We now return to $\Ps{R}{X}$-sets or, more generally, to so-called $\C$-sets, where $\C = (\C_n)_{n \in \NN}$ is a collection of subrings $\C_n$ of $C^\infty_n$ obtained as follows: for every polyradius $r \in (0,\infty)^n$, we assume being given a subring $\C_r$ of $C^\infty_r$ such that,
- (C1)
- for $f \in \C_r$, there exists $s > r$ and $g \in \C_s$ with $f = g \rest{B(r)}$,
and such that $\C_n$ is the set of all germs at $0$ of functions in $\bigcup_{r \in (0,\infty)^n} \C_r$. We assume, in addition, that
- (C2)
- the restriction of the Taylor map $T$ to $\C_n$ is injective (quasianalyticity),
- (C3)
- for $f \in \C_r$ and $a \in B(r)$, the germ $f_a$ at $0$ of the function $x \mapsto f(a+x)$ belongs to $\C_n$,
and that
- (C4)
- the images $\D_n:= T(\C_n)$ satisfy conditions (D1)–(D5).
Examples
- Each $\C_r$ is obtained from $\Pc{R}{X}_r$ as defined here.
- Certain Denjoy-Carleman classes; see the introduction of this paper for details.
In analogy to the definition of $\Pc{R}{X}$-sets, we define $\C$-sets as follows: for our purposes, a neighbourhood of the origin is any set $W \subseteq \RR^n$ such that $0 \in \ir(W)$.
Definition
Let $f_1, \dots, f_k \in \C_n$ and set $f:= (f_1, \dots, f_k)$.
- A polyradius $r \in (0,\infty)^n$ is $f$-admissible if there exists $s > r$ such that each $f_i$ has a representative in $\C_s$.
- For an $f$-admissible polyradius $r$ and a sign condition $\sigma:\{1, \dots, k\} \into \{-1,0,1\}$, the set $$B(r,f,\sigma):= \set{x \in B(r):\ \sign(f_i) = \sigma(i) \text{ for } i=1, \dots, k}$$ is a basic $\C$-set.
- A $\C$-set is a finite union of basic $\C$-sets.
- A set $S \subseteq \RR^n$ is a semi-$\C$-set if, for every $a \in \RR^n$, there exists a polyradius $r$ such that $B(r) \cap (S-a)$ is a $\C$-set.
The reason for Assumption (C3) is the following:
Exercise
Show that every $\C$-set is a semi-$\C$-set.
To illustrate how we use the Normalization Corollary, we now show:
Theorem
Every bounded semi-$\C$-set has finitely many connected components.
By the definition of semi-$\C$-set, it suffices to prove the following: given $f = (f_1, \dots, f_k) \in \C_n^k$, we call a neighbourhood $W$ of the origin $f$-admissible if there is an $f$-admissible polyradius $r$ such that $W \subseteq B(r)$ and, in this situation, we set $$B(W,f,\sigma):= \set{x \in W:\ \sign(f_i) = \sigma(i) \text{ for } i=1, \dots, k}.$$
Proposition
Let $f = (f_1, \dots, f_k) \in \C_n^k$ be such that $f_i \ne 0$ for each $i$, and let $\sigma:\{1, \dots, k\} \into \{-1,0,1\}$. Then there exists an $f$-admissible neighbourhood $W$ of the origin such that the set $B(W,f,\sigma)$ has finitely many connected components.
Proof. We set $Tf:= T(f_1 \cdots f_k)$; note that $f_1 \cdots f_k \ne 0$, so Assumption (C2) implies that $Tf \ne 0$. We proceed by induction on $h_n(Tf)$; if $h_n(Tf) = 0$ then, by Exercise 15, each $Tf_j$ is normal and the proposition holds by this corollary and the Taylor expansion theorem. So we assume $h_n(Tf) \gt 0$ and distinguish four cases according to the Normalization Corollary:
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 exists an $(f \circ l_{i,c})$-admissible neighbourhood $V$ of the origin such that the set $B(V,f \circ l_{i,c},\sigma)$ has finitely many connected components. Since $l_{i,c}$ is a homeomorphism, the set $W:= l_{i,c}(V)$ is an $f$-admissible neighbourhood of the origin and $$B(W,f,\sigma) = l_{i,c}\left(B(V,f \circ l_{i,c},\sigma)\right);$$ in particular, $B(W,f,\sigma)$ has finitely many connected components.
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 $x \mapsto (x_1, \dots, x_{i-1}, x_i^q, x_{i+1}, \dots, x_n)$ (not a homeomorphism in general). By the inductive hypothesis, for $\ast \in \{+,-\}$, there exists an $\left(f \circ p^*_{i,q}\right)$-admissible neighbourhood $V^\ast$ of the origin such that $B\left(V^\ast,f \circ p^*_{i,q},\sigma\right)$ has finitely many connected components. Then $$W:= p^+_{i,q}\left(V^+\right) \cup p^-_{i,q}\left(V^-\right)$$ is an $f$-admissible neighbourhood of the origin and $$B(W,f,\sigma) = p^+_{i,q}\left(B\left(V^+,f \circ p^+_{i,q}, \sigma\right)\right) \cup p^-_{i,q}\left(B\left(V^-,f \circ p^-_{i,q}, \sigma\right)\right),$$ so the proposition follows in this case.
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 (not a homeomorphism). By the inductive hypothesis, for $\lambda \in \RR \cup \{\infty\}$, there exists an $\left(f \circ \bl^{i,j}_\lambda\right)$-admissible neighbourhood $V^\lambda$ of the origin such that $B\left(V^\lambda,f \circ \bl^{i,j}_\lambda,\sigma\right)$ has finitely many connected components.
Since $V^\lambda$ is a neighbourhood of the origin, there exists a polyradius $r_\lambda$ such that $B(r_\lambda) \subseteq V^\lambda$. By this exercise, there exist $\lambda_1, \cdots, \lambda_k \in \RR \cup \{\infty\}$ such that $\bl^{i,j}_{\lambda_1}(B(r_{\lambda_1})) \cup \cdots \cup \bl^{i,j}_{\lambda_k}(B(r_{\lambda_k}))$ is a neighbourhood of the origin. Therefore, the set $$W:= \bl^{i,j}_{\lambda_1}\left(V^{\lambda_1}\right) \cup \cdots \cup \bl^{i,j}_{\lambda_k}\left(V^{\lambda_k}\right)$$ is a neighbourhood of the origin and, since $$B(W,f,\sigma) = \bigcup_{l=1}^k\bl^{i,j}_{\lambda_l}\left(B\left(V^{\lambda_l},f \circ \bl^{i,j}_{\lambda_l},\sigma\right)\right),$$ the proposition follows in this case. $\qed$