma, we have $h_n\left(p^{\ast}_{i,q} T(f\hat{}x_i)\right) \lt h_n(Tf)$ as well, for $\ast \in \{+,-\}$. Since the restriction of $p^\ast_{i,q}$ to the set $\set{x:\ \sign(x_i) = \sigma(\mu+1)}$ is a diffeomorphism, the proposition follows by arguing as in Case 1 in each of the charts $p^+_{i,q}$ and $p^-_{i,q}$.
Case 4: there exist $1 \le i \lt j \le n$ such that, for $\lambda \in \RR \cup \{\infty\}$, we have $\,h_n\left(\bl^{i,j}_\lambda Tf\right) \lt h_n(Tf)$. Let $\bl^{i,j}_\lambda:\RR^n \into \RR^n$ be the corresponding map, which is not a diffeomorphism. However, it suffices to prove $(\ast)$ with $f\hat{}x_i$ in place of $f$ in each chart $\bl^{i,j}_\lambda$ with $\lambda \in \RR$, and with $f\hat{}x_j$ in place of $f$ in the chart $\bl^{i,j}_\infty$. We conclude using part 2 of this lemma and the facts that the restriction of $\bl^{i,j}_\lambda$ to $\set{x:\ \sign(x_i) = \sigma(\mu+1)}$ and the restriction of $\bl^{i,j}_\infty$ to $\set{x:\ \sign(x_j)=\sigma(\mu+1)}$ are diffeomorphisms. $\qed$
Definition
A set $M \subseteq \RR^n$ is a semi-$\C$-manifold (of class $C^p$) if it is a semi-$\C$-set as well as a manifold (of class $C^p$), and it is a trivial semi-$\C$-set if there exist $a \in \RR^n$ and a $\C$-trivial $N \subseteq \RR^n$ such that $M = N+a$.
Since the closure of a bounded set is compact, we obtain the following from Step 1:
Corollary
Let $A \subseteq \RR^{n}$ be a bounded semi-$\C$-set. Then there are $k \in \NN$ and $n_i \geq n$ and trivial semi-$\C$-sets $N_i \subseteq \RR^{n_i}$, for $i=1,\dots,k$, such that $$A = \Pi_{n}(N_1) \cup \dots \cup \Pi_{n}(N_k)$$ and, for each $i$, the set $\Pi_{n}(N_i)$ is a manifold and $\Pi_{n}\rest{N_i} : N_i \longrightarrow \Pi_{n}(N_i)$ is a diffeomorphism. In particular, $A$ has dimension. $\qed$