Proof. Let X be a definable subset of $M^n$ . The definable set X is discrete if and only if the projection image $\pi (X)$ has an empty interior for all the coordinate projections $\pi :M^n \rightarrow M$ by the property (a). Therefore, X is discrete if and only if $\dim X=0$ . A discrete definable set is always closed by Lemma 2.4.⊣

Proof. Permuting the coordinates if necessary, we may assume that $\pi $ is the projection onto the first d coordinates. Set

$$ \begin{align*}S=\{x \in \pi(X)\;|\; \text{the fiber }X \cap \pi^{-1}(x) \text{ is not discrete}\}\text{.}\end{align*} $$

We have $S=\{x \in \pi (X)\;|\; \dim (X \cap \pi ^{-1}(x))>0\}$ by Proposition 3.2. We want to show that S has an empty interior. Assume the contrary. Let $\rho _j:M^n \rightarrow M$ be the coordinate projections onto the j-th coordinate for all $d<j \leq n$ . Set

$$ \begin{align*}S_j=\{x \in \pi(X)\;|\; \rho_j(X \cap \pi^{-1}(x)) \text{ contains an open interval}\}\text{.}\end{align*} $$

We have $S=\bigcup _{d<j \leq n}S_j$ by the definition of dimension. The definable set $S_j$ has a nonempty interior by the property (b) for some $d<j \leq n$ . Fix such j. Let $\Pi :M^n \rightarrow M^{d+1}$ be the coordinate projection given by $\Pi (x)=(\pi (x),\rho _j(x))$ . The definable set $T=\{x \in M^d \;|\; \text {the fiber }(\Pi (X))_x \text { contains an open interval}\}$ contains $S_j$ and it has a nonempty interior. Therefore, the projection image $\Pi (X)$ has a nonempty interior by Lemma 2.8. It contradicts the assumption that $\dim (X)=d$ . We have shown that S has an empty interior. Since $\pi (X)$ has a nonempty interior, there exists a definable open subset U of $\pi (X)$ with $U \cap S=\emptyset $ by the property (b).⊣

Proof. Permuting the coordinates if necessary, we may assume that $\pi $ is the projection onto the first d coordinates. Let $\rho _j:M^n \rightarrow M$ be the coordinate projections onto the j-th coordinate for all $d<j \leq n$ . The fiber $Y_x$ is discrete for any $x \in U$ by the assumption. Since $X_x$ is a definable subset of $Y_x$ , $X_x$ is also a discrete set. There exists a definable map $g:U \rightarrow X$ such that the composition $\pi \circ g$ is the identity map on U by the property (d). Note that $\rho _j(Y_x)$ is discrete and closed by the property (a) and Lemma 2.4. Consider the definable functions $\kappa _j^+:U \rightarrow M \cup \{+\infty \}$ defined by

$$ \begin{align*} \kappa_j^+(x)=\left\{ \begin{array}{ll} \inf \{t \in \rho_j(Y_x)\;|\; t>\rho_j(g(x))\} & \text{if } \{t \in \rho_j(Y_x)\;|\; t>\rho_j(g(x))\}\not=\emptyset \text{,}\\ +\infty & \text{otherwise}, \end{array} \right. \end{align*} $$

for all $d<j \leq n$ . We define $\kappa _j^-:U \rightarrow M \cup \{{-}\infty \}$ similarly. Then, we have

$$ \begin{align*} \pi^{-1}(x) \cap Y \cap (\{x\} \times (\kappa_{d+1}^-(x),\kappa_{d+1}^+(x)) \times \cdots \times (\kappa_{n}^-(x),\kappa_{n}^+(x))) = \{g(x)\}, \end{align*} $$

for all $x \in U$ . There exists a definable open subset V of U such that the restriction f of g to V and the restrictions of $\kappa _j^-$ and $\kappa _j^+$ to V are all continuous by the properties (b) and (c). Set $W=\{(x,y_{d+1},\ldots , y_n) \in V \times M^{n-d}\;|\; \kappa _j^-(x) < y_j < \kappa _j^+(x) \text { for all } d<j \leq n\}$ . The definable sets V and W and a definable continuous map f satisfy the requirements.⊣

Proof. Immediate from the definition of dimension and Lemmas 3.3 and 3.4.⊣

Proof. We may assume that C is an open box without loss of generality. The lemma is obvious when $n=0$ . We assume that $n>0$ . We lead to a contradiction assuming that $f(C)$ has an empty interior. Set $d=\dim f(C)$ . We have $0 \leq d <n$ . When $d=0$ , the set $f(C)$ is discrete by Proposition 3.2. The image $f(C)$ is a singleton by [Reference Miller8, Proposition 1.6] because the open box C is definably connected. Contradiction to the assumption that f is injective.

We next consider the case in which $d \not =0$ . Applying Lemma 3.5, we can take a coordinate projection $\pi :M^n \rightarrow M^d$ and a definable open set W of $M^n$ such that the restriction of $\pi $ to $f(C) \cap W$ is injective and its image is a definable open set. We may assume that the restriction of $\pi $ to $f(C)$ is injective by considering $f^{-1}(W)$ instead of C. Since f is injective and continuous by the assumption, the composition $\pi \circ f$ is also injective and continuous.

Take an open box B contained in C. Let $B_1$ and $B_2$ be the open boxes in $M^d$ and $M^{n-d}$ with $B=B_1 \times B_2$ , respectively. Take $c \in B_2$ . Consider the definable map $g:B_1 \rightarrow M^d$ given by $g(x)=\pi (f(x,c))$ . It is injective and continuous. There exists an open box D in $M^d$ with $D \subset g(B_1)$ by the induction hypothesis. Take a point $x_0 \in B_1$ with $g(x_0) \in D$ and a point $c' \in B_2$ sufficiently close to c with $c' \not =c$ . We have $\pi (f(x_0,c')) \in D$ because $\pi \circ f$ is continuous. There exists a point $x_1 \in B_1$ with $\pi (f(x_0,c'))=g(x_1)=\pi (f(x_1,c))$ because $D \subset g(B_1)$ . It contradicts the fact that $\pi \circ f$ is injective.⊣

Proof. We lead to a contradiction assuming that $m>n$ . Take a definable continuous injective map $f:B \rightarrow C$ and $c \in M^{m-n}$ . Consider the definable map $g:B \rightarrow C \times M^{m-n}$ given by $g(x)=(f(x),c)$ . It is obviously continuous and injective. The image $g(B)$ has a nonempty interior by Lemma 3.6. Contradiction.⊣

Proof. The assertions (1) and (2) are obvious. We omit the proofs.

We demonstrate the assertion (3). Assume that X and Y are definable subsets of $M^m$ and $M^n$ , respectively. Set $d=\dim (X)$ , $e=\dim (Y)$ , and $f=\dim (X \times Y)$ . We first show that $d+e \leq f$ . In fact, let $\pi :M^m \rightarrow M^d$ and $\rho :M^n \rightarrow M^e$ be coordinate projections such that both $\pi (X)$ and $\rho (Y)$ have nonempty interiors. The definable set $(\pi \times \rho )(X \times Y)$ has a nonempty interior. Therefore, we have $d+e \leq f$ . We show the opposite inequality. Let $\Pi :M^{m+n} \rightarrow M^f$ be a coordinate projection with $\operatorname{int} (\Pi (X \times Y)) \not =\emptyset $ . There exist coordinate projections $\pi _1:M^m \rightarrow M^{f_1}$ and $\pi _2:M^n \rightarrow M^{f_2}$ with $\Pi =\pi _1 \times \pi _2$ . In particular, we get $f=f_1+f_2$ . Since $\Pi (X \times Y)$ has a nonempty interior, there exist open boxes $C \subset M^{f_1}$ and $D \subset M^{f_2}$ with $C \times D \subset \Pi (X \times Y)$ . We get $C \subset \pi _1(X)$ and $D \subset \pi _2(Y)$ . Hence, we have $d \geq f_1$ and $e \geq f_2$ . We finally obtain $d+e \geq f_1+f_2 =f$ .

We next show the assertion (4). The inequality $\dim (X \cup Y) \geq \max \{\dim (X),\dim (Y)\}$ is obvious by the assertion (1). We show the opposite inequality. Set $d=\dim (X \cup Y)$ . There exists a coordinate projection $\pi :M^n \rightarrow M^d$ such that $\pi (X \cup Y)$ has a nonempty interior by the definition of dimension. At least one of $\pi (X)$ and $\pi (Y)$ has a nonempty interior by the property (b) because $\pi (X \cup Y)=\pi (X) \cup \pi (Y)$ . We may assume that $\pi (X)$ has a nonempty interior without loss of generality. We have $d \leq \dim (X)$ by the definition of dimension. We have demonstrated that $\dim (X \cup Y) \leq \max \{\dim (X),\dim (Y)\}$ .

The next target is the assertion (5). Let X be a definable subset of $M^m$ . The notation $\Gamma (f)$ denotes the graph of the map f. We first demonstrate that $\dim (\Gamma (f))=\dim (X)$ . In fact, the inequality $\dim (X) \leq \dim (\Gamma (f))$ is obvious because X is the projection image of $\Gamma (f)$ . Set $d=\dim (\Gamma (f))$ and $e=\dim (X)$ .

Applying Lemma 3.5 to the graph $\Gamma (f)$ , we can take a coordinate projection $\pi :M^{m+n} \rightarrow M^d$ , an open box V contained in $\pi (\Gamma (f))$ , and a definable continuous map $\tau :V \rightarrow \Gamma (f)$ such that the composition $\pi \circ \tau $ is the identity map on V. In particular, the map $\tau $ is injective.

Let $\Pi :M^{m+n} \rightarrow M^m$ be the projection onto the first m-coordinate. The restriction of $\Pi $ to the graph $\Gamma (f)$ is obviously injective. Applying Lemma 3.5 to the set X, we can take a coordinate projection $\rho :M^m \rightarrow M^e$ and a definable open subset W of $M^m$ such that the restriction of $\rho $ to $W \cap X$ is injective. The inverse image $(\Pi \circ \tau )^{-1}(W)$ contains an open box because $\Pi \circ \tau $ is continuous. Replacing V with the open box, we may assume that the restriction of $\rho $ to $\Pi (\tau (V))$ is injective. We finally get the definable continuous injective map $\rho \circ \Pi \circ \tau : V \rightarrow M^e$ . We have $d \leq e$ by Corollary 3.7. We have shown that $\dim X = \dim \Gamma (f)$ .

It is now obvious that $\dim f(X) \leq \dim \Gamma (f) = \dim X$ because $f(X)$ is the projection image of $\Gamma (f)$ .

We demonstrate the assertion (6). Let X be a definable subset of $M^m$ . We lead to a contradiction assuming that $d=\dim X=\dim \mathcal D(f)$ . By Lemma 3.5, there exist a coordinate projection $\pi :M^m \rightarrow M^d$ , definable open subsets $V \subset \pi (\mathcal D(f))$ and $W \subset M^m$ , and a definable continuous function $g:V \rightarrow \mathcal D(f)$ such that $\pi (W)=V$ , $X \cap W=g(V)$ , and $\pi \circ g$ is the identity map on V. Shrinking V and replacing W with $W \cap \pi ^{-1}(V)$ if necessary, we may assume that $f \circ g$ is continuous by the property (c). Since g is a definable homeomorphism onto its image, the function f is continuous on $g(V)= X \cap W$ . On the other hand, f is discontinuous everywhere on $X \cap W$ because $X \cap W$ is open in X and $X \cap W =g(V)$ is contained in $\mathcal D(f)$ . Contradiction. We have demonstrated the assertion (6).

The remaining task is to demonstrate the assertion (7). Take distinct elements $c,d \in M$ . Consider the definable function $f:\overline {X} \rightarrow M$ given by

$$ \begin{align*} f(x)= \left\{ \begin{array}{ll} c & \text{if } x \in X, \text{ and }\\ d & \text{otherwise.} \end{array} \right. \end{align*} $$

It is obvious that $\mathcal D(f)$ contains $\partial X$ . The assertion (7) follows from the assertions (1) and (6).⊣

Proof. Let X be a definable set of dimension d. We prove that X is constructible by induction on d. When $d=0$ , the definable set X is discrete and closed by Proposition 3.2. In particular, it is constructible. When $d>0$ , the frontier $\partial X$ is of dimension smaller than d by Theorem 3.8(7). It is constructible by the induction hypothesis. Therefore, $X= \overline {X} \setminus \partial X$ is also constructible.⊣

Proof. Let $d'$ be the maximum of nonnegative integers e satisfying the condition given in the theorem. We first demonstrate $d' \leq d$ . In fact, let B be an open box contained in $M^{d'}$ and $\varphi :B \rightarrow X$ be a definable injective continuous map homeomorphic onto its image. We have $\dim \varphi (B)=\dim B=d'$ by Theorem 3.8(5). We get $d = \dim X \geq \dim (\varphi (B)) = d'$ by Theorem 3.8(1).

We next demonstrate $d \leq d'$ . Applying Lemma 3.5 to the definable set X, we can get a coordinate projection $\pi :M^n \rightarrow M^d$ , a definable open box U in $\pi (X)$ , and a definable continuous map $\tau :U \rightarrow X$ such that $\pi \circ \tau $ is the identity map on U. In particular, $\tau $ is a definable continuous injective map homeomorphic onto its image. Therefore, we have $d \leq d'$ by the definition of $d'$ .⊣

Proof. Set $d=\dim (X)$ . There exists an open box U in $M^d$ and a definable continuous injective map $\varphi :U \rightarrow X$ homeomorphic onto its image by Theorem 3.11. Take an arbitrary point $t \in U$ and set $x=\varphi (t)$ . For any open box B containing the point x, the inverse image $\varphi ^{-1}(B)$ is a definable open set. Take an open box V with $t \in V \subset \varphi ^{-1}(B)$ . The restriction $\varphi |_{V} V \rightarrow X \cap B$ is a definable continuous injective map homeomorphic onto its image. Hence, we have $\dim (X \cap B) \geq d$ by Theorem 3.11. The opposite inequality follows from Theorem 3.8(1).⊣

Proof. Let X and Y be definable subsets of $M^m$ and $M^n$ , respectively. Set $d=\dim (X)$ and $e=\dim (Y)$ .

We first assume that $\varphi $ is continuous. We have $d \geq e$ by Theorem 3.8(5). We demonstrate the opposite inequality. We first reduce to the case in which X is a definable open subset of $M^m$ . There exist a definable open subset U of $R^d$ and a definable continuous injective map $\sigma :U \rightarrow X$ homeomorphic onto its image by Theorem 3.11. If the lemma holds true for the composition $\varphi \circ \sigma $ , we have $\dim X = d = \dim U = \dim \varphi \circ \sigma (U) \leq \dim Y=e$ by Theorem 3.8(1). The lemma is also true for the original $\varphi $ . Hence, we may assume that X is open in $M^m$ . In particular, we have $m=d$ .

Let $\Pi :M^{m+n} \rightarrow M^n$ be the projection onto the last n coordinates. Consider the graph $\Gamma (\varphi )$ of $\varphi $ . Note that $\Pi ^{-1}(y) \cap \Gamma (\varphi )$ are discrete for all $y \in Y$ . Take a coordinate projection $\pi :M^n \rightarrow M^e$ such that $\pi (Y)$ has a nonempty interior. The definable set $(\pi \circ \Pi )^{-1}(z) \cap \Gamma (\varphi )$ is discrete and closed if $\pi ^{-1}(z) \cap Y$ is discrete for $z \in M^e$ by Lemma 2.5. By Lemmas 3.3 and 3.4, there exist definable open subsets $V \subset \pi (Y)$ and $W \subset M^{m+n}$ and a definable continuous map $\tau :V \rightarrow \Gamma (\varphi )$ such that $\pi \circ \Pi (W)=V$ , $W \cap \Gamma (\varphi )=\tau (V)$ , and $\pi \circ \Pi \circ \tau $ is the identity map on V. In particular, the restriction of $\pi \circ \Pi $ to $W \cap \Gamma (\varphi )$ is injective.

Let $\iota :X \rightarrow \Gamma (\varphi )$ be the natural injection. The map $\iota $ is continuous because $\varphi $ is continuous. We may assume that $\pi \circ \Pi \circ \iota $ is injective replacing X with an open box contained in the definable open set $\iota ^{-1}(W)$ . We finally obtain the definable continuous injective map from an open box in $M^d$ to $M^e$ . We get $d \leq e$ by Corollary 3.7.

We next demonstrate the lemma when $\varphi $ is not necessarily continuous by induction on d. When $d=0$ , the definable set X is discrete and closed by Proposition 3.2. In particular, the definable map $\varphi $ is continuous. Therefore, the lemma holds true in this case. We next consider the case in which $d>0$ . Let $\mathcal D(\varphi )$ be the set of points at which $\varphi $ is discontinuous. We have $\dim \mathcal D(\varphi ) < \dim X$ by Theorem 3.8(6). We get $\dim \varphi (\mathcal D(\varphi ))=\dim \mathcal D(\varphi )$ by the induction hypothesis. We obtain $\dim (X \setminus \mathcal D(\varphi )) = \dim \varphi (X \setminus \mathcal D(\varphi ))$ because $\varphi $ is continuous on $X \setminus \mathcal D(\varphi )$ . We finally get $\dim \varphi (X)= \max \{\dim \varphi (X \setminus \mathcal D(\varphi )), \dim \varphi (\mathcal D(\varphi ))\}= \max \{\dim (X \setminus \mathcal D(\varphi )), \dim (\mathcal D(\varphi ))\} = \dim (X)$ by Theorem 3.8(4).⊣

Proof. Let X and Y be definable subsets of $M^m$ and $M^n$ , respectively. Set $d=\dim (\varphi ^{-1}(y))$ , $e=\dim (Y)$ , and $f=\dim (X)$ . We first reduce to the case in which there exists a coordinate projection $\pi :M^m \rightarrow M^d$ such that $\pi (\varphi ^{-1}(y))$ have nonempty interiors for all $y \in Y$ . In fact, consider the set $\Pi _{m,d}$ of all the coordinate projections of $M^m$ onto $M^d$ . Set $Y_{\pi }=\{y \in Y\;|\; \pi (\varphi ^{-1}(y)) \text { has a nonempty interior}\}$ . We get $Y=\bigcup _{\pi \in \Pi _{m,d}} Y_\pi $ by the assumption. Assume that the theorem is true for the restrictions of $\varphi $ to $\varphi ^{-1}(Y_{\pi })$ for all $\pi \in \Pi _{m,d}$ . We have

$$ \begin{align*}\dim X = \max_{\pi \in \Pi_{m,d}} \dim \varphi^{-1}(Y_{\pi}) = d + \max_{\pi \in \Pi_{m,d}} \dim Y_{\pi} = d + \dim(Y)\end{align*} $$

by Theorem 3.8(4). The theorem holds true for the original $\varphi $ . We may assume that there exists a coordinate projection $\pi :M^m \rightarrow M^d$ such that $\pi (\varphi ^{-1}(y))$ have nonempty interiors for all $y \in Y$ . We fix such a $\pi $ through the proof.

We next show that $d+e \leq f$ . By Lemma 3.5, we can get a coordinate projection $p:M^n \rightarrow M^e$ , a definable open subset W of $M^e$ contained in $p(Y)$ , and a definable continuous injective map $\tau :W \rightarrow Y$ which is homeomorphic onto its image such that $p \circ \tau $ is the identity map and $p^{-1}(w) \cap Y$ is discrete for any $w \in W$ . Consider the definable set

$$ \begin{align*}T=\{(w,v) \in W \times M^d\;|\;v \in \pi(\varphi^{-1}(\tau(w))) \text{ and } \pi^{-1}(v) \cap \varphi^{-1}(\tau(w)) \text{ is discrete}\} \text{.}\end{align*} $$

The fiber $T_w$ has a nonempty interior for any $w \in W$ by Lemma 3.3. Therefore, the set T has a nonempty interior by Lemma 2.8. In particular, we have $\dim (T)=d+e$ .

Consider the definable subset $S=(p \times \pi )^{-1}(T) \cap \Gamma '(\varphi ) \cap (\tau (W) \times M^m)$ of $M^m \times M^n$ , where $\Gamma '(\varphi )$ denotes the reversed graph of the definable map $\varphi $ given by $\Gamma '(\varphi )=\{(y,x) \in Y \times X\;|\;y=\varphi (x)\}$ . It is obvious that $(p \times \pi )(S)=T$ and $S \cap (p \times \pi )^{-1}(w,v)$ are discrete for all $(w,v) \in T$ . Apply the property (d) to S and the projection $p \times \pi $ . We can get a definable map $\psi ':T \rightarrow S$ such that $(p \times \pi ) \circ \psi '$ is the identity map on T. Set $\psi =\pi \circ \psi ':T \rightarrow X$ . It is obviously injective. We have $d+e = \dim (T)=\dim \psi (T)\leq f$ by Lemma 3.13 and Theorem 3.8(1).

We next demonstrate the opposite inequality $d+e \geq f$ . There exist a coordinate projection $q:M^m \rightarrow M^f$ , a definable open subset U of $M^f$ contained in $q(X)$ , and a definable continuous injective map $\sigma :U \rightarrow X$ by Lemma 3.5. The notation $\mathcal D(\varphi )$ denotes the set of points at which $\varphi $ is discontinuous. Since $\dim \mathcal D(\varphi ) < \dim X=f$ by Theorem 3.8(6), the projection image $q(\mathcal D(\varphi ))$ has an empty interior. The difference $U \setminus q(\mathcal D(\varphi ))$ has a nonempty interior by the property (b). Shrinking U if necessary, we may assume that $\varphi $ is continuous on $\sigma (U)$ . Take a coordinate projection $p:M^n \rightarrow M^e$ and a definable set W as in the proof of the inequality $d+e \leq f$ . Set

$$ \begin{align*}Z=\{(v,w) \in M^d \times W\;|\;\pi^{-1}(v) \cap (p \circ \varphi)^{-1}(w) \text{ is discrete}\}\text{.}\end{align*} $$

We demonstrate that the set Z has a nonempty interior. Fix a point $w \in W$ . We have only to demonstrate that $Z_w=\{v \in M^d\;|\;\pi ^{-1}(v) \cap (p \circ \varphi )^{-1}(w) \text { is discrete}\}$ has a nonempty interior for any $w \in W$ by Lemma 2.8.

For any $z \in p^{-1}(w) \cap Y$ , set $B(z)=\{v \in M^d\;|\;\pi ^{-1}(v) \cap \varphi ^{-1}(z) \text { is not discrete}\}$ . We have $\dim B(z)<d$ for any $z \in p^{-1}(w)$ by Lemma 2.8. Consider the set $D=\{(v,z) \in M^d \times M^n\;|\;v \in B(z) \text { and } z \in p^{-1}(w) \cap Y\}$ . We get $\dim D =\sup _{z \in p^{-1}(w) \cap Y} \dim B(z)<d$ by Theorem 3.11 because $p^{-1}(w) \cap Y$ is discrete. The definable set $\bigcup _{z \in p^{-1}(w) \cap Y} B(z)$ is the projection image of D, and it is of dimension smaller than d by Theorem 3.8(5). In particular, it has an empty interior. Consider the definable set $ Z^{\prime }_w = \bigcup _{z \in p^{-1}(w) \cap Y} \pi (\varphi ^{-1}(z)) \setminus \left ( \bigcup _{z \in p^{-1}(w) \cap Y} B(z) \right )$ . The set $Z^{\prime }_w$ has a nonempty interior by the property (b) because $\bigcup _{z \in p^{-1}(w) \cap Y} \pi (\varphi ^{-1}(z))$ has a nonempty interior by the definition of $\pi $ . On the other hand, the set $Z_w$ contains the set $Z^{\prime }_w$ . In fact, take a point $v \in Z^{\prime }_w$ . Consider the restriction of $\varphi $ to $\pi ^{-1}(v) \cap (p \circ \varphi )^{-1}(w)$ . The image is contained in $p^{-1}(w) \cap Y$ , and it is discrete. The fiber at $z \in p^{-1}(w) \cap Y$ is $\pi ^{-1}(v) \cap \varphi ^{-1}(z)$ and it is also discrete by the definition of $B(z)$ and $Z^{\prime }_w$ . Finally, the definable set $\pi ^{-1}(v) \cap (p \circ \varphi )^{-1}(w)$ is discrete by applying Lemma 2.5 to the restriction of $\varphi $ . We have demonstrated that $Z_w$ has a nonempty interior. Therefore, the definable set Z has a nonempty interior.

Take an open box V contained in Z. Consider the definable continuous map $\Phi :U \rightarrow M^d \times M^e$ given by $\Phi (x)=(\pi \circ \sigma (x), p \circ \varphi \circ \sigma (x))$ . Replacing the open definable set U with the definable open set $\Phi ^{-1}(V)$ if necessary, we may assume that $\Phi (U) \subset Z$ . By the definition of Z, the fiber $\Phi ^{-1}(v,w)$ is discrete for any $(v,w) \in Z$ . Therefore, we have $f=\dim U = \dim (\Phi (U)) \leq d+e$ by Lemma 3.13 and Theorem 3.8(1). We have finished the proof of the theorem.⊣

Proof. It is easy to prove that $X(d)$ is definable. We omit the proof. Apply Theorem 3.14 to the restriction of the projection $\Pi :M^{m+n} \rightarrow M^m$ to the set $\bigcup _{x \in X(d)} \{x\} \times X_x$ , then we get the corollary.⊣

Proof. We first reduce to the case in which the fibers $X \cap \pi ^{-1}(x)$ are equi-dimensional for all $x \in \pi (X)$ . In fact, set $Y_k=\{x \in \pi (X)\;|\; \dim (X \cap \pi ^{-1}(x))=k\}$ and $X_k=X \cap \pi ^{-1}(Y_k)$ for all $1 \leq k \leq n$ . They are definable because of the definition of dimension. Since we have $\dim (\pi (X_k \cap U)) \leq \dim (\pi (X \cap U))$ for any open box U by Theorem 3.8(1), the conditions in the corollary are satisfied for $X_k$ . Assume that the corollary holds true for $X_k$ . We have $\dim (Y_k)=\dim \pi (X_k) \leq d$ . We obtain $\dim (\pi (X))=\max _{1\leq k \leq n}\dim (Y_k) \leq d$ by Theorem 3.8(4). The corollary is also true for X. We have succeeded in reducing to the case in which the fibers are equi-dimensional.

Set $e=\dim (\pi (X))$ and $f=\dim (X \cap \pi ^{-1}(x))$ for $x \in \pi (X)$ . We have $\dim (X)=e+f$ by Theorem 3.14. We can take a point b in $R^{m+n}$ such that $\dim (X \cap V)=e+f$ for any open box V containing the point b by Corollary 3.12. Choose an open box U containing the point b so that $\dim (\pi (X \cap U)) \leq d$ , which exists by the assumption. Set $X'=X \cap U$ . It is obvious that the fibers $X' \cap \pi ^{-1}(x)$ are of dimension not greater than f for all $x \in \pi (X \cap U)=\pi (X')$ . Set $Y^{\prime }_k=\{x \in \pi (X')\;|\; \dim (X' \cap \pi ^{-1}(x))=k\}$ and $X^{\prime }_k=X' \cap \pi ^{-1}(Y^{\prime }_k)$ for $1 \leq k \leq f$ . Since we have $X'=\bigcup _{k=1}^f X^{\prime }_k$ , we get $\dim (X^{\prime }_l)=\dim (X')=e+f$ for some $1 \leq l \leq f$ by Theorem 3.8(4). Again by Theorems 3.14 and 3.8(1), we get $e+f = \dim \pi (X^{\prime }_l)+l\leq \dim (\pi (X \cap U)) + l \leq d+l$ . We finally obtain $e \leq d$ because $0 \leq l \leq f$ .⊣

Proof. We may assume that $\pi $ is the projection onto the first d coordinates without loss of generality. It is obvious that $\pi (X)$ is open because X is locally the graph of a continuous map. We fix a point $c \in \pi (X)$ . Note that the fiber $X_c = X \cap \pi ^{-1}(c)$ is discrete by the assumption. The fiber $X_c$ is also closed by Lemma 2.4. Let $p_e: M^d \rightarrow M^e$ be the projection onto the first e coordinates for all $0 \leq e \leq d$ . We demonstrate the following claim. The lemma is obvious from the claim for $e=d$ .

Claim. Let e be a nonnegative integer with $0 \leq e \leq d$ . There exists an open box $U_e$ in $M^e$ containing the point $p_e(c)$ such that, for any $y \in X_c$ , there exist an open box $V_{e,y}$ in $M^{d-e}$ and an open box $W_{e,y}$ in $M^n$ such that $y \in W_{e,y}$ , $\pi (W_{e,y})=U_e \times V_{e,y}$ , and the intersection of X with $W_{e,y}$ is the graph of a continuous map defined on $U_e \times V_{e,y}$ .

We prove the claim by induction on e. The claim follows from the assumption that all the points $x \in X$ are ( $X,\pi $ )-normal when $e=0$ . Consider the case in which $e>0$ . Let $c_e$ be the e-th coordinate of the point c. Take an element $d_{+,e} \in M$ with $c_e < d_{+,e}$ . For any $y \in X_c$ , let $\varphi _+(y)$ be the supremum of the point $x' \in M$ satisfying

1. (i) $c_e < x' < d_{+,e}$ , and

2. (ii) that there exist $a \in M$ with $a < c_e$ , an open box $V^{\prime }_{e,y}$ in $M^{d-e}$ , and an open box $W^{\prime }_{e,y}$ in $M^n$ such that

   • • $y \in W^{\prime }_{e,y}$ ,

   • • $\pi (W^{\prime }_{e,y})=U_{e-1} \times (a,x') \times V^{\prime }_{e,y}$ , and

   • • the intersection of X with $W^{\prime }_{e,y}$ is the graph of a continuous map defined on $\pi (W^{\prime }_{e,y})$ .

Such $x'$ exists and the value $\varphi _+(y)$ is larger than $c_e$ by the induction hypothesis. We get a definable function $\varphi _+:X_c \rightarrow M$ . The image $\varphi _+(X_c)$ is discrete by the property (a) because the fiber $X_c$ is discrete. It is closed by Lemma 2.4. Set $b^{\prime }_{e,+}=\inf \{z \in \varphi _+(X_c)\}$ . We have $b^{\prime }_{e,+}> c_e$ because $\varphi _+(X_c)> c_e$ . Take $b_{e,+}$ so that $c_e<b_{e,+}<b^{\prime }_{e,+}$ .

Take an element $d_{-,e} \in M$ with $c_e> d_{-,e}$ . For any $y \in X_c$ , we define $\varphi _-(y)$ as the infimum of the point $x \in M$ satisfying

• (i’) $c_e> x > d_{-,e}$ , and

• (ii’) that there exist an open box $V_{e,y}$ in $M^{d-e}$ and an open box $W_{e,y}$ in $M^n$ such that

  • • $y \in W_{e,y}$ ,

  • • $\pi (W_{e,y})=U_{e-1} \times (x,b_{e,+}) \times V_{e,y}$ , and

  • • the intersection of X with $W_{e,y}$ is the graph of a continuous map defined on $\pi (W_{e,y})$ .

We can take a point x satisfying the above conditions (i’) and (ii’). In fact, we can take a, $V^{\prime }_{e,y}$ , and $W^{\prime }_{e,y}$ satisfying the condition (ii) by putting $x'=b_{e,+}$ . Let x be an element satisfying the inequality $x \geq a$ and the condition (i’), then $V_{e,y}=V^{\prime }_{e,y}$ and $W_{e,y}=W^{\prime }_{e,y} \cap \pi ^{-1}(U_{e-1} \times (x,b_{e,+}) \times V_{e,y})$ satisfy the condition (ii’).

In the same way as above, the supremum $b^{\prime }_{e,-}=\sup \{z \in \varphi _-(X_c)\}$ satisfies the inequality $b^{\prime }_{e,-}<c_e$ . We take $b_{e,-}$ so that $b^{\prime }_{e,-}<b_{e,-}<c_e$ . Set $U_e=U_{e-1} \times (b_{e,-}, b_{e,+})$ . It is now obvious that $U_e$ satisfies the claim. We have finished the proofs of both the claim and the lemma.⊣

Proof. We first define the full dimension of a definable subset X of $M^n$ . Set $d=\dim X$ . The notation $\Pi _{n,d}$ denotes the set of all the coordinate projections of $M^n$ onto $M^d$ . The set $\Pi _{n,d}$ is a finite set. The full dimension $\operatorname{fdim} (X)$ is $(d,e)$ by definition if $d=\dim (X)$ and e is the number of elements in $\Pi _{n,d}$ under which the projection image of X has a nonempty interior. The pairs $(d,e)$ are ordered by the lexicographic order.

We prove the theorem by induction on $\operatorname{fdim} (X)$ . When $\dim (X)=0$ , X is closed and discrete by Proposition 3.2. The definable set X is obviously a quasi-special submanifold in this case.

We consider the case in which $\dim (X)>0$ . Set $(d,e)=\operatorname{fdim} (X)$ . Take a coordinate projection $\pi :M^n \rightarrow M^d$ such that $\pi (X)$ has a nonempty interior. Set $G=\{x \in X\;|\; x \text { is } (X,\pi ) \text {-normal}\}$ and $B=X \setminus G$ . It is obvious that any point $x \in G$ is $(G,\pi )$ -normal. The definable set G is $\pi $ -quasi-special submanifold by Lemma 4.2.

We demonstrate that $\pi (B)$ has an empty interior. Assume the contrary. There exists an open box U such that the fibers $B_x=\pi ^{-1}(x) \cap B$ are discrete for all $x \in U$ by Lemma 3.3. We can take a definable map $\tau :U \rightarrow B$ with $\pi (\tau (x))=x$ for all $x \in U$ because the structure $\mathcal M$ possesses the property (d) in Definition 2.6. The dimension of points at which the map $\tau $ is discontinuous is of dimension smaller than d by Theorem 3.8(6). We may assume that the restriction of $\tau $ to U is continuous shrinking U if necessary.

Set $Z=\partial (X \setminus \tau (U))$ . We get $\dim Z = \dim \partial (X \setminus \tau (U)) < \dim (X \setminus \tau (U)) \leq \dim X=d$ by Theorem 3.8(1,7). We have $\dim \overline {Z} = \dim Z <d$ again by Theorem 3.8(4,7). On the other hand, we have $d=\dim U =\dim \pi (\tau (U)) \leq \dim \tau (U) \leq \dim X =d$ by Theorem 3.8(1,5). We get $\dim (\tau (U))=d$ . It means that $\tau (U) \not \subset \overline {Z}$ by Theorem 3.8(1).

Take a point p in $\tau (U) \setminus \overline {Z}$ . Take a sufficiently small open box V containing the point p. We have $X \cap V=\tau (U) \cap V$ by the definition of Z and p. Since the restriction of $\tau $ to U is continuous, there exists an open box $U'$ contained in $U \cap \tau ^{-1}(V)$ . Consider the open box $V'=V \cap \pi ^{-1}(U')$ . It is obvious that $X \cap V'=\tau (U) \cap V'$ is the graph of the restriction of $\tau $ to $U'$ by the definition. Any point $\tau (U) \cap V'$ is $(X,\pi )$ -normal, but it contradicts to the definition of B and the inclusion $\tau (U) \subset B$ . We have shown that $\pi (B)$ has an empty interior. In particular, we get $\operatorname{fdim} (B) < \operatorname{fdim} (X)$ .

There exists a decomposition $B=C_1 \cup \cdots \cup C_k$ of B satisfying the conditions in the lemma by the induction hypothesis. The decomposition $X=G \cup C_1 \cup \cdots \cup C_k$ is the desired decomposition of X.

It is obvious that the number of quasi-special submanifolds N is not greater than

$$ \begin{align*}\sum_{d=0}^n (\text{the cardinality of }\Pi_{n,d})=\sum_{d=0}^n \left(\begin{array}{c}n\\d\end{array}\right)=2^n\text{.}\end{align*} $$

⊣

Proof. Set $X_i^0=X_i$ and $X_i^1=M^n \setminus X_i$ for all $1 \leq i \leq m$ . For any $\sigma \in \{0,1\}^m$ , the notation $\sigma (i)$ denotes the i-th component of $\sigma $ . Set $X_\sigma = \bigcap _{i=1}^m X_i^{\sigma (i)}$ for any $\sigma \in \{0,1\}^m$ . The family $\{X_\sigma \}_{\sigma \in \{0,1\}^m}$ is mutually disjoint and satisfies the equality $M^n=\bigcup _{\sigma \in \{0,1\}^m} X_\sigma $ . For all $\sigma \in \{0,1\}^m$ , there exist families $\{C_{\sigma ,j}\}_{j=1}^{N_\sigma }$ of mutually disjoint quasi-special submanifolds with $X_\sigma = \bigcup _{j=1}^{N_\sigma } C_{\sigma ,j}$ and $N_\sigma \leq 2^n$ by Lemma 4.3. The family $\bigcup _{\sigma \in \{0,1\}^m} \{C_{\sigma ,j}\}_{j=1}^{N_\sigma }$ gives the decomposition we are looking for.⊣

Proof. By reverse induction on d, we construct a decomposition $\{C_{\lambda }\}_{\lambda \in \Lambda _d}$ of $M^n$ into quasi-special submanifolds partitioning $\{X_i\}_{i=1}^m$ such that the closures of all the quasi-special submanifolds of dimension not smaller than d are the unions of subfamilies of the decomposition.

When $d=n$ , take a decomposition $\{D_\lambda \}_{\lambda \in \Lambda }$ of $M^n$ into quasi-special submanifolds partitioning $\{X_i\}_{i=1}^m$ by Theorem 4.4. Set $\Lambda _n'=\{\lambda \in \Lambda \;|\;\dim (D_{\lambda })=n\}$ . Get a decomposition $\{E_\lambda \}_{\lambda \in \widetilde {\Lambda _n}}$ of $M^n$ into quasi-special submanifolds partitioning the family $\{D_\lambda \}_{\lambda \in \Lambda } \cup \{\overline {D_{\lambda }}\setminus D_{\lambda }\}_{\lambda \in \Lambda _n'}$ . Consider the set

$$ \begin{align*}\widetilde{\Lambda_n}'=\{\lambda \in \widetilde{\Lambda_n} \;|\; E_\lambda \text{ is not contained in any }D_{\lambda'}\text{ with }\lambda' \in \Lambda_n'\}\text{.} \end{align*} $$

We always have $\dim (E_\lambda )<n$ for all $\lambda \in \widetilde {\Lambda _n}'$ by Theorem 3.8(7). Hence, the family $\{D_\lambda \}_{\lambda \in \Lambda _n'} \cup \{E_\lambda \}_{\lambda \in \widetilde {\Lambda _n}'}$ is trivially a decomposition of $M^n$ into quasi-special submanifolds partitioning $\{X_i\}_{i=1}^m$ we are looking for.

We next consider the case in which $d < n$ . Let $\{D_{\lambda }\}_{\lambda \in \Lambda _{d+1}}$ be a decomposition of $M^n$ into quasi-special submanifolds partitioning $\{X_i\}_{i=1}^m$ such that the closures of all the quasi-special submanifolds of dimension not smaller than $d+1$ are the unions of subfamilies of the decomposition. It exists by the induction hypothesis. Set $\Lambda _d'=\{\lambda \in \Lambda _{d+1}\;|\;\dim (D_{\lambda })=d\}$ and $\Lambda _d"=\{\lambda \in \Lambda _{d+1}\;|\;\dim (D_{\lambda }) \geq d\}$ . Get a decomposition $\{E^d_\lambda \}_{\lambda \in \widetilde {\Lambda _d}}$ of $M^n$ into quasi-special submanifolds partitioning the family $\{D_\lambda \}_{\lambda \in \Lambda _{d+1}} \cup \{\overline {D_{\lambda }}\setminus D_{\lambda }\}_{\lambda \in \Lambda _d'}$ . Set $\widetilde {\Lambda _d}'=\{\lambda \in \widetilde {\Lambda _d} \;|\; E_\lambda \text { is not contained in any }D_{\lambda '}\text { with }\lambda ' \in \Lambda _d"\}\text {.} $ The family $\{D_\lambda \}_{\lambda \in \Lambda _d"} \cup \{E_\lambda \}_{\lambda \in \widetilde {\Lambda _d}'}$ is a decomposition of $M^n$ into quasi-special submanifolds partitioning $\{X_i\}_{i=1}^m$ we want to construct.

The ‘furthermore’ part of the theorem is obvious from the proof.⊣