Proof. Let \(y\in \cl(A)_x\). Then there is a sequence \(\big((x_i,y_i)\big)_{i\in \NN}\) of elements in \(A\) such that \(\lim_{i\to \infty} (x_i,y_i)= (x,y)\). Since \(B\) is open, we may assume that \(x_i \in B\) for \(i\in \NN\). Thus, \(y \in \cl\big((B \times \RR) \cap A\big)_x\). The other inclusion is immediate. ◻

Proof. Let \(z\in (B\times \RR)\cap A\). Since \(A\) is étale at \(z\), there is an open box \(U\subseteq \RR^{n+1}\) around \(z\) such that \(\pi|_{A\cap U}\) is a homeomorphism onto its image. Therefore, \(\pi|_{(B\times \RR)\cap A \cap U}\) is also homeomorphic onto its image and thus \((B\times \RR)\cap A\) is étale at \(z\). ◻

Proof. Statements (1) and (2) are well known results. For (3), observe that since \(B\) is both closed and open in \(A\) then \(B^{[\lambda]}=A^{[\lambda]}\cap B\) for any ordinal \(\lambda\). ◻

Proof. Since \(\R\) is not o-minimal, there is an infinite \(X\subseteq \RR\) definable in \(\R\) with empty interior. Because \(\R\) is noiseless, the set \(X\) is nowhere dense. Thus its topological closure \(\cl(X)\) is definable, infinite, and nowhere dense. Hence the open core \(\R^{\circ}\) is not o-minimal. By [13] we know that \(\R\) defines an infinite, closed, and discrete set \(E \subseteq \RR\). Note that \(E\) is necessarily unbounded. By taking a subset and, if necessary, multiplying the elements of \(E\) by \(-1\), we can assume that \(E \subseteq \RR_{\geq 1}\). Then \[D = \{1\} \cup \{ e^{-1} \;: \;e \in E\}\] is the desired sequence set. ◻

Proof. Statements (1)-(3) follow easily from the definition of the dimension. Statement (4) is [5] and Statement (5) is [5]. ◻

Proof. We proceed by induction on \(n\). Suppose \(n=0\). First, consider the case that \[\max_{d \in D} \dim(A_d)=-\infty.\] Then \(A_d\) is empty for all \(d\in D\). Thus \[\dim(\bigcup_{d\in D} A_d)=\dim \emptyset = -\infty.\] Now suppose that \[\max_{d \in D} \dim(A_d)=0.\] Then, for each \(d\in D\), the set \(A_d\) is either empty or a singleton set for each \(d\in D\). Thus \(\bigcup_{d\in D} A_d\) does not have interior by the Baire Category Theorem. Since there is \(d\in D\) such that \(A_d\neq \emptyset\), we get that \(\dim(\bigcup_{d\in D} A_d)=0\).
Now suppose that \(n\in \NN_{\geq 1}\). By [fact:naive](2), it is enough to show that \[\dim\left ( \bigcup_{d\in D} A_d \right) \leq \max_{d \in D} \dim(A_d).\] If \(\max_{d\in D} \dim A_d=n\), then this is trivial. So we can reduce to the case that \[\max_{d \in D} \dim (A_d)<n.\] Towards a contradiction, suppose that there is \(k\in \NN\) such that \[\dim(\bigcup_{d\in D} A_d)=k>\max_{d \in D} \dim (A_d).\] We first show that \(k<n\). Let \(d\in D\). Since \(n \geq k > \dim A_d\), we know that \(A_d\) does not have interior and hence is nowhere dense by [fact:sbctforn]. Thus, its topological closure \(\cl(A_d)\) is nowhere dense. By the Baire Category Theorem \(\bigcup_{d\in D}\cl(A_d)\) does not have interior, and hence is nowhere dense by [fact:sbctforn]. Thus \[k = \dim(\bigcup_{d\in D} A_d)\leq \dim(\bigcup_{d\in D} \cl(A_d))<n.\] Let \(\pi: \RR^n \to \RR^k\) be a coordinate projection such that \(\pi(\bigcup_{d\in D} A_d)\) has interior in \(\RR^k\). Therefore \[\dim(\bigcup_{d\in D} \pi(A_d))\geq k.\] Since \(\dim(A_d)<k\) for each \(d \in D\), we conclude by induction that \[\dim(\bigcup_{d\in D} \pi(A_d))<k,\] a contradiction. ◻

Proof. Statement (1) follows as in the proof of [15].

For (2), it is enough to show that \(A\) is locally closed by [fact:D-Sigma]. We proceed by induction on \(n\). The case \(n=1\) is immediate.

Suppose that (2) holds for \(n-1\). Let \(x \in A\), and let \(U\subseteq \RR^{n-1}\) be an open box such that \(\pi(x)\in U\) and \(\pi(A) \cap U\) is closed in \(U\).
Suppose that \(w_n=0\). Then there is \(f \in \mathcal{C}(f)\) such that \(A = \Gamma (f)\). Set \(V:= U \times \RR\). It follows from the continuity of \(f\) that \(A \cap V\) is closed in \(V\).
Suppose that \(w_n=1\). Let \(f,g \in \mathcal{C}_{\infty}(\pi(A))\) be such that \(f < g\) and \(A= \big(f,g\big)\). Let \(J\) be an open subinterval of \(\big(f(\pi(x)),g(\pi(x))\big)\). Since \(f,g \in \mathcal{C}_{\infty}(\pi(A))\), there is an open box \(B\subseteq U\) such that for all \(b\in B\) \[J \subseteq \big(f(\pi(b)),g(\pi(b))\big).\] Now set \(V:=B\times J\), and observe that \(A\cap V=V\), and, thus \(A\cap V\) is closed in \(V\). This completes the proof of (2).
For (3), we also proceed by induction on \(n\). First, suppose that \(n = 1\). If \(w_1=1\), the statement follows, since \(\pi_{(1)}\) is the identity. If \(w_1=0\), observe that \(A\) is a singleton set. Now suppose that (3) holds for \(n-1\).

Suppose that \(w_n=0\). Let \(X \subseteq \RR^{n-1}\) be a \((w_1, \dots, w_{n - 1})\)-cell and let \(f \in\mathcal{C}(X)\) be such that \(\Gamma(f) = A\). Observe that the function from \(X\) to \(A\) mapping \(x\in X\) to \((x, f(x))\) is a homeomorphism with inverse \(\pi_{(1, 1, \dots, 1, 0)}\vert_A\). Thus \[\pi_w = \pi_{(w_1, \dots, w_{n - 1})} \circ \pi_{(1, 1, \dots, 1, 0)}\vert_A\] is a homeomorphism onto an open set.

Suppose that \(w_{n}=1\). Let \(X\) be a \((w_1, \dots, w_{n - 1})\)-cell and let \(f,g \in \mathcal{C}_{\infty}(X)\) be such that \(A = \big(f, g\big)\). Then \[\begin{align} \big(f, g\big) &\to \RR^{\dim X} \times \RR\\ (x, y) &\mapsto (\pi_{(w_1, \dots, w_{n - 1})}(x), y) \end{align}\] is a homeomorphism onto an open set by our induction hypothesis, and is equal to the projection \(\pi_{(w_1, \dots, w_{n - 1}, 1)}\vert_A\).

For (4), we again use induction on \(n\). Suppose that \(n=1\). There are two cases. If \(A\) is a \(0\)-cell, then it is a point and thus \(\dim(A)=0\). If \(A\) is a \(1\)-cell, then it is an interval in \(\RR\), and thus \(A\) has interior. Therefore \(\dim(A)=1\).
Now suppose that \(n>1\). Set \[B \mathrel{\vcenter{:}}= \pi_{(1, 1, \dots, 1, 0)}(A).\] By induction, \[\dim B = w_1 + \cdots + w_{n - 1}.\] First consider the case that \(w_n=1\). Then for every \(b \in B\) \[\dim ((\pi_{(1, 1, \dots, 1, 0)}|_A)^{-1}(b)) = \dim A_b = 1.\] By [fact:DSigDim] \[\dim A = \dim ((\pi_{(1, 1, \dots, 1, 0)}|_A)^{-1}(B)) = \dim B + 1 = w_1+\cdots + w_n.\] Now consider the case that \(w_n=0\). Then for every \(b \in B\) \[\dim ((\pi_{(1, 1, \dots, 1, 0)}|_A)^{-1}(b)) = \dim A_b = 0.\] By [fact:DSigDim] \[\dim A = \dim ((\pi_{(1, 1, \dots, 1, 0)}|_A)^{-1}(B)) = \dim B = w_1+\cdots + w_n.\] ◻

Proof. We first show (1). Suppose that \(Z\) is somewhere dense in \(A\). Then \(\pi_w(Z)\) is somewhere dense in \(\pi_w(A)\). Since \(\pi_w(A)\) is open by [lem:proj-cell], the projection \(\pi_w(Z)\) is somewhere dense in \(\RR^{\dim(A)}\). Thus \(\pi_w(Z)\) has interior in \(\RR^{\dim(A)}\) by [fact:sbctforn], and hence has interior in \(\pi_w(A)\). Since \(\pi_w|_A\) is a homeomorphism, it follows that \(Z\) has interior in \(A\).

For (2), suppose that \(Z\) is nowhere dense in \(A\). Let \(Y\) be the topological closure of \(Z\) in \(A\). Note that \(Y\) is \(\cl(Z)\cap A\), and hence \(Y\) is \(\DSig\) by [lem:celldsig] and [fact:D-Sigma](2). Since \(Z\) is nowhere dense in \(A\), so is \(Y\). Since \(\pi_w|_A\) is a homeomorphism by [lem:proj-cell], we have that \(\pi_w(Y)\) is nowhere dense and closed in the open set \(\pi_w(A)\). Thus by [fact:naive](1) \[\dim \pi_w(Y) < w_1+\dots + w_n = \dim A.\] Observe that \(\pi_w(Y)\) is \(\DSig\) by [fact:D-Sigma](5). By [fact:DSigDim] and using that \(\pi_w|_A\) is a homeomorphism, \[\dim Z \leq \dim Y = \dim ((\pi_w|_A)^{-1}(\pi_w(Y))) = \dim \pi_w(Y) < \dim A.\] ◻

Proof. By [lem:proj-cell], we know that \(\pi_w(A)\) is open and that \(\pi_w|_A\) is a homeomorphism. By [fact:genericcnt] there is an open subset \(V\subseteq \pi_w(A)\) such that \(V\) is dense in \(\pi_w(A)\) and \(\big(f \circ (\pi_w|_A)^{-1}\big)|_V\) is continuous. Thus \((\pi_w|_A)^{-1}(V)\) is an open and dense subset of \(A\) and the restriction of \(f\) to \((\pi_w|_A)^{-1}(V)\) is continuous. ◻

Proof. We proceed by induction on \(n\). If \(n=0\), this is immediate. Now assume that the statement holds for all \(k< n\).

Suppose \(w_1+\dots + w_n=n\). Then \(\pi(A)\) is open by [lem:proj-cell]. Thus the statement follows from [fact:fiberlemma].

Now consider the case that \(w_1+\dots+w_n <n\). By [lem:proj-cell], the projection \(\pi_w|_{\pi(A)}\) is a homeomorphism. Set \[B:=\{x \in \pi(A) : \cl(A_x)\neq \cl(A)_x\}\] and \[C:= \{ (\pi_w(x),z) : x \in \pi(A), (x,z)\in A\}.\] Since \(\pi_w|_{\pi(A)}\) is a bijection, we have that for all \(x \in \pi(A)\) \[A_x = C_{\pi_w(x)}.\] Thus \(\cl(A_x)=\cl(C_{\pi_w(x)})\). Since \(\pi_w|_{\pi(A)}\) is a homeomorphism, \[\cl(A)_x = \cl(C)_{\pi_w(x)}.\] Hence \[\pi_w(B) = \{ y \in \pi_w(\pi(A)) \;: \;\cl(C_y) \neq \cl(C)_{y} \}.\] Since \(w_1+\dots+w_n <n\), we obtain by induction that \(\pi_w(B)\) is nowhere dense in \(\pi_w(\pi(A))\). Since \(\pi_w|_{\pi(A)}\) is a homeomorphism, we conclude that \(B\) is nowhere dense in \(\pi(A)\). ◻

Proof. Statement (3) follows from (1) and (2). We prove the other statements by simultaneous induction on \(n\).

For \(n=0\) and \(n=1\), this follows immediately from the definition of a stack in \(\RR\). Now assume that \(n>1\) and that (1)-(2) holds for \(n-1\).

By the inductive hypothesis, \(\pi(A)\) has finitely many connected components, each of which is a \((w_1,\dots,w_{n-1})\)-cell and clopen in \(\pi(A)\). Thus, we may assume that \(\pi(A)\) is a \((w_1,\dots,w_{n-1})\)-cell, and, in particular, connected. We split into two cases.
Suppose that \(w_n=0\). Let \(f_1,\ldots ,f_m \in \mathcal{C}(\pi(A))\) witness that \(A\) is a \(w\)-stack. Since \(f_1<\ldots<f_m\), we know that for each \(i,j\in\{1,\dots,m\}\) with \(i\neq j\) \[\Gamma(f_i)\cap \Gamma(f_j)=\emptyset.\] Moreover, since each \(f_i\) is continuous and \(\pi(A)\) is connected, it follows that each \(\Gamma(f_i)\) is connected and clopen. Thus the collection of graphs \(\Gamma(f_1),\ldots,\Gamma(f_m)\) are the connected components of \(A\). Each \(\Gamma(f_i)\) is a \(w\)-cell by definition, and its image under \(\pi\) is \(\pi(A)\).
Suppose that \(w_n=1\). Let \(f_1,\ldots f_m,g_1,\ldots,g_m \in \mathcal{C}_{\infty}(\pi(A))\) witness that \(A\) is a \(w\)-stack. Since either \(g_i < f_{i+1}\) or \(g_i=f_{i+1}\) for each \(i=1,\dots,m-1\), then \[(f_i,g_i)\cap (f_j,g_j)=\emptyset\] for all \(i,j\in\{1,\dots,m\}\) with \(i\neq j\). Since \(\pi(A)\) is connected, it follows that each \((f_i,g_i)\) is connected and clopen in \(A\). Thus the sets \((f_1,g_1),\dots,(f_m,g_m)\) are the connected components of \(A\). Again, each \((f_i,g_i)\) is a \(w\)-cell by definition, and its image under \(\pi\) is \(\pi(A)\). ◻

Proof. For (1), it is enough to show that \(A\) is locally closed. Let \(a \in A\) and let \(C\) be the connected component of \(A\) containing \(a\). By [lem:fin-CC], we know that \(C\) is a \(w\)-cell and open in \(A\). Let \(U\subseteq \RR^n\) be open such that \(C = A \cap U\). By [lem:celldsig](2), there is an open neighborhood \(V\) of \(x\) such that \(C\cap V\) is closed in \(V\). Thus \(A\cap (U\cap V) = C\cap (U \cap V)\) is closed in \(U\cap V\).

For (2), let \(a \in A\) and let \(C\) be the connected component of \(A\) containing \(a\). By [lem:fin-CC], we know that \(C\) is a \(w\)-cell and there is an open box \(U\subseteq \RR^n\) containing \(a\) such that \(A\cap U\subseteq C\). Thus, \(\pi_w|_C\) is a homeomorphism and \(\pi_w(C)\) is open by [lem:proj-cell]. Hence \(\pi_w|_A\) is a local homeomorphism. Since \[\pi_w(A) = \bigcup_{C \text{ conn. comp. of A}} \pi_w(C),\] we conclude that \(\pi_w(A)\) is open.

For (3), observe that \(A\) has finitely many connected components \(C_1,\ldots,C_k\) by [lem:fin-CC], each of which is a \(w\)-cell. By [cor:cell-dim], we know that \(\dim(C_i)=w_1+\cdots+w_n\). Thus, by (1) amd [lem:dim], we conclude that \[\dim(A)=\dim\left(\bigcup_{i=1}^k C_i\right)=w_1+\cdots+w_n\] as desired. Hence (3) holds. ◻

Proof. For (1), suppose that \(Z\) is somewhere dense in \(A\). Then \(Z\) is somewhere dense in some connected component \(C\) of \(A\) by [lem:fin-CC]. Since \(C\) is a \(w\)-cell, we know that \(Z\) has interior in \(C\) by [lem:BC-cell]. Thus \(Z\) has interior in \(A\), because \(C\) is open in \(A\).

For (2), suppose that \(Z\) is nowhere dense in \(A\). Let \(C_1,\dots,C_m\subseteq A\) be the connected components of \(A\), each of which is a \(w\)-cell. Let \(i\in \{1,\dots,m\}\). By [lem:fin-CC], we have that \(Z\cap C_i\) is nowhere dense in \(C_i\), and thus by [cor:cellnowheredense] \[\dim Z \cap C_i < \dim C_i = \dim A.\] Putting this together with [fact:naive], we conclude that \[\dim Z = \max_{i\in\{1,\dots,m\}} \dim(Z \cap C_i) < \dim A.\] ◻

Proof. By [lem:fin-CC], there are \(w\)-cells \(C_1,\dots,C_k\) such that these are the connected components of \(A\) and each is clopen in \(A\). By [cor:cellcont], for each \(i=1,\dots,k\) there is a definable set \(B_i\subseteq C_i\) such that \(B_i\) is open and dense in \(C_i\), and \(f|_{B_i}\) is continuous. Since the \(C_i\)’s are open in \(A\), it follows that \(\bigcup_{i=1}^k B_i\) is open and dense in \(A\), and the restriction of \(f\) to \(\bigcup_{i=1}^k B_i\) is continuous. ◻

Proof. By Lemma [lem:fin-CC], the connected components of \(\pi(A)\) are \(w\)-cells \(C_1,\ldots, C_k\subseteq \RR^{n}\), each of which is clopen in \(\pi(A)\). Thus by [fact:opensubset] we have that for each \(i\in \{1,\dots,k\}\) and each \(x\in C_i\) \[\cl(A)_x = \cl\big((C_i \times \RR)\cap A)\big)_xand\cl(A_x) = \cl\big((C_i \times \RR)\cap A)_x\big).\] Therefore \[\begin{align} \{x\in \pi(A) &: \cl(A_x)\neq \cl(A)_x \} \\ &= \bigcup_{i=1}^k \{x\in C_i : \cl\big((C_i \times \RR)\cap A)_x\big)\neq \cl\big((C_i \times \RR)\cap A)\big)_x \} \end{align}\] By Lemma [lem:fl-cell] the set \[\{x\in C_i : \cl\big((C_i \times \RR)\cap A)_x\big)\neq \cl\big((C_i \times \RR)\cap A)\big)_x \}\] is nowhere dense in \(C_i\) for all \(i\in\{1,\dots,k\}\). ◻

Proof. We will prove the statement only for \(w_n=0\), as the proof for \(w_n=1\) is similar. Since \(C\) is connected, it is contained in some connected component \(C'\) of \(\pi(A)\). Let \(f_1,\ldots,f_m\in \mathcal{C}(C')\) witness that \(A\) is a stack. Then \(f_1|_C,\ldots,f_m|_C \in \mathcal{C}(C)\) witness that \(A \cap [C \times \RR]\) is a stack. ◻

Proof. Let \(C\) be a connected component of \(\pi(A)\). For \(m \in \NN\), set \[C(m) \mathrel{\vcenter{:}}= \{a \in C \;: \;|A_a| = m\}.\] We first prove that there is \(m \in \NN\) such that \(C=C(m)\). Since \(C\) is connected, it suffices to show that \(C(m)\) is open. Pick an arbitrary point \(a \in C(m)\), and let \(b_1, \dots, b_m \in A_a\) be such that \(b_1 < \cdots < b_m\). Pick open intervals \(J_1, \dots, J_m\subseteq \RR\) such that for all \(i \in \{1,\dots,m\}\)

• \(b_i \in J_i\),

• \(J_i \cap J_j = \emptyset\) for all \(j \in \{1,\dots, i\}\) with \(j \neq i\).

By (3), we can pick a nonempty open box \(B_0 \subseteq \RR^n\) and continuous functions \(f_1, \dots, f_m \colon (B_0\cap C) \to \RR\) such that \(a\in B_0\) and for all \(i \in \{1,\dots,m\}\) \[\label{eq:lem411} A \cap (B_0 \times J_i) = \Gamma(f_i).\tag{1}\] In particular, we have that \(f_i(a) = b_i\).

Towards a contradiction, suppose that for every open box \(B \subseteq B_0\) containing \(a\) there is \(c \in B\) such that \[A_c \neq \{f_1(c), \dots, f_m(c)\}.\] Then there is \(\ell \in \{0, \dots,m\}\) and a sequence \((a_i)_{i\in \NN}\) of elements in \(C\) such that \(\lim_{i\to \infty} a_i= a\) and for each \(i\in \NN\) \[A_{a_i} \cap \Big(f_\ell(a_i), f_{\ell + 1}(a_i)\Big) \neq \emptyset.\] Here, we use the convention that \(f_0(c)=-\infty\) and \(f_{m + 1}(c)=\infty\) for \(c\in C\). For each \(i\in \NN\) pick \(u_i\in \RR\) such that \[u_i\in A_{a_i} \cap \Big(f_\ell(a_i), f_{\ell + 1}(a_i)\Big).\] Since \(A\) is vertically bounded, the sequence \(((a_i,u_i))_{i\in \NN}\) is bounded. Thus, there is a convergent subsequence \(((a_{i_k},u_{i_k}))_{k\in \NN}\). Let \(u\in \RR\) be such that \[\lim_{k\to \infty } (a_{i_k},u_{i_k}) = (a,u).\] Since \(\cl(A_a)= \cl(A)_a\) and \(A_a\) is finite, we conclude that \(u \in A_a\). Thus \[\lim_{k\to \infty } u_{i_k} \in \{f_\ell(a), f_{\ell + 1}(a)\}.\] However, by 1 , we have that for each \(k\in \NN\) \[u_{i_k} \notin (J_\ell \cup J_{\ell+1}).\] This contradicts that \(u \in (J_\ell \cup J_{\ell+1})\).

Let \(m\in \NN\) be such that \(C(m) = C\). For \(i \in \{1,\dots,m\}\), let \(h_i \colon C \to \RR\) be the function \[a \mapsto \text{(i-th element of A_a)}.\] Then \(h_1 < \dots < h_m\) and \[\bigcup_{i = 1}^{n} \Gamma(h_i) = A.\] The continuity of each \(h_i\) follows from (3). ◻

Proof. Let \(C\) be a connected component of \(\pi(A)\), and let \(f_1,\dots,f_m,g_1,\dots,g_m \in \mathcal{C}_{\infty}(C)\) witness that \(A\) is a stack. For \(i\in \{1,\dots,m\}\), set \(h_i:=\Mid(f_i,g_i)\). It is easy to check that \(h_1,\dots,h_m \in \mathcal{C}(C)\), \[h_1 < \dots < h_m,and\Mid(A) \cap [C \times \RR] = \bigcup_{i=1}^m \Gamma(h_i).\] Thus \(\Mid(A)\) is a \((w_1,\dots,w_{n},0)\)-stack. The stated equivalence follows. ◻

Proof. We only prove the case that \(w_n=0\) as the proof in the case that \(w_n=1\) is similar. Let \(C\) be a connected component of \(\pi(A)\) and let \(f_1,\ldots,f_{m_C} \in \mathcal{C}(C)\) be such that \[A \cap [C \times \RR] = \bigcup_{i=1}^{m_C} \Gamma(f_i).\] Then each \(\Gamma(f_i)\) is a connected component of \(A\), and hence a connected component of \(B\). Thus \(C=\pi(\Gamma(f_i))\) is a connected component of \(\pi(B)\) by [lem:fin-CC](2). ◻

Proof. Let \(C\subseteq \RR^n\) be a connected component of \(A\). Since \(A\) is a \(w\)-stack, there is \(f \in \mathcal{C}(\pi(C))\) such that \(\pi(C)\) is a connected component of \(\pi(A)\) and \(C = \Gamma(f)\). Hence \(\pi(C)\) a connected component of \(\pi(B)\). Since \(B\) is a \(w\)-stack, there are \(f_1,\dots,f_m \in \mathcal{C}(C)\) such that \[B \cap (C\times \RR) = \bigcup_{i=1}^m \Gamma(f_i).\] Let \(i\in \{1,\dots,m\}\) be such that \(\Gamma(f_i)\cap \Gamma(f)\neq \emptyset\). Since \(\Gamma(f_i)\) is a connected component of \(B\) and \(A\subseteq B\), we get that \(\Gamma(f)\subseteq \Gamma(f_i)\). Since the domains of \(f\) and \(f_i\) are the same the functions must be the same. Thus \(C\) is a connected component of \(B\). ◻

Proof. We will prove this only for \(w_{n+1}=0\) as the proof for \(w_{n+1}=1\) is similar. Let \(C_0\) be a connected of \(A\cap [A' \times \RR]\). By (4) there is a connected component \(C'\) of \(A'\cap \pi(A)\) and \(f_0\in \mathcal{C}(C')\) such that \(C_0=\Gamma(f_0)\). Let \(C\) be the connected component of \(\pi(A)\) containing \(C'\). Then there is \(f\in \mathcal{C}(C)\) such that \(f|_{C'} = f_0\) and \(\Gamma(f)\) is a connected component of \(A\). By (1), we know that \(\Gamma(f)\) is a connected component of \(B\) as well. Since \(C'\) is a connected component of \(A'\cap \pi(A)\), it is a connected component of \(A'\) by (3), and thus a connected component of \(B'\) by (2). Hence \[C_0 \subseteq \Gamma(f_0|_{C'}) \subseteq B \cap [B' \times \RR].\] Towards a contradiction, suppose that \(C_0\) is not a connected component of \(B \cap [B'\times \RR]\). Then there is \(C''\subseteq \RR^{n}\) containing \(C'\) such that \(C''\neq C'\) and \(\Gamma(f|_{C''})\) is a connected component of \(B \cap [B' \times \RR]\). However, this contradicts that \(C'\) is a connected component of \(B'\). ◻

Proof. We need only prove that \(A\) is a stack, because then it follows that \(A\) is a substack of \(B\). We proceed by induction on \(n\). The case \(n=0\) is immediate.

Let \(n\geq 1\), and suppose that the statement holds for \(n-1\). We only consider the case when \(w_n=0\), since the proof for \(w_n=1\) is similar. Let \(C\) be a connected component of \(\pi(A)\). Then there is a connected component \(C'\) of \(\pi(B)\) such that \(C\subseteq C'\). Because \(B\) is a \(w\)-stack, there are \(f_1,\ldots,f_{m_{C'}} \in \mathcal{C}(C')\) such that \[B\cap[C'\times \RR] = \bigcup_{i=1}^{m_{C'}} \Gamma(f_i).\] Since \(A\subseteq B\), \[A\cap [C\times \RR] \subseteq B \cap [C'\times \RR].\] Because each connected component of \(A\) is a connected component of \(B\), there is \(I\subseteq \{1,\ldots,m_ {C'}\}\) such that \[A\cap [C\times \RR] = \bigcup_{i\in I} \Gamma(f_i|_{C}).\] If \(C\neq C'\), then \(\Gamma(f_i|_{C})\) is a connected component of \(A\) for \(i\in I\), but not a connected component of \(B\), a contradiction. Thus \(C=C'\). By induction, we conclude that \(\pi(A)\) is a stack and a substack of \(\pi(B)\). Then the functions \(f_i\) with \(i\in I\) witness that \(A\) is a stack. ◻

Proof. Let \(d,e\in D\) with \(d>e\) and let \(C\) be a connected component of \(\Mid(A)\). By [lem:stackmid] there are \(f,g \in \mathcal{C}_{\infty}(\pi(C))\) such that \((f,g)\) is a connected component of \(A\) and \[C = \Gamma(\Mid(f,g)).\] Then \((f,g)\) is a connected component of \(B\), because \(A\) is substack of \(B\). Thus \(\Gamma(\Mid(f,g))\) is a connected component of \(\Mid(B)\) by [lem:stackmid]. ◻

Proof. We first show that \[\surf(\es)=\bigcup_{d\in D_2} S_{\lambda_{D_1}(d)}.\] The inclusion \(\bigcup_{d\in D_2} S_{\lambda_{D_1}(d)} \subseteq \surf(\es)\) is immediate from the definitions. Thus, we only need to prove that \(\surf(\es)\subseteq \bigcup_{d\in D_2} S_{\lambda_{D_1}(d)}\). Let \(d\in D_1\). Since \(D_1\) and \(D_2\) have limit points at \(0\), there is \(e\in D_2\) with \(e<d\). But then \(\lambda_{D_1}(e)\leq d\) as well, and therefore \(S_d\subseteq S_{\lambda_{D_1}(e)}\).
Let \(d,e\in D_2\) be such that \(e<d\). We show that \(S_{\lambda_{D_1}(d)}\) is a substack of \(S_{\lambda_{D_1}(e)}\). Let \(C\) be a connected component of \(S_{\lambda_{D_1}(d)}\). Since \(e<d\), we have that \(\lambda_{D_1}(e)\leq \lambda_{D_1}(d)\). Thus \(C\) is a connected component of \(S_{\lambda_{D_1}(e)}\), because \(\es\) is a \((w,D_1)\)-grid.
We now prove by induction on \(n\) that \(\es|_{D_2}\) is a \((w,D_2)\)-grid. The case \(n=0\) is trivial. Now suppose that \(n>1\) and the claim holds for \(n-1\). Since \[\pi(\es|_{D_2}) = \pi(\es)|_{D_2},\] we get from the induction hypothesis that \(\pi(\es|_{D_2})\) is a \((w_1,\ldots,w_{n-1},D_2)\)-grid. If \(w_n=0\), then \(\bigcup_{d\in D_2} S_{\lambda_{D_1}(d)}\) is étale, because \(\bigcup_{d\in D} S_d\) is. Thus, together with the above observations, we conclude that \(\es|_{D_2}\) is a \((w,D_2)\)-grid and \(\surf(\es) = \surf(\es|_{D_2}).\) ◻

Proof. Since \(S_d\) has finitely many connected components for each \(d\in D\) by [lem:fin-CC], the third statement follows from the second. Set \(A:=\surf(\mathcal{S})\). We prove (1) and (2) by induction on \(n\). The base case that \(n=0\) is trivial.

Now suppose the statement holds for \(n-1\). Let \(C\) be a connected component of \(A\). Then there is a connected component \(C'\) of \(\pi(A)\) such that \(\pi(C)\subseteq C'\). By our inductive assumption, there is \(d\in D\) such that \(C'\) is a connected component of \(\pi(S_d)\). Let \(x\in C\) and let \(e \in D_{\leq d}\) be maximal such that \(x \in S_e\). We will show that \(C\) is a connected component of \(S_e\). Since \(\pi(S_d)\) is a substack of \(\pi(S_e)\), we have that \(C'\) is a connected component of \(\pi(S_e)\).
Suppose first that \(w_n=0\). Let \(f_1,\ldots, f_m \in \mathcal{C}(C')\) witness that \(S_e\) is a \(w\)-stack. Then \[S_e\cap [C' \times \RR]=\bigcup_{i=1}^m \Gamma(f_i).\] Let \(j\in \{1,\dots,m\}\) be such that \(x \in \Gamma(f_j).\) Since \(\Gamma(f_j)\) is connected and \(x\in C\), we have that \(\Gamma(f_j)\subseteq C\). Thus, it is enough to show that \(\Gamma(f_j)\) is a connected component of \(A\). Because \(\Gamma(f_j)\) is connected, it is enough to show that \(\Gamma(f_j)\) is maximal with this property. Since \(C'\) is closed in \(\pi(A)\), it follows that \(\Gamma(f_j)\) is closed in \(A\). Let \(y \in \Gamma(f_j)\). Since \(A\) is étale at \(y\), there is an open box \(U\subseteq \RR^n\) containing \(x\) such that \(\pi|_{U\cap A}\) is a homeomorphism onto its image. Thus, \[U\cap A = U\cap \Gamma(f_j).\] Hence \(\Gamma(f_j)\) is clopen in \(A\), and hence a connected component of \(A\).

Now suppose that \(w_n=1\). Let \(f_1,\ldots,f_m,g_1,\ldots,g_m \in \mathcal{C}_{\infty}(C')\) witness that \(S_e\) is a \(w\)-stack. Then \[S_e\cap [C' \times \RR] = \bigcup_{i=1}^m \big(f_i,g_i\big).\] Let \(j\in \{1,\dots,m\}\) be such that \(x \in \big(f_j,g_j\big).\) Since \((f_j,g_j)\) is connected and \(x\in C\) we have that \((f_j,g_j)\subseteq C\). Thus, it is enough to show that \((f_j,g_j)\) is a connected component of \(A\). Because \((f_j,g_j)\) is connected, it is enough to show that \((f_j,g_j)\) is maximal with this property. Since \(C'\) is open in \(\pi(A)\), we have that \((f_j,g_j)\) is open in \(A\). Let \(y \in A \setminus (f_j,g_j)\), and let \(e' \in D_{\geq e}\) be such that \(y \in S_{e'}\). Since \(S_{e'}\) is a \(w\)-stack, there are definable continuous functions \(f,g : C'\to \RR\) such that \(y \in (f,g)\). Since \(S_e\) is a subgrid of \(S_{e'}\) and \(y\notin (f_j,g_j)\), we have that \((f_j,g_j) \cap (f,g) =\emptyset\). Since \(C'\) is closed in \(\pi(A)\), it follows that \((f_j,g_j)\) is closed in \(A\). Hence \((f_j,g_j)\) is clopen in \(A\), and hence a connected component of \(A\).
We have established (2) and the forward direction of (1). We now prove the backwards direction of (1). Let \(d\in D\), and let Let \(C\) be a connected component of \(S_d\). Let \(C'\) be a connected subset of \(A\) such that \(C\subseteq C'\).

Suppose first that \(w_n=0\). Then there is \(f\in \mathcal{C}(\pi(C))\) such that \(C=\Gamma(f)\) and \(\pi(C)\) is a connected component of \(\pi(S_d)\). Then \(\pi(C)\subseteq\pi(C')\), and thus, by our inductive hypothesis, \(\pi(C')=\pi(C)\). By the forward direction of \((1)\), there is \(e\in D\) such that \(C'\) is a connected component of \(S_e\). Thus, there is a continuous definable function \(g:\pi(C)\to \RR\) such that \(C'=\Gamma(g)\). But then \(f=g\), and so \(C=C'\). Hence, \(C\) is a connected component of \(A\).
Suppose now that \(w_n=1\). Let \(f,g \in \mathcal{C}(\pi(C))\) be such that \(C=\big(f,g\big)\) and \(\pi(C)\) is a connected component of \(\pi(S_d)\). Similarly to above, we conclude that \(\pi(C')=\pi(C)\). By the forward direction of \((1)\) there is \(e\in D\) such that \(C'\) is a connected component of \(S_e\). Thus, there are \(\hat{f},\hat{g}\in\mathcal{C}(\pi(C))\) such that \(C'=\big(\hat{f},\hat{g}\big)\). Since \(S_d\) is a substack of \(S_e\), we conclude that \(f=\hat{f}\) and \(g=\hat{g}\). Thus, \(C=C'\) and \(C\) is a connected component of \(A\). ◻

Proof. We proceed by induction on \(n\). The case \(n=0\) is trivial. Now suppose that \(n\geq 1\) and that the statement holds for \(n-1\).

Let \(x\in\surf(\es)\), and let \(d\in D\) be such that \(x\in S_d\). By our induction hypothesis, \(\{\pi(x)\}\) is a connected component of \(\pi(S_d)\). Thus \(\{x\}\) is a connected component of \(S_d\). So \(\{x\}\) is a connected component of \(\surf(\es)\) as well and is clopen in \(\surf(\es)\) by [lem:CC0-grid]. Therefore, \(\surf(\es)\) is discrete. ◻

Proof. Set \(A:= \surf(\mathcal{S})\). For (1), it is enough to show that \(A\) is locally closed. Let \(a \in A\) and let \(C\) be the connected component of \(A\) containing \(a\). By [lem:fin-CC] and [lem:CC0-grid], we know that \(C\) is a \(w\)-cell and clopen in \(A\). Let \(U\subseteq \RR^n\) be open such that \(C = A \cap U\). By [lem:celldsig](2), there is an open neighborhood \(V\) of \(x\) such that \(C\cap V\) is closed in \(V\). Thus \(A\cap (U\cap V) = C\cap (U \cap V)\) is closed in \(U\cap V\).

For (2), let \(a\in A\) and let \(C\) be the connected component of \(A\) containing \(a\). By Lemma [lem:CC0-grid] we know that \(C\) is clopen in \(A\) and there is \(d\in D\) such that \(C\) is a connected component of \(S_d\). Let \(U\subseteq \RR^n\) be an open ball containing \(a\) such that \(A\cap U\subseteq C\cap U\). Since \(C\subseteq S_d\), \[S_d \cap U = A \cap U = C\cap U.\] By Lemma [lem:proj-stack] we get that \(\pi_w|_{S_d\cap U}\) is a local homeomorphism and \(\pi_w(S_d)\) is open. Therefore \(\pi_w|_A\) is a local homeomorphism. Since \[\pi_w(A) = \bigcup_{d \in D} \pi_w(S_d),\] we conclude that \(\pi_w(A)\) is open.

By [lem:dim] \[\dim(A)=\dim\left(\bigcup_{d\in D} S_d\right) = \max_{d\in D}\dim(S_d)\] and \(\dim(S_d)=w_1+\cdots+w_n\) by [cor:dim-stack]. Thus (3) holds. ◻

Proof. Set \(A:=\surf(\mathcal{S})\). For (1), suppose that \(Z\) is somewhere dense in \(A\). By [lem:CC0-grid](2) we can assume that \(Z\) is dense in some connected component \(C\) of \(A\). Then, by [lem:CC0-grid](1), there is \(d\in D\) such that \(C\) is a connected component of \(S_d\). Thus \(Z\) is somewhere dense in \(S_d\), and has interior in \(S_d\) by [lem:BC-stacks]. Therefore \(Z\) has interior in \(A\) by [lem:CC0-grid](2).

For (2), suppose that \(Z\) is nowhere dense in \(A\). By [lem:CC0-grid], we know that \(Z\cap S_d\) is nowhere dense in \(S_d\). Thus, by [cor:stacknowheredense] \[\dim (Z \cap S_d) < \dim S_d \leq \dim A.\] By [lem:dim] \[\dim Z = \dim \bigcup_{d\in D} (Z\cap S_d) = \max_{d\in D} \dim (Z\cap S_d) < \dim A.\]

For (3), note that the boundary of \(Z\) in \(\surf(\mathcal{S})\) does not have interior in \(\surf(\mathcal{S})\), and hence is nowhere dense in \(\surf(\mathcal{S})\) by (1). ◻

Proof. Let \(Y\subseteq \surf(\es)\) be the set of points at which \(f\) is not continuous. Let \(Z\) be the topological closure of \(Y\) in \(\surf(\es)\). Set \(B:= \surf(\es)\setminus Z\). By definition, \(B\) is open and \(f|_B\) is continuous. It is left to show that \(B\) is dense in \(\surf(\es)\). Let \(U\subseteq \RR^n\) be an open box such that \(\surf(\es)\cap U \neq\emptyset\). By [lem:CC0-grid], we may shrink \(U\) such that there is \(d\in D\) with \[\label{eq:gridcont} U\cap \surf(\es) = U \cap S_d.\tag{2}\] By [cor:stackcont] there is \(B'\subseteq S_d\) such that \(B'\) is open and dense in \(S_d\) and the restriction \(f|_{B'}\) is continuous. From 2 , it follows that there is \(x \in U \cap \surf(\es)\) at which \(f\) is continuous. ◻

Proof. Towards a contradiction, suppose that \(Z_i\) does not have interior in \(\surf(\es)\) for \(i=1,\dots,m\). By [lem:BC-grids], each \(Z_i\) is nowhere dense in \(\surf(\es)\). Then \(\bigcup_{i=1}^k Z_i\) is nowhere dense in \(\surf(\es)\), contradicting our assumption. ◻

Proof. Let \(\mathcal{S}=(S_d)_{d\in D}\) be a grid such that \(\surf(\mathcal{S})=\pi(A)\). Set \[B:=\{x\in \pi(A): \cl(A_x)\neq \cl(A)_x \}.\] By [lem:BC-grids] either \(B\) has interior or is nowhere dense in \(\pi(A)\). Suppose that \(B\) has interior in \(\pi(A)\). Then, by [lem:CC0-grid](2), there is a connected component \(C\) of \(\pi(A)\) such that \(B\) has interior in \(C\). Since \(C\) is clopen, we have that for each \(x\in C\) \[\cl(A)_x = \cl\big((C \times \RR)\cap A)\big)_xand\cl(A_x) = \cl\big((C \times \RR)\cap A)_x\big).\] Therefore \[\begin{align} B\cap C &= \{x\in C : \cl(A_x)\neq \cl(A)_x \} \\ &= \{x\in C : \cl\big((C \times \RR)\cap A)_x\big)\neq \cl\big((C \times \RR)\cap A)\big)_x \}. \end{align}\] By [lem:CC0-grid](1) there is \(d\in D\) such that \(C\) is a connected component of \(S_d\). Thus \(C\) is a cell by [lem:fin-CC]. By Lemma [lem:fl-cell] \[\{x\in C : \cl\big((C \times \RR)\cap A)_x\big)\neq \cl\big((C \times \RR)\cap A)\big)_x \}\] is nowhere dense in \(C\). This contradicts that \(B\) has interior in \(C\). ◻

Proof. For each \(d\in D\), define \(S_d':= S_d \times \RR\), and set \(\es' := (S_d')_{d\in D}\). Clearly, \(\pi(S_d')_{d\in D}\) is just \((S_d)_{d\in D}\), and hence a \(w\)-grid. It is also easy to see that \(S'_d\) is a \((w,1)\)-stack for every \(d \in D\). Let \(d,e\in D\) be such that \(d>e\). Then every connected component of \(S_d'\) is of the form \(C\times \RR\), where \(C\) is a connected component of \(S_d\). Since \(S_d\) is a substack of \(S_e\), the set \(C\) is a connected component of \(S_e\) as well. Thus \(C\times \RR\) is a connected component of \(S_e'\). Hence \(S_d'\) is a substack of \(S'_e\). Therefore \((S_d')_{d\in D}\) is a \((w,1)\)-grid. ◻

Proof. Let \(\es = (S_d)_{d \in D}\) be the \(w\)-grid such that \(\pi(A) = \surf(\es)\). Let \[\begin{align} \lambda : A &\to \RR\\ (x,y) &\mapsto \max \Big( (\RR\setminus A_x) \cap (-\infty,y) \Big) \end{align}\] and let \[\begin{align} \mu : A &\to \RR\\ (x,y) &\mapsto \min \Big( (\RR\setminus A_x) \cap (y,\infty)\Big). \end{align}\] Since \((x,y) \in A\), then \[\label{eq:lemgrco3} \big(\lambda(x,y),\mu(x,y)\big) \subseteq A_x.\tag{3}\] Set \[S^*_{d} := \{ (x,y) \in (S_d \times \RR) \cap A \;: \;\bigvee_{i,j \in \{1,\dots,m\}} \big( (x,\lambda(x,y)) \in S_{i,d} \wedge (x,\mu(x,y)) \in S_{j,d}\big)\}.\] Since \(A_x\) is open for every \(x \in \pi(A)\), it is easy to check that \[A = \bigcup_{d\in D} S^*_d.\] Thus, it is left to show that \((S^*_d)_{d\in D}\) is a \((w,1)\)-grid.
Let \(C\subseteq \RR^{n}\) be a connected component of \(S_d\). For \(i\in \{1,\dots,m\}\) and \(d\in D\), let \(f_{i,d,1},\dots,f_{i,d,m_{i,d}} \in \mathcal{C}(C)\) be such that

1. \(f_{i,d,1}<\cdots<f_{i,d,m_{i,d}}\),

2. \(S_{i,d}\cap (C\times \RR) = \bigcup_{j=1}^{m_{i,d}} \Gamma(f_{i,d,j})\).

Since the surfaces of \(\es_1,\dots,\es_m\) are disjoint, \[\Gamma(f_{i,d,k}) \cap \Gamma(f_{j,d,\ell}) = \emptyset,\] for all \(i,j \in \{1,\dots,m\}\) and \(k,\ell\in \NN\) whenever \(i\neq j\) or \(k\neq \ell\). Since \(C\) is connected, we get in such a situation that \[\label{eq:lemgrco1} eitherf_{i,d,k} < f_{j,d,\ell}orf_{j,d,\ell} < f_{i,d,k}.\tag{4}\]

Let \(x_0\in C\). Since \(S_{1,d},\dots,S_{m,d}\) are \((w,0)\)-stacks, there are \(y_1,\dots,y_{2p} \in \RR\) such that \[\label{eq:lemgrco2} y_1<\dots < y_{2p}, \text{ and } (S_d^*)_{x_0} = \bigcup_{i=0}^{p-1} \big(y_{2i+1},y_{2i+2}\big).\tag{5}\] Let \(k_1,\dots,k_{2p},\ell_1,\dots,\ell_{2p} \in \NN\) be such that for \(i=1,\dots,2p\) \[y_i = f_{k_i,d,\ell_i}(x).\] By 4 , \[f_{k_1,d,\ell_1} < \dots < f_{k_{2p},d,\ell_{2p}}.\] To show that \(S_d^*\) is a \((w,1)\)-stack, it is enough to show that \[S_d^*\cap (C\times \RR) = \bigcup_{i=0}^{p-1} \big(f_{k_{2i+1},d,\ell_{2i+1}}, f_{k_{2i+2},d,\ell_{2i+2}}).\]

Let \(i\in \{0,\dots,p-1\}\). First we show that \[\big(f_{k_{2i+1},d,\ell_{2i+1}},f_{k_{2i+2},d,\ell_{2i+2}}\big) \subseteq S_d^*\cap (C\times \RR) .\] Let \((x,y) \in \big(f_{k_{2i+1},d,\ell_{2i+1}},f_{k_{2i+2},d,\ell_{2i+2}}\big)\). It is enough to show that \[\lambda(x,y) = f_{k_{2i+1},d,\ell_{2i+1}}(x)and\mu(x,y) = f_{k_{2i+2},d,\ell_{2i+2}}(x).\] Suppose not. Then there is \(u \in \{1,\dots,m\}\) and \(v \in \{1,\dots,m_{u,e}\}\) such that \(\surf(\es_u)\subseteq B_1\) and \[f_{k_{2i+1},d,\ell_{2i+1}}(x) < f_{u,d,v}(x) < f_{k_{2i+2},d,\ell_{2i+2}}(x).\] By 4 , \[f_{k_{2i+1},d,\ell_{2i+1}}< f_{u,d,v} <f_{k_{2i+2},d,\ell_{2i+2}}.\] But then \(f_{u,d,v}(x_0) \in (y_{2i+1},y_{2i+2})\), contradicting 5 .
Let \((x,y) \in S_d^*\cap (C\times \RR)\). Let \(u_1,u_2 \in \{1,\dots,m\}\) and let \(v_1,v_2 \in \NN\) be such that \[\lambda(x,y) = f_{u_1,d,v_1}(x)and\mu(x,y) = f_{u_2,d,v_2}(x).\] It suffices to show that \[\big(f_{u_1,d,v_1}(x_0),f_{u_2,d,v_2}(x_0)\big)\subseteq A_{x_0}.\] Suppose not. Since \(f_{u_1,d,v_1}(x)\) and \(f_{u_2,d,v_2}(x)\) are boundary points of \(A_x\), we get from 3 that \[\frac{f_{u_1,d,v_1}(x)+f_{u_2,d,v_2}(x)}{2}\in \Midp(A_x).\] Thus there is \(u_3 \in \{1,\dots,m\}\) and \(v_3 \in \NN\) such that \[f_{u_3,d,v_3}(x) = \frac{f_{u_1,d,v_1}(x)+f_{u_2,d,v_2}(x)}{2}.\] By 4 , \[f_{u_1,d,v_1}< f_{u_3,d,v_3} < f_{u_2,d,v_2}.\] Hence \[\big(f_{u_1,d,v_1}(x_0),f_{u_2,d,v_2}(x_0)\big)\cap A_{x_0} \neq \emptyset.\] Thus \[\big(f_{u_1,d,v_1}(x_0),f_{u_2,d,v_2}(x_0)\big)\cap \operatorname{bd}(A_{x_0}) \neq \emptyset.\] So there is some \(u_4 \in \{1,\dots,m\}\) and \(v_4 \in \NN\) such that \(\Gamma(f_{u_4,d,v_4}) \subseteq B_1\), and \[f_{u_1,d,v_1}(x_0)< f_{u_4,d,v_4}(x_0) < f_{u_2,d,v_2}(x_0).\] By 4 , \[f_{u_1,d,v_1}(x)< f_{u_4,d,v_4}(x) < f_{u_2,d,v_2}(x).\] Since \(f_{u_4,d,v_4}(x) \in (B_1)_x\), this contradicts 3 .
We now show that \((S_d^*)_{d\in D}\) is a \((w,1)\)-grid. We have already seen that \(\pi(S_d^*)=\pi(A)\) and thus that \((\pi(S_d^*))_{d\in D}\) is a grid. Let \(d,e\in D\) be such that \(d>e\). It is left to show that \(S_d^*\) is a substack of \(S_e^*.\) For this, let \(C'\) be a connected component of \(S_d^*\). We know that there are \(i,j \in \{1,\dots,m\}\) and \(k,\ell \in \NN\) such that \[C'=\big(f_{i,d,k},f_{j,d,\ell}\big)\] We claim that \(C'\) is a connected component of \(S_e^*\). We first show that \(C'\subseteq S_e^*\). Let \((x,y)\in C'\). Then \[\lambda(x,y)=f_{i,d,k}(x)and\mu(x,y)=f_{j,d,\ell}(x).\] Since \(\Gamma(f_{i,d,k})\) is a connected component of \(S_{i,d}\), it is also a connected component of \(S_{i,e}\). Therefore \(f_{i,d,k}=f_{i,e,k'}\) for some \(k'\in \NN\), and \(f_{i,e,k'}(x)=\lambda(x,y)\). Similarly, we find \(\ell'\in \NN\) such that \(f_{j,d,\ell}=f_{j,e,\ell'}\) and \(f_{j,e,\ell'}(x)=\mu(x,y)\). Thus \((x,y)\in S_e^*\).
Towards a contradiction, suppose that \(C'\) is not a connected component of \(S_e^*\). Then there are \(i',j'\in \{1,\dots,m\}\) and \(k',\ell'\in \NN\) such that \[\big(f_{i,d,k},f_{j,d,\ell}\big) = C'\subsetneq \big(f_{i',e,k'},f_{j',e,\ell'}\big).\] Thus either \(f_{i',e,k'}<f_{i,d,k}\) or \(f_{j',e,\ell'}>f_{j,d,\ell}\). Then either \[\Gamma(f_{i,d,k}) \subseteq \big(f_{i',e,k'},f_{j',e,\ell'}\big)or\Gamma(f_{j,d,\ell}) \subseteq \big(f_{i',e,k'},f_{j',e,\ell'}\big).\] However, this contradicts that \[\Gamma(f_{i,d,k}) \cap A = \emptysetand\Gamma(f_{j,d,\ell}) \cap A = \emptyset.\] Thus \(C'\) is a connected component of \(S_e^*\). ◻

Proof. Let \(\es=(S_d)_{d\in D}\) be a \((0,\dots,0)\)-grid in \(\RR^n\) such that \(\pi(A)=\surf(\es)\). For every \(d\in D\), set \[A(d):=\fisol(A,d) \cap \big(\RR^n \times (-\frac{1}{d},\frac{1}{d})\big).\] Since the fiber \(A_a\) is discrete for all \(a\in \pi(A)\) and \(D \cup \{0\}\) is closed, we have that \[\bigcup_{d\in D} A(d) = A.\] Moreover, the family \((A(d))_{d\in D}\) is decreasing and each \(A(d)\) is vertically bounded. Thus, for all \(d\in D\) and \(a\in \pi(A)\) the fiber \((A(d))_x\) is finite. Set \[S_d':= A(d) \cap [S_d \times \RR] \text{ and } D':=\{d\in D: \pi(A(d))\cap S_d\neq\emptyset\}.\] We claim that \((S_d')_{d\in D'}\) is a \(((0,\ldots,0),D')\)-grid.
First, we show that \(S_d'\) is a stack for all \(d\in D'\). Let \(d \in D'\). By construction \(\pi(S_d')=S_d\) and hence is a stack. Since \(\dim S_d = 0\), every connected component of \(S_d\) is a singleton set. Let \(x \in S_d\) be such that \(\{x\}\) is a connected component of \(S_d\). Since \(A(d)_x\) is finite, we may write \(A(d)=\{y_1,\ldots,y_k\}\) for some \(k\in \NN\) and \(y_1<\cdots<y_k\). For each \(i=1,\dots, k\), let \(f_i\) be the function mapping \(x\) to \(y_i\). Then \(f_1,\dots,f_k\) witness that \(S_d'\) is a stack.
Next, we show that \((S_d')_{d\in D'}\) is a grid. Observe that \(\pi(S_d')_{d\in D'}=(S_d)_{d\in D'}\), the latter being a grid by [lem:changeseqs]. Let \(d,e\in D'\) with \(d>e\). Let \(C\) be a connected component of \(S_d'\). Then there are \((x,y) \in C\) such that \(x\in S_d\) and \(C = \{(x,y)\}.\) Since \(S_d\) is a substack of \(S_e\), the set \(\{x\}\) is a connected component of \(S_e\). Moreover, since \((x,y)\in A(d)\), we know that \((x,y)\in A(e)\) as well. Thus \(\{(x,y)\}\) is a connected component of \(S_e'\).
Lastly, we must show that \(\bigcup_{d\in D'} S_d'\) is étale. Let \(d\in D'\), and let \((x,y)\in S'_d\). Since \(S_d\) is discrete there’s some open box \(U\subseteq \RR^n\) such that \(S_d \cap U=\{x\}\). Similarly, since \(A_x\) is discrete there is an open interval \(I\subseteq \RR^n\) such that \(A_x \cap I=\{y\}\). Then \(U\times I\) is an open box in \(\RR^{n+1}\) and since \((U \times I) \cap A = \{(x,y)\}\) this box witnesses that \(A\) is étale at \((x,y)\). Thus \((S_d')_{d\in D'}\) is a grid and \(\surf((S_d')_{d\in D'})=A\). ◻

Proof. We proceed by induction on \(n\). For \(n=0\), the statement is immediate. Now suppose that \(n\geq 1\) and that the statement holds for \(n-1\). We first show that \((\pi(S_d'))_{d\in D}\) is a grid. By the assumptions and [lem:projsubstack], for every \(d,e\in D\) with \(d>e\), the stack \(\pi(S'_d)\) is a substack of \(\pi(S_d)\) and \(\pi(S'_e)\). By induction, it follows that \((\pi(S_d'))_{d\in D}\) is a grid.
For every \(d,e\in D\) with \(d>e\), the set \(S_d'\) is a substack of \(S_e'\) by assumption. If \(w_n=0\), the surface \(\bigcup_{d\in D} S_d\) is étale. In this situation the union \(\bigcup_{d\in D} S_d'\) is étale as well, since \(\bigcup_{d\in D} S_d'\subseteq \bigcup_{d\in D} S_d\). Thus \((S_d')_{d\in D}\) is a grid. ◻

Proof. By [lem:stackmid] we know that \(\Mid(S_d)\) is a \((w,0)\)-stack with \(\pi(\Mid(S_d))=\pi(S_d).\) Thus \(\pi(\Mid(\es))=\pi(\es)\), and hence is a \(w\)-grid. For \(d,e \in D\) with \(d>e\), we have that \(S_d\) is a substack of \(S_e\), and hence \(\Mid(S_d)\) is a substack of \(\Mid(S_e)\) by Lemma [lem:midsubstack].
It is left to show that \(\bigcup_{d\in D} \Mid(S_d)\) is étale. Let \((x,y)\in \Mid(S_d)\) for some \(d\in D\). Let \(C\) be the connected component of \(\pi(S_d)\) containing \(x\), and let \(f,g\in C_{\infty}(C)\) be functions such that \((f,g)\) is the connected component of \(S_d\) containing \((x,y)\). Then \(\Gamma(\Mid(f,g))\) is the connected component of \(S_d\) containing \((x,y)\). Since \((f,g)\) is open in \(\surf(\es)\), there exists some open box \(U\) around \((x,y)\) such that \[U\cap \surf(\es)\subseteq (f,g).\] Therefore \[\bigcup_{d\in D} \Mid(S_d) \cap U = \Mid(S_d)\cap U\subseteq \Gamma(\Mid(f,g)).\] Hence \(\pi\) maps \((\bigcup_{d\in D} S_d) \cap U\) homeorphically onto its image. Thus \(\bigcup_{d\in D} \Mid(S_d)\) is étale, and hence \((\Mid(S_d))_{d\in D}\) is a \((w,0)\)-grid. ◻

Proof. For each \(d\in D\), set \[S_d' := S_d \cap (\Mid(\pi(S_d)) \times \RR).\] Let \(d\in D\). We claim that \(S_d'\) is a \((w,0,0)\)-stack. First observe that \[\label{eq:lemucb} \pi(S_d')=\pi(S_d) \cap \Mid(\pi(S_d))=\Mid(\pi(S_d)).\tag{6}\]

Thus \(\pi(S_d')\) is a \((w,0)\)-stack. Let \(C\) be a connected component of \(\Mid(\pi(S_d))\). Then \(C\) is contained in a connected component \(C'\) of \(\pi(S_d)\). Thus there are \(f_1,\ldots,f_m \in \mathcal{C}(C')\) such that \[S_d \cap [C' \times \RR] = \bigcup_{i=1}^m \Gamma(f_i).\] It follows that \[S_d' \cap [C \times \RR] = \bigcup_{i=1}^m \Gamma(f_i|_C).\] Therefore the functions \(f_1|_C, \ldots,f_m|_C\) witness that \(S_d'\) is a \((w,0,0)\)-stack.
We claim that \((S_d')_{d\in D}\) is a \((w,0,0)\)-grid. By [lem:midgrid] it follows from 6 that \((\pi(S_d)')_{d \in D}\) is a \((w,0)\)-grid. Let \(d,e\in D\) be such that \(d>e\). We now show that \(S_d'\) is a substack of \(S_e'\). Let \(C_0\) be a connected component of \(S_d'\). Then there is a connected component \(C\) of \(\Mid(\pi(S_d))\), a connected component \(C'\) of \(\pi(S_d)\), and \(f \in \mathcal{C}(C')\) such that \(\Gamma(f)\) is a connected component of \(S_d\) and \(\Gamma(f|_C)=C_0.\) Then \(\Gamma(f)\) is also a connected component of \(S_e\), the set \(C'\) is a connected component of \(\pi(S_e)\), and \(C\) is a connected component of \(\Mid(\pi(S_e))\). Thus \(\Gamma(f|_C)\) must also be a connected component of \(S_e'\), as its connected components are precisely of this form.
It is left to show that \(\bigcup_{d\in D} S_d'\) is étale. Observe that \[\bigcup_{d\in D} S_d' = \bigcup_{d\in D} \Big(S_d \cap (\Mid(\pi(S_d)) \times \RR)\Big) \subseteq \bigcup_{d\in D} S_d.\] Since \(\bigcup_{d\in D} S_d\) is étale, we conclude that \(\bigcup_{d\in D} S_d'\) is étale as well by [fact:etale]. ◻

Proof. The statement follows immediately from the definitions of the functions \(\mu\) and \(\nu\). ◻

Proof. Towards a contradiction, suppose that there are \(v_1 \in C\) and \(v_2\in B_{v_1}\) such that \[\lim_{d \to 0} f_{d,\mu(d,u)}(v_1)< v_2 <\lim_{d \to 0} f_{d,\nu(d,u)}(v_1).\] Set \(v:=(v_1,v_2)\). Since \(v_2 \in B_{v_1}\), there is \(e \in D\) such that one of the following two statements holds:

1. \(\lim_{d \to 0} f_{d,\mu(d,u)}(v_1) < f_{e,\mu(e,v)}(v_1) \leq v_2\), or

2. \(v_2 \leq f_{e,\nu(e,v)}(v_1) <\lim_{d \to 0} f_{d,\nu(d,u)}(v_1).\)

In both cases, we can deduce that for all \(d\in D\) \[f_{d,\mu(d,u)}(v_1) < f_{e,\mu(e,v)}(v_1)< f_{d,\nu(d,u)}(v_1).\] Thus for all \(d \in D\) \[f_{d,\mu(d,u)}(u_1) < f_{e,\mu(e,v)}(u_1)< f_{d,\nu(d,u)}(u_1).\] In particular, \[f_{e,\mu(e,u)}(u_1) < f_{e,\mu(e,v)}(u_1)< f_{e,\nu(e,u)}(u_1).\] By the maximality of \(\mu(e,u)\), we obtain that \[f_{e,\mu(e,u)}(u_1) \leq u_2 < f_{e,\mu(e,v)}(u_1) < f_{e,\nu(e,u)}(u_1).\] This is a contradiction against the minimality of \(\nu(e,u)\). ◻

Proof. This is again immediate from the definitions of \(\mu\) and \(\nu\). ◻

Proof. Set \(u:=(u_1,u_2)\), and for ease of notation, set \[u_2^- := \lim_{d \to 0} f_{d,\mu(d,u)}(u_1) \text{ and } u_2^+ := \lim_{d \to 0} f_{d,\nu(d,u)}(u_1)\] Similarly, set \[v_2^- := \lim_{d \to 0} f_{d,\mu(d,u)}(v_1) \text{ and } v_2^+ := \lim_{d \to 0} f_{d,\nu(d,u)}(v_1).\] Set \(v^-:=(v_1,v_2^-)\) and \(v^+:=(v_1,v_2^+)\). We now show that \[v^{-} \sim_{\es} u \sim_{\es} v^+.\] By [lem:emptybetween], \[\big(v_2^-,v_2^+\big) \cap B_{v_1} = \emptyset.\] Thus by [lem:emptyequiv], \(v_2^{-} \sim_{\es,v_1} v_2^+\). Hence it is enough to show that for all \(d\in D\) \[\mu(d,u)=\mu(d,v^-), \text{ and } \nu(d,u)=\nu(d,v^+).\] We first show that \(\mu(d,u)=\mu(d,v^-)\) for all \(d\in D\). Towards a contradiction, suppose that there is \(e\in D\) such that \(\mu(e,u)\neq \mu(e,v^-)\). Then \[f_{e,\mu(e,u)}(v_1) < f_{e,\mu(e,v^-)}(v_1) \leq v_2^-,\] yet, \[f_{e,\mu(e,u)}(u_1) \leq u_2 < f_{e,\mu(e,v^-)}(u_1).\] Thus for all \(d\in D\) \[f_{d,\mu(d,u)}(u_1) < f_{e,\mu(e,v^-)}(u_1),\] and hence \[f_{d,\mu(d,u)}(v_1) < f_{e,\mu(e,v^-)}(v_1).\] Thus \(v_2^- = f_{e,\mu(e,v^-)}(v_1)\), contradicting \(C\subseteq \fisol(A)\).

Similarly, one can show that \(\nu(d,u)=\nu(d,v^+)\) for all \(d\in D\). ◻

Proof. We proceed by induction on \(\lambda\). For \(\lambda = 0\), this is immediate. Now suppose that the claim holds for \(\lambda\), and suppose that \(u_2 \in B_{u_1}^{[\lambda+1]}\). Thus there is a sequence \((s_i)_{i\in \NN}\) of elements in \(B_{u_1}^{[\lambda]}\) such that \(\lim_{i\to \infty} s_{i} = u_2\). Reducing to a subsequence, we can assume that either

• \((s_i)_{i\in \NN}\) is strictly increasing and \(s_i < u_2\) for all \(i\in \NN\), or

• \((s_i)_{i\in \NN}\) is strictly decreasing and \(s_i > u_2\) for all \(i\in \NN\).

Both cases can be handled similarly. Let us assume the first case holds; that is \(s_i< u_2\) for all \(i\in \NN\). By the induction hypothesis, there is a sequence \((t_i)_{i \in \NN}\) of elements of \(B_{v_1}^{[\lambda]}\) such that \((u_1,s_i) \sim_{\es} (v_1,t_i)\) for each \(i\in \NN\).

Set \(v_2:=\lim_{d \to 0} f_{d,\mu(d,u)}(v_1)\) and \(v:=(v_1,v_2)\). By Lemma [lem:equiv2] we directly get \(u\sim_{\es} v\). Thus it is enough to show that \(\lim_{i \to \infty} t_i = v_2.\)

We first show that \(t_i < v_2\) for all \(i \in \NN\). Let \(i\in \NN\). Since \(s_i < u_2\), there is \(e \in D\) such that \[s_i < f_{e,\mu(e,u)}(u_1) \leq u_2.\] Thus \[f_{e,\mu(e,(u_1,s_i))}(u_1) < f_{e,\mu(e,u)}(u_1).\] Since \(u\sim_{\es} v\) and \((u_1,s_i)\sim_{\es} (v_1,t_i)\), then \[f_{e,\mu(e,(v_1,t_i))}(v_1) < f_{e,\mu(e,v)}(v_1).\] Thus \(t_i<v_2\).

Let \(\varepsilon >0\). We need to show that there is \(j\in \NN\) such that \(|v_2 - t_i|< \varepsilon\) for all \(i\in \NN_{\geq j}\). Let \(d \in D\) be such that \(d < \varepsilon\). Since \(v_2\) is the supremum of a subset of \(B_{v_1}\), there is \(e\in D\) such that \[v_2 - f_{e,\mu(e,v)}(v_1) < d.\] Since \(\lim_{i \to \infty} s_i=u_2\), there is \(j\in \NN\) such that for all \(i\in \NN_{\geq j}\) \[u_2 > s_i \geq f_{e,\mu(e,u)}(u_1).\] Thus for all \(i\in \NN_{\geq j}\) \[\mu(e,u)=\mu(e,(u_1,s_i)).\] Since \(u\sim_{\es} v\) and \((u_1,s_i)\sim_{\es} (v_1,t_i)\), we get that for all \(i\in \NN_{\geq j}\) \[\mu(e,v)=\mu(e,(v_1,t_i)).\] Thus \[t_i \geq f_{e,\mu(e,v)}(v_1).\] Hence \(|v_2 - t_i|<\varepsilon\). ◻

Proof of [prop:boundfortowers]. Let \(u_1,v_1\in C\). Towards a contradiction, suppose that \(\rank(B_{u_1})<\rank(B_{v_1})\). Let \(N := \rank(B_{u_1})\). Let \(\lambda\geq N\). Then \(B_{u_1}^{[\lambda]}=\emptyset\). Since \(\rank(B_{v_1})>N\), we have \(B_{v_1}^{[\lambda]}\neq \emptyset\). Let \(v_2 \in B_{v_1}^{[\lambda]}\). By [lem:typeimplies] there is \(u_2\in B_{u_1}^{[\lambda]}\). Contradiction. ◻

Proof of [thm:uniform2]. Let \(A\subseteq \RR^{n+1}\) be definable. Set \[A' := \{ (x,y) \in \RR^n \times \RR \;: \;y \in A_x \setminus \inter(A_x)\}.\] Note that \(A'_x\) has empty interior for all \(x \in \RR^n\). Thus, by [thm:mainthm] there is \(k\in \NN\) such that for all \(x\in \pi(A)\) \[\rank (\cl(A'_x))\leq k.\] Let \(x \in \RR^n\). Suppose that \(A_x\) has empty interior. By d-minimality, \(A_x\) is a finite union of discrete sets, and hence countable. It is enough to show that \(A_x\) is the union of \(k\) discrete sets. Since \(\inter(A_x)=\emptyset\), we have that \(A_x = A_x'\). Thus, \(\rank(\cl(A_x))\leq k\), and \(\cl(A_x)\) is the union of \(k\) discrete sets by [fact:RK]. Hence, since a subset of a discrete set is discrete, we conclude that \(A_x\) is the union of \(k\) discrete sets. ◻

Proof. Let \(\mathcal{G}_0\) be a grid decomposition of \(\RR^{n+1}\) compatible with the graph \(\Gamma(f)\). Set \[\mathcal{G} := \{ \pi(\es_0) \;: \;\es_0 \in \mathcal{G}_0\}.\] Let \(\es \in \mathcal{G}\), and let \(\es_0=(S_d)_{d \in D} \in \mathcal{G}_0\) be such that \(\pi(\es_0)=\es\). We now show that \(f|_{\surf(\es)}\) is continuous. Since all connected components of \(\surf(\es)\) are clopen in \(\surf(\es)\), it is enough to show that the restriction to a connected component is continuous.

Let \(C\subseteq \surf(\es)\) be a connected component of \(\surf(\es)\), and let \(d \in D\) be such that \(C\) is a connected component of \(\pi(S_d)\). Let \(f_1,\dots,f_m \in \mathcal{C}(C)\) witness that \(S_d\) is a stack over \(\pi(S_d)\). Since \(\surf(\es_0)\subseteq \Gamma(f)\), we know that \(m=1\). Thus \[\Gamma(f_1) = S_d \cap [C \times \RR] \subseteq \Gamma(f).\] Thus \(f|_C=f_1\), and hence continuous. ◻

Proof. For (1), observe that \(A\) is a finite disjoint union of surfaces of grids by [thm:mainthm], and thus a finite disjoint union of locally closed sets by [lem:proj-grid]. Hence \(A\) is constructible.

For (2), first observe that by [lem:proj-grid] and [fact:naive](2) \[\dim A \geq \max \{ w_1+\dots+w_n \;: \;A \text{ contains the surface of a (w_1,\dots w_n)-grid}\}.\] By [thm:mainthm], there are grids \(\es_1,\dots,\es_m\) in \(\RR^n\) such that \(A = \bigcup_{i=1}^m \surf(\es_i)\). By [fact:naive](4), \[\dim A = \max \{\dim(\surf(\es_1)),\dots,\dim(\surf(\es_m))\}.\] Thus (2) follows from [lem:proj-grid]. ◻

Proof. Let \(\es_1=(S_{1,d})_{d\in D},\dots,\es_m=(S_{m,d})_{d\in D}\) be grids in \(\RR^n\) such that \[A = \bigcup_{i=1}^m \surf(\es_i).\] For \(d \in D\), set \[A_d := \bigcup_{i=1}^m S_{i,d}.\] Since stacks have finitely many connected components, so does \(A_d\). We observe that \[\begin{align} \dim A &= \max \{ \dim \surf(\es_i) \;: \;i=1,\dots, m\}\\ &= \max \{ \dim S_d \;: \;i=1,\dots, m\}\\ &= \dim A_d. \end{align}\] Since grids are definable families, so is \((A_d)_{d\in D}\). ◻

Proof. We proceed by induction on \(n\). The case \(n=0\) is immediate. Let \(n>0\). It is clear that \(\mathcal{D}\) is partition. Thus it is left to show that \(\{ \pi(C) \;: \;C \in \mathcal{D}\}\) is a countable cell decomposition of \(\RR^{n-1}\). By definition, \(\{ \pi(\es) \;: \;\es \in \mathcal{G}\}\) is a grid decomposition of \(\RR^{n-1}\). By induction, \[\mathcal{D}':=\{ C'\subseteq \RR^{n-1} \;: \;\es \in \mathcal{G}, \;C' \text{ is a connected component of } \surf(\pi(\es))\}\] is a countable cell decomposition. It remains to argue that for \(C'\in \mathcal{D}'\) there is \(C\in \mathcal{D}\) with \(\pi(C)=C'\). Let \(C'\in \mathcal{D}'\) and let \(\es=(S_d)_{d\in D} \in \mathcal{G}\) be such that \(C'\) is a connected component of \(\surf(\pi(\es))\). By [lem:CC0-grid], there is \(d \in D\) such that \(C'\) is a connected component of \(\pi(S_d)\). By [lem:fin-CC] there is a connected component \(C\) of \(S_d\) such that \(\pi(C)=C'.\) Since \(C\) is a connected component of \(\surf(\es)\) by [lem:CC0-grid], we conclude that \(C\in \mathcal{D}\). ◻

Proof. Let \(\mathcal{G}\) be a grid decomposition compatible with \(\mathcal{A}\). Then \[\mathcal{D} := \{ C\subseteq \RR^n \;: \;\es \in \mathcal{G}, \;C \text{ is a connected component of } \surf(\es)\}\] is compatible with \(\mathcal{A}\), and by [lem:ccd] a countable cell decomposition. ◻

Proof. By (UCB)\(_{n}\), there is \(M\in \NN\) such that for all \(x\in \pi(A)\) \[\rank(\cl(A_x))\leq M.\] For each \(j=1,\ldots,M,\) define a set \[X_j := \{(x,y)\in \RR^{n-1}\times \RR: y\in A_x \cap \isol(\cl(A_x)^{[j]})\}.\] Observe that for all \(j,\ell \in \{1,\dots,M\}\)

1. \(A=\bigcup_{j=1}^{M} X_j\),

2. \(X_j\) is definable,

3. \(X_{j}\cap X_{\ell} = \emptyset\), whenever \(j\neq \ell\), and

4. \((X_j)_x\) is discrete for all \(x\in \pi(X_j)\).

By (PC)\(_{n}\), the set \(X_j\) is a finite disjoint union of surfaces of grids for \(j\in \{1,\dots,M\}\). Taking the union of all such surfaces, we conclude that \(A\) is a finite disjoint union of surfaces of grids. ◻

Proof. Let \(\mathcal{A}=\{A_1,\dots,A_k\}\) be a collection of definable subsets of \(\RR^n\) that partition \(\RR^n\). For each \(i=1,\ldots,k\) let \(\es_{i,1},\ldots, \es_{i,p(i)}\) be the grids such that \[\bigcup_{j=1}^{p(i)} \surf(\es_{i,j})=A_i.\]

By (GD)\(_{n-1}\), there is a grid decomposition \(\G\) of \(\RR^{n-1}\) compatible with \[\{ \surf(\pi(\es_{i,j}))\;: \;i\in \{1,\ldots,k\}, j\in \{1,\ldots,p(i)\}\}.\]

Let \(i\in \{1,\ldots,k\}\) and \(j\in \{1,\ldots,p(i)\}\). For ease of notation, we write \(\es=(S_d)_{d\in D}\) for \(\es_{i,j}\). Now let \(\es'= (S'_d)_{d\in D}\in \G\) be such that \[\surf(\es') \subseteq \surf(\pi(\es)).\] Let \(w=(w_1,\ldots,w_n)\in \{0,1\}^n\) be such that \(\es\) is a \(w\)-grid, and let \(w'=(w'_1,\ldots,w'_{n-1})\in \{0,1\}^{n-1}\) be such that \(\es'\) is a \(w'\)-grid. Set \[D':=\{d\in D: \pi(S_d) \cap S_d' \neq \emptyset \}\] and for \(d\in D'\), set \[S^*_d:=S_d \cap [S_d' \times \RR].\] We claim that \((S^*_d)_{d\in D'}\) is a \((w_1',\ldots,w_{n-1}',w_n)\)-grid in \(\RR^n\).
Let \(d\in D\). We first show that \(S_d^*\) is \((w_1',\ldots,w_{n-1}',w_n)\)-stack. By [lem:fin-CC] and [lem:restrict-stack], it is enough to prove that \(\pi(S^*_d)\) is a substack of \(S_d'\). By [lem:substack-CC] it is sufficient to establish that every connected component of \(\pi(S^*_d)\) is a connected component of \(S_d'\).
Let \(C\subseteq \RR^{n-1}\) be a connected component of \(\pi(S^*_d)=\pi(S_d)\cap S_d'\). Then there is a connected component \(C'\) of \(S_d'\) containing \(C\). Since \(\surf(\es')\subseteq \surf(\pi(\es))\), there is a connected component \(C''\) of \(\surf(\pi(\es))\) containing \(C'\). By [lem:CC0-grid] there is \(e\in D\) such that \(C''\) is a connected component of \(\pi(S_e)\). If \(d\leq e\), then \(C''\) is a connected component of \(\pi(S_d)\) as well. On the other hand, if \(e<d\), then either \(C''\) is a connected component of \(\pi(S_d)\) or \(C''\) is disjoint from \(\pi(S_d)\). Since \(C''\) is not disjoint from \(\pi(S_d)\), we must have that \(C''\subseteq \pi(S_d)\). Therefore \(C'\subseteq \pi(S_d)\) as well. Thus \(C'\subseteq \pi(S_d)\cap S_d'\) and so \(C'=C\). Hence \(C\) is a connected component of \(S'_d\).
Let \(d,e\in D'\) with \(e<d\). We now show that \(S_d^*\) is a substack of \(S_e^*\). By [lem:substack2], it is enough to show that \(\pi(S_d^*)\) is a substack of \(\pi(S_e^*)\). Let \(C\) be a connected component of \(\pi(S_d^*)\). Suppose towards a contradiction that \(C\) is not a connected component of \(\pi(S_e^*)\). Thus \(C\) is strictly contained in a connected component \(C'\) of \(\pi(S_e^*)\). By the above, the set \(C'\) is also a connected component of \(S_e'\). Since \(C\) is a connected component of \(\pi(S_d^*)\) it is a connected component of \(S_d'\), and thus a connected component of \(S_e'\). Thus \(C=C'\), a contradiction.
By [lem:subgrid-CC] we have that \((\pi(S_d*))_{d\in D'}\) is a \((w_1',\ldots,w_{n-1}')\)-grid. Lastly, suppose that that \(w_n=0\). Since \(\bigcup_{d\in D'} S_d^* \subseteq \bigcup_{d\in D} S_d\) and the latter being étale, we conclude that \(\bigcup_{d\in D'} S_d^*\) is étale as well. Therefore \((S_d^*)_{d\in D'}\) is a grid.
Observe that \(\surf((S_d^*)_{d\in D'})=\surf(S)\cap [\surf(S')\times \RR]\), and thus the collection of all such grids has the desired properties. ◻

Proof. Let \[Y:= \{(x,y)\in A: A \text{ is not \'etale at } (x,y)\},\] and note that \(Y\) is definable. Suppose towards a contradiction that \(\pi(Y)\) is somewhere dense in \(\pi(A)\). Then \(\pi(Y)\) has interior in \(\pi(A)\) by [cor:boundary](1). Applying (GD)\(_{n}\), we obtain a grid decomposition of \(\RR^n\) compatible with \(\pi(A)\) and \(\pi(Y)\). Therefore, we may write \(\pi(Y)\) as a finite union of surfaces of grids. Since \(\pi(Y)\) has interior in \(\pi(A)\), by Lemma [lem:grid-inter] there is a grid \(\es\) in \(\RR^n\) whose surface is contained in \(\pi(Y)\) and has interior in \(\pi(A)\). Let \(f : \surf(\es) \to \RR\) be the definable function which maps \(x \in \surf(\es)\) to the minimum of \(Y_x\). Since \(Y_x\) is finite for all \(x \in \pi(A)\), this function \(f\) is well-defined. By [cor:gridcont], there is a dense, definable, open set \(Z\subseteq \surf(\es)\) such that \(f|_Z\) is continuous. Applying (GD)\(_{n}\) again, we obtain a grid decomposition of \(\RR^n\) compatible with \(Z\) and \(\pi(A)\). Since \(Z\) has interior in \(\pi(A)\), there is a grid \(\es'\) by Lemma [lem:grid-inter] such that \(\surf(\es')\subseteq Z\) and \(\surf(\es')\) has interior in \(\pi(A)\).

Now define sets \(A'_0, A'_a, A'_b\), and \(A'_{ab}\) as follows: \[\begin{align} A'_0 &:= \{x \in \surf(\es') \;: \;A_x=\{f(x)\} \},\\ A'_a &:= \{x\in \surf(\es') : \forall y\in A_x \;y\geq f(x)\}, \\ A'_b &:= \{x\in \surf(\es'): \forall y\in A_x \; y \leq f(x)\},\\ A'_{ab} &:= \{x\in \surf(\es') : \exists y_1, y_2 \in A_x \;(y_1<f(x)) \land (y_2>f(x))\}. \end{align}\] Each of these sets is definable and \[\surf(\es')=A'_0\cup A'_a \cup A'_b \cup A'_{ab}.\] Using that \(\surf(\es')\) has interior in \(\pi(A)\), apply (GD)\(_{n}\) and Lemma [lem:grid-inter] in the same way as before to obtain a grid \(\es''\) whose surface is contained in one of \(A'_0,A'_a, A'_b,\) or \(A'_{ab}\) and has interior in \(\pi(A)\). We will only consider the case when \(\surf(\es'')\subseteq S_{ab}\), as the other cases follow similarly.
Let \(f_a,f_b : \surf(\es'')\to \RR\) be definable function such that for all \(x\in \surf(\es'')\) \[\begin{align} f_a(x) &:= \min{\{y':y'>\min{Y_x}\}}),\\ f_b(x) &:= \max{\{y':y'<\min{Y_x}\}}). \end{align}\] By [cor:gridcont], there are dense, definable, open sets \(U_a, U_b \subseteq \surf(\es'')\) such that the restriction \(f_a|_{U_a}\) and \(f_b|_{U_b}\) are continuous. Set \(U:=U_a \cap U_b\) and observe that \(U\) is a dense, definable, and open subset of \(\surf(\es'')\) on which \(f,f_a\) and \(f_b\) are continuous.

Let \(x\in U\). We claim that \(A\) is étale at \((x,f(x))\). By construction, the graphs of \(\Gamma(f|_U), \Gamma(f_a|_U)\), and \(\Gamma(f_b|_U)\) do not intersect. Therefore, there is an open box \(B\subseteq \RR^{n+1}\) centered at \((x,f(x))\) such that \[A\cap B\subseteq \Gamma(f)\cap [U \times \RR].\] Thus \(\pi\) maps \(A\cap B\) homeomorphically onto an open set, and \(A\) is étale at \((x,f(x))\). However, since \((x,f(x))\in Y\), this is a contradiction. ◻