# Analytic finiteness for preparation

*Written by Claude Opus 5.5 (Anthropic), October 2026; self-checked by the writing AI. Original text: public domain (CC0). The theorems and the route of proof are those of Guillaume Valette, [On subanalytic geometry](https://arxiv.org/abs/2507.23622v1), arXiv:2507.23622v1, Chapter 1 and §2.3. Global subanalytic sets are defined by the projective product convention of Jean-Marie Lion and Jean-Philippe Rolin, [Théorème de préparation pour les fonctions logarithmico-exponentielles](https://www.numdam.org/item/10.5802/aif.1583.pdf).*

This reading derives the finiteness theorems of global subanalytic geometry from Weierstrass division. Division makes the ring of convergent real power series Noetherian, and the Artin–Rees lemma and Krull's intersection theorem follow. A linear approximation theorem then shows that a real analytic function of \((u,z)\) is controlled, near a point, by finitely many of its Taylor coefficients in \(z\):

\[
\psi(u,z)=\sum_{i=0}^{d}c_i(u)(z-z_0)^iA_i(u,z),
\qquad A_i(u_0,z_0)=1,
\]

with analytic functions \(A_i\). Dividing by the largest coefficient produces a family whose order in \(z\) is bounded uniformly. This is the analytic input to the preparation theorem: on the cells of a finite decomposition, every function built from restricted analytic functions and rational powers becomes a base coefficient times a power of a distance \(|z-\theta(x)|\) times an analytic unit. The cell decomposition theorem for globally subanalytic sets, Gabrielov's complement theorem, convergent Puiseux expansions and the dimension theory of the last section are consequences.

## Real analytic division and finite generation

Let \(R_n=\mathbb R\{x_1,\ldots,x_n\}\) be the ring of germs at \(0\in\mathbb R^n\) of real analytic functions, that is, of real power series with a positive radius of convergence, and let \(\mathfrak m_n\subset R_n\) consist of the germs vanishing at \(0\). Translation transfers every statement below to germs at an arbitrary point.

The ring \(R_n\) is local with maximal ideal \(\mathfrak m_n\). A germ with \(f(0)\ne0\) has an analytic reciprocal near \(0\), and a germ with \(f(0)=0\) has no inverse. The ideal \(\mathfrak m_n\) is generated by \(x_1,\ldots,x_n\). To see this, give each monomial \(x^\alpha\ne1\) of a series without constant term to the smallest index \(i\) with \(\alpha_i>0\), and write it as \(x_i x^{\alpha-e_i}\). The monomials given to \(i\) form \(x_ig_i\), where \(g_i\) is a subseries of an absolutely convergent series, so convergent on the same polydisc.

The complex preparation and division theorems are proved in [Preparation by the zeros in one fibre](../weierstrass-preparation-and-division.html#preparation-by-the-zeros-in-one-fibre) and [Division, finite generation and elementary algebra](../weierstrass-preparation-and-division.html#division-finite-generation-and-elementary-algebra), Theorem 1 and Corollary 3 there. They descend to real germs because the factors are unique. Call \(g\in R_n\) regular of order \(s\) in \(x_n\) if \(g(0,\ldots,0,x_n)=x_n^s\,h(x_n)\) with \(h(0)\ne0\). Regard \(g\) as a complex germ and write \(g=uP\), with \(u\) a unit and \(P=x_n^s+\sum_{k=1}^sa_k(x')x_n^{s-k}\) a Weierstrass polynomial. Replacing every coefficient of the power series of \(u\) and \(P\) by its complex conjugate gives a second such factorization of the real germ \(g\). By uniqueness the two coincide, so \(u\) and \(P\) have real coefficients. In the same way, for real \(f\) the unique division

\[
f=qP+\sum_{j=0}^{s-1}r_j(x')x_n^j
\]

has real \(q\) and real \(r_j\in R_{n-1}\). Thus a real germ regular of order \(s\) in \(x_n\) is a real unit times a real Weierstrass polynomial of degree \(s\), and every \(f\in R_n\) is uniquely \(qP\) plus a polynomial of degree less than \(s\) in \(x_n\) with coefficients in \(R_{n-1}\).

Regularity can be produced by a real linear change of coordinates. A nonzero real polynomial \(F\) does not vanish on all of \(\mathbb R^n\). Indeed, view \(F\) as a polynomial in \(x_n\) whose coefficients are polynomials in the other variables. By induction one of these coefficients is nonzero at some real point, and there the resulting nonzero polynomial in \(x_n\) has a non-root. If \(g\ne0\) has order \(m\) and lowest homogeneous part \(g_m\), choose \(v\in\mathbb R^n\) with \(g_m(v)\ne0\) and a linear isomorphism \(L\) with \(L(e_n)=v\). Then \(g(L(0,\ldots,0,t))=g(tv)=t^mg_m(v)+O(t^{m+1})\), so \(g\circ L\) is regular of order \(m\) in \(x_n\). Applying this to a product of finitely many germs makes all of them regular at once.

We use the module fact established in the proof of Theorem 6 of the division reading: over a Noetherian ring every submodule of a finite free module is finitely generated.

**Noetherianity of real convergent germs.** The ring \(R_n\) is Noetherian.

**Proof.** We induct on \(n\); the field \(R_0=\mathbb R\) is Noetherian. Let \(J\subset R_n\) be a nonzero ideal. A linear change of coordinates is a ring automorphism, so we may assume that \(J\) contains a germ \(g\) regular of some order \(s\) in \(x_n\). Write \(g=uP\) as above. Then \(P=u^{-1}g\in J\). If \(s=0\), \(P=1\) and \(J=R_n\). Otherwise divide every \(f\in J\) by this one polynomial: \(f=qP+r(f)\). The remainder \(r(f)=f-qP\) lies in \(J\), and by uniqueness of division the map \(f\mapsto r(f)\) is \(R_{n-1}\)-linear. Its image is an \(R_{n-1}\)-submodule of the polynomials of degree less than \(s\), a free module of rank \(s\). By induction and the module fact it is generated by finitely many remainders \(r(f_1),\ldots,r(f_k)\). Every \(f\in J\) is then \(qP\) plus an \(R_{n-1}\)-combination of these, so \(J=(P,r(f_1),\ldots,r(f_k))\). \(\square\)

## Polynomial rings and stable filtrations

**Hilbert basis theorem.** If \(A\) is a Noetherian ring, so is \(A[T]\).

**Proof.** Let \(J\subset A[T]\) be an ideal. The leading coefficients of the elements of \(J\), together with \(0\), form an ideal of \(A\): multiplying by a power of \(T\) aligns degrees before adding. Pick finitely many elements \(f_1,\ldots,f_k\) of \(J\) such that their leading coefficients generate that ideal, and let \(D\) be the largest of their degrees. If \(f\in J\) has degree \(N\ge D\) and leading coefficient \(\sum c_i\operatorname{lc}(f_i)\), then \(f-\sum c_iT^{N-\deg f_i}f_i\) lies in \(J\) and has smaller degree. Repeating, every element of \(J\) is a combination of the \(f_i\) plus an element of \(J\) of degree less than \(D\). Those elements form a submodule of \(A^D\), finitely generated by the module fact. So \(J\) is finitely generated. By induction, polynomial rings in finitely many variables over \(A\) are Noetherian, and so are their quotients, since ideals of a quotient pull back to ideals. \(\square\)

Let \(A\) be Noetherian, \(I\subset A\) an ideal and \(M\) a finitely generated \(A\)-module. A decreasing sequence of submodules \(M=M_0\supset M_1\supset\cdots\) is an **\(I\)-filtration** if \(IM_i\subset M_{i+1}\) for all \(i\). It is **stable** if \(IM_i=M_{i+1}\) for all large \(i\). The Rees algebra and the Rees module of the filtration are

\[
A^\ast=\bigoplus_{i\ge0}I^iT^i\subset A[T],\qquad
M^\ast=\bigoplus_{i\ge0}M_iT^i,
\]

with \(I^0=A\). Since \(I^jM_i\subset M_{i+j}\), \(M^\ast\) is a module over \(A^\ast\). If \(a_1,\ldots,a_\ell\) generate \(I\), then \(Z_j\mapsto a_jT\) maps \(A[Z_1,\ldots,Z_\ell]\) onto \(A^\ast\). By the Hilbert basis theorem \(A^\ast\) is Noetherian.

**Stability criterion.** The module \(M^\ast\) is finitely generated over \(A^\ast\) if and only if the filtration is stable.

**Proof.** Suppose \(IM_i=M_{i+1}\) for \(i\ge e\). Then \(M_{e+j}=I^jM_e\) for all \(j\ge0\). Each \(M_i\) is a submodule of the finitely generated module \(M\), hence finitely generated, and generators of \(M_0,\ldots,M_e\), placed in their degrees, generate \(M^\ast\). Conversely, a finite generating set may be replaced by the homogeneous components of its elements. Let these be \(m_kT^{e_k}\) with \(m_k\in M_{e_k}\), and \(e=\max e_k\). For \(i\ge e\), every element of \(M_{i+1}\) is a sum \(\sum_ka_km_k\) with \(a_k\in I^{i+1-e_k}=I\cdot I^{i-e_k}\). Since \(I^{i-e_k}m_k\subset M_i\), it lies in \(IM_i\). So \(M_{i+1}\subset IM_i\), and the reverse inclusion holds for every filtration. \(\square\)

**Artin–Rees lemma.** For every submodule \(N\subset M\) there is an \(e\) with

\[
N\cap I^{i+1}M=I\,(N\cap I^iM)\qquad(i\ge e),
\]

and consequently \(N\cap I^iM=I^{i-e}(N\cap I^eM)\) for \(i\ge e\).

**Proof.** The filtration \((I^iM)_i\) is stable, so its Rees module is finitely generated over the Noetherian ring \(A^\ast\). The sets \(N\cap I^iM\) form an \(I\)-filtration of \(N\), whose Rees module is an \(A^\ast\)-submodule of the former. It is therefore finitely generated, and the stability criterion gives the assertion. \(\square\)

The lemma holds for every ideal of a Noetherian ring. Below, \(I\) will be the maximal ideal of a local Noetherian ring.

## Krull intersection for modules

Let \(A\) be a Noetherian local ring with maximal ideal \(\mathfrak m\). Its non-units are exactly the elements of \(\mathfrak m\).

**Nakayama's lemma.** If a finitely generated module \(L\) satisfies \(L=\mathfrak mL\), then \(L=0\).

**Proof.** Suppose \(L\ne0\) and let \(e_1,\ldots,e_p\), \(p\ge1\), be a generating set of minimal size. Write \(e_p=\sum_jb_je_j\) with all \(b_j\in\mathfrak m\). Then \((1-b_p)e_p=\sum_{j<p}b_je_j\). Since \(1-b_p\notin\mathfrak m\), it is a unit. So \(e_p\) lies in the span of \(e_1,\ldots,e_{p-1}\), contradicting minimality; for \(p=1\) it gives \(e_1=0\) and \(L=0\). \(\square\)

**Krull intersection theorem.** If \(M\) is finitely generated, then \(\bigcap_{i\ge0}\mathfrak m^iM=0\).

**Proof.** The intersection \(L\) of all \(\mathfrak m^iM\) is a submodule of \(M\). For \(i\) large, the Artin–Rees lemma with \(N=L\) gives \(L=L\cap\mathfrak m^{i+1}M=\mathfrak m(L\cap\mathfrak m^iM)=\mathfrak mL\). Nakayama's lemma gives \(L=0\). \(\square\)

Applied to \(M/N\), whose submodules \(\mathfrak m^i(M/N)\) are the images of \(\mathfrak m^iM\), the theorem says that every submodule of a finitely generated module is closed in the \(\mathfrak m\)-adic sense:

\[
\bigcap_{i\ge0}\left(N+\mathfrak m^iM\right)=N.
\]

For an ideal \(J\subset A\) this reads \(\bigcap_i(J+\mathfrak m^i)=J\). Locality cannot be dropped. In \(\mathbb R[x]\), with \(J=(x-1)\) and \(I=(x)\), the identity \(1=x^i-(x^i-1)\) shows \(J+I^i=\mathbb R[x]\) for every \(i\ge1\), although \(J\) is a proper ideal.

## Formal linear solutions have convergent replacements

Let \(A=R_n\), let \(\widehat A=\mathbb R[[x_1,\ldots,x_n]]\) be the ring of formal power series, and let \(H\) be a \(b\times a\) matrix with entries in \(A\). It defines linear maps \(A^a\to A^b\) and \(\widehat A^a\to\widehat A^b\).

**Linear approximation theorem.** Let \(v\in A^b\), and suppose \(H\widehat f=v\) for some \(\widehat f\in\widehat A^a\). Then \(Hf=v\) for some \(f\in A^a\). For every \(q\ge0\) the solution \(f\) can be chosen to agree with \(\widehat f\) in all Taylor coefficients of degree less than \(q\).

**Proof.** Let \(p_k\) be the vector of polynomials obtained by truncating each component of \(\widehat f\) after degree \(k\). Then \(v-Hp_k=H(\widehat f-p_k)\) has no terms of degree at most \(k\), because multiplication by a power series cannot lower the order. It is also a vector of convergent series. A convergent series without terms of degree at most \(k\) belongs to \(\mathfrak m^{k+1}\). To see this, assign each of its monomials to one of the finitely many monomials of degree \(k+1\) dividing it, and collect: the series becomes a finite sum of such monomials times convergent subseries. Hence \(v\in H(A^a)+\mathfrak m^{k+1}A^b\) for every \(k\), and the closedness of submodules gives \(v\in H(A^a)\).

For the second statement let \(q\ge1\) and let \(p\) be the truncation of \(\widehat f\) below degree \(q\). Grouping as before, \(\widehat f=p+\sum_{|\alpha|=q}x^\alpha\widehat g_\alpha\) with formal vectors \(\widehat g_\alpha\). So the linear system \(\sum_{|\alpha|=q}x^\alpha Hg_\alpha=v-Hp\), in the unknowns \(g_\alpha\in A^a\), has a formal solution. By the first part it has a convergent solution \(g_\alpha\), and \(f=p+\sum x^\alpha g_\alpha\) solves \(Hf=v\) with the required low-degree coefficients. \(\square\)

The theorem replaces a formal solution by a convergent one. It says nothing about convergence of the given formal solution, and it concerns linear systems only.

## A finite Taylor family with analytic units

**Finite Taylor family.** Let \(\psi\) be real analytic near \((u_0,z_0)\in\mathbb R^{n-1}\times\mathbb R\), and let \(c_j(u)=\frac1{j!}\partial_z^j\psi(u,z_0)\) be its Taylor coefficients in \(z\). There are \(d\ge0\) and analytic germs \(A_0,\ldots,A_d\) at \((u_0,z_0)\) with \(A_i(u_0,z_0)=1\) such that, near \((u_0,z_0)\),

\[
\psi(u,z)=\sum_{i=0}^{d}c_i(u)(z-z_0)^iA_i(u,z).
\]

**Proof.** Translate so that \((u_0,z_0)=(0,0)\); then \(\psi=\sum_{j\ge0}c_j(u)z^j\) with \(c_j\in R_{n-1}\). The ideals \((c_0,\ldots,c_k)\) increase with \(k\), so by Noetherianity they are constant from some \(d\) on. For \(j>d\) choose \(b_{ij}\in R_{n-1}\) with \(c_j=\sum_{i\le d}b_{ij}c_i\). In \(\mathbb R[[u,z]]\) put \(\widehat F_i=\sum_{j>d}b_{ij}(u)z^{j-i-1}\). A given monomial \(u^\beta z^\gamma\) receives a contribution only from \(j=\gamma+i+1\), so \(\widehat F_i\) is a well-defined formal series, and

\[
\sum_{i=0}^dc_iz^{i+1}\widehat F_i=\sum_{j>d}\Bigl(\sum_{i\le d}b_{ij}c_i\Bigr)z^j=\psi-\sum_{i=0}^dc_iz^i.
\]

The linear equation \(\sum_{i\le d}c_iz^{i+1}F_i=\psi-\sum_{i\le d}c_iz^i\) in the unknowns \(F_0,\ldots,F_d\in R_n\) therefore has a formal solution. By the linear approximation theorem it has a convergent solution \(G_0,\ldots,G_d\). Then \(A_i=1+zG_i\) satisfies \(A_i(0,0)=1\) and gives the identity. If \(\psi=0\), take \(d=0\) and \(A_0=1\). \(\square\)

The infinitely many \(b_{ij}\) need not admit a common convergence estimate, so the formal series \(\widehat F_i\) may diverge. The linear approximation theorem is what produces analytic \(A_i\). Near \((u_0,z_0)\) all \(A_i\) are nonzero.

## Signed normalization and a uniform order bound

Suppose that near \((u_0,z_0)\)

\[
\psi(u,z)=\sum_{i=0}^dc_i(u)(z-z_0)^iA_i(u,z),\qquad A_i(u_0,z_0)\ne0.
\]

For \(0\le m\le d\) let \(K_m=\{t\in[-1,1]^{d+1}:t_m=1\}\) and

\[
Q(t,u,z)=\sum_{i=0}^dt_i(z-z_0)^iA_i(u,z).
\]

**Uniform regularity.** There are a neighbourhood \(V\) of \((u_0,z_0)\) and \(\delta>0\) such that

\[
\max_{0\le k\le m}\bigl|\partial_z^kQ(t,u,z)\bigr|\ge\delta\qquad(t\in K_m,\ (u,z)\in V).
\]

Consequently, for \(t\in K_m\) and \((u,z')\in V\), the function \(z\mapsto Q(t,u,z)\) has order at most \(m\) at \(z'\).

**Proof.** At \((u_0,z_0)\) the \(k\)-th \(z\)-derivative of \((z-z_0)^iA_i\) vanishes for \(k<i\) and equals \(i!A_i(u_0,z_0)\ne0\) for \(k=i\). Hence \(\partial_z^kQ(t,u_0,z_0)\), \(k\le m\), depends only on \(t_0,\ldots,t_k\), through a triangular linear system with nonzero diagonal. As \(t_m=1\), these \(m+1\) derivatives are not all zero. Their maximum is a positive continuous function on the compact set \(K_m\), with a minimum \(2\delta>0\). The derivatives are uniformly continuous on \(K_m\) times a closed box around \((u_0,z_0)\), which gives a neighbourhood \(V\) on which the maximum stays at least \(\delta\). \(\square\)

Here is how the bound is used. Let \(\phi\) be a map from a set \(B\) to the parameter space and put \(g_i=c_i\circ\phi\). On a part of \(B\) where \(g_m\ne0\) and \(|g_i|\le|g_m|\) for every \(i\),

\[
\psi(\phi(x),z)=g_m(x)\,Q\Bigl(\frac{g_0(x)}{g_m(x)},\ldots,\frac{g_d(x)}{g_m(x)},\phi(x),z\Bigr),
\]

and the normalized vector lies in \(K_m\) whatever the signs of the \(g_i\). Where all \(g_i\) vanish, \(\psi(\phi(x),z)=0\). Arranging the \(g_i\) into pieces of this kind is part of the cell induction below and is not assumed here. If \(\psi\) is analytic near a compact set of parameters, finitely many of the neighbourhoods above cover it, and the induction refines their preimages compatibly.

## How this supplies the geometric argument

The analytic chain proved so far is: real division gives Noetherianity of convergent germs; the Rees algebra gives the Artin–Rees lemma; Artin–Rees and Nakayama give Krull's theorem and the closedness of submodules; closedness gives convergent solutions of formal linear systems; these give the finite Taylor family, whose normalization has a uniformly bounded order. The remaining sections build the cell induction on this input. They prove the preparation of functions built from restricted analytic functions and rational powers, the cell theorem, the complement theorem and convergent Puiseux expansions under the projective product definition. The [curve-selection and Łojasiewicz reading](../curve-selection-and-lojasiewicz.html) uses the cell theorem and the Puiseux expansions proved here.

## A convergent split before restricting to a hyperbola

On the hyperbola \(c=zw\) a monomial \(z^iw^j\) can be written as a power of \(c\) times a pure power of \(z\) or of \(w\), according to whether \(i>j\) or \(i\le j\). Summing these rewrites gives a decomposition of an analytic function of \((u,z,w)\) into two functions, each of which depends on only one of the coordinates \(z,w\) besides \(c\).

**Splitting lemma.** Let \(\psi(u,z,w)\) be real analytic near \((a,0,0)\), with \(u\in\mathbb R^k\). There are \(R>0\) and functions \(\psi_+(u,c,z)\), \(\psi_-(u,c,w)\), analytic for \(|u_\nu-a_\nu|<R\), \(|c|<R^2\) and \(|z|,|w|<R\), such that

\[
\psi(u,z,w)=z\,\psi_+(u,zw,z)+\psi_-(u,zw,w)
\qquad(|u_\nu-a_\nu|<R,\ |z|,|w|<R).
\]

**Proof.** The power series of \(\psi\) at \((a,0,0)\) defines a holomorphic function on a complex polydisc around that point. Shrinking, we may take a common radius \(2R\) for that polydisc with \(|\psi|\le M\) on the closed polydisc of radius \(R\). Write

\[
\psi(u,z,w)=\sum_{\alpha\in\mathbb N^k}\ \sum_{i,j\ge0}b_{\alpha ij}(u-a)^\alpha z^iw^j .
\]

Cauchy's estimate in each variable gives \(|b_{\alpha ij}|\le MR^{-|\alpha|-i-j}\). Define

\[
\psi_+(u,c,z)=\sum_\alpha\sum_{i>j}b_{\alpha ij}(u-a)^\alpha c^j z^{i-j-1},\qquad
\psi_-(u,c,w)=\sum_\alpha\sum_{i\le j}b_{\alpha ij}(u-a)^\alpha c^i w^{j-i}.
\]

In the first sum, \((\alpha,i,j)\mapsto(\alpha,j,i-j-1)\) is a bijection onto \(\mathbb N^k\times\mathbb N\times\mathbb N\), and the absolute value of a term is at most

\[
\frac MR\prod_\nu\Bigl(\frac{|u_\nu-a_\nu|}R\Bigr)^{\alpha_\nu}\Bigl(\frac{|c|}{R^2}\Bigr)^j\Bigl(\frac{|z|}R\Bigr)^{i-j-1}.
\]

So the sum converges absolutely, uniformly on smaller closed polydiscs, and defines an analytic function in the stated range. The second sum is treated in the same way with the bijection \((\alpha,i,j)\mapsto(\alpha,i,j-i)\) and the bound \(M\prod_\nu(|u_\nu-a_\nu|/R)^{\alpha_\nu}(|c|/R^2)^i(|w|/R)^{j-i}\). If \(|z|,|w|<R\), then \(c=zw\) satisfies \(|c|<R^2\). Substituting it, the terms of \(z\psi_+\) become \(b_{\alpha ij}(u-a)^\alpha z^iw^j\) for \(i>j\), and those of \(\psi_-\) become the same expression for \(i\le j\). All series converge absolutely, so they can be recombined, which gives \(\psi\). \(\square\)

For example, \(\psi=1+z^2+zw+w^2\) gives \(\psi_+=z\) and \(\psi_-=1+c+w^2\): the term \(z^2\) has \(i>j\), while \(1\), \(zw\) and \(w^2\) have \(i\le j\). The identity holds for all signs of \(z\) and \(w\), and also at \(z=0\). Where \(z\ne0\), substituting \(w=c/z\) gives

\[
\psi(u,z,c/z)=z\,\psi_+(u,c,z)+\psi_-(u,c,c/z).
\]

## Localize on the actual image, rather than a larger product

Throughout the cell arguments, points of \(\mathbb R^n\) are written \((x,z)\), with \(x\in\mathbb R^{n-1}\) the base variables and \(z\) the last coordinate. Let \(C\) be a set of points with \(z\ne0\). Let \(\phi\) and \(c\) be functions of \(x\), with values in \(\mathbb R^k\) and \(\mathbb R\), such that

\[
\Phi(x,z)=\bigl(\phi(x),z,c(x)/z\bigr)
\]

is bounded on \(C\). Let \(\psi\) be analytic on an open neighbourhood \(U\) of the compact set \(K=\overline{\Phi(C)}\). Write \(w=c/z\), so that \(c=zw\) on \(C\).

**Localization lemma.** There are finitely many open sets \(V_1,\ldots,V_N\) covering \(K\) such that, on each \(C\cap\Phi^{-1}(V_l)\), the function \(\psi\circ\Phi\) has one of the following forms.

1. \(F(\phi,c,z)\), where \(F(u,c,z)=\psi(u,z,c/z)\) is analytic near a closed box containing all values of \((\phi,c,z)\) on the piece, and \(|z|\) is bounded below by a positive constant there.
2. \(F(\phi,c,w)\), where \(F(u,c,w)=\psi(u,c/w,w)\) is analytic near a closed box containing all values of \((\phi,c,w)\) on the piece, and \(|w|\) is bounded below by a positive constant there.
3. \(z\,\psi_+(\phi,c,z)+\psi_-(\phi,c,w)\), with the functions of the splitting lemma at a point \((a,0,0)\in K\), all values of \((\phi,c,z)\) and \((\phi,c,w)\) lying in closed boxes inside their domains of analyticity.

**Proof.** Let \(p=(a,z_0,w_0)\in K\). Suppose first \(z_0\ne0\). The map \((u,c,z)\mapsto(u,z,c/z)\) is analytic where \(z\ne0\) and sends \((a,z_0w_0,z_0)\) to \(p\in U\), so \(F(u,c,z)=\psi(u,z,c/z)\) is analytic near \((a,z_0w_0,z_0)\). Choose a closed box around this point inside the domain of \(F\), on which \(|z|\ge|z_0|/2\). The continuous map \((u,z,w)\mapsto(u,zw,z)\) sends \(p\) into the interior of the box, and therefore also some neighbourhood \(V_p\) of \(p\). On \(C\cap\Phi^{-1}(V_p)\) we have \(\psi(\phi,z,c/z)=F(\phi,c,z)\), and form 1 holds. If \(w_0\ne0\), the same argument with \((u,c,w)\mapsto(u,c/w,w)\) gives form 2. If \(z_0=w_0=0\), then \(p=(a,0,0)\) lies in \(K\subset U\), so \(\psi\) is analytic near \(p\). The splitting lemma applies, and a small \(V_p\) keeps \((u,zw,z)\) and \((u,zw,w)\) inside closed boxes in the domains of \(\psi_+\) and \(\psi_-\). This gives form 3. Compactness of \(K\) selects finitely many \(V_p\). \(\square\)

Only actual points of \(K\) are used. The function \(\psi\) need not be analytic at \((a,0,0)\) for every \(a\) in the closure of \(\phi(C)\), and no analytic extension of \(\psi\) beyond \(U\) is required.

## The signed balanced-coordinate calculation

In the remaining case of the two-coordinate argument below, both summands in form 3 are comparable, and the coordinate \(z\) is comparable to a function of the base. The following lemma turns the function into an analytic composition in a single new coordinate.

**Balanced substitution lemma.** Keep the notation of the localization lemma, and assume moreover \(c\ne0\) on \(C\). Let \(b>0\) be a function of \(x\) with \(\kappa^{-1}b\le|z|\le\kappa b\) on \(C\), for a constant \(\kappa\ge1\). Put \(q=z/b\) and \(d=c/b\). Then the map \(\Gamma=(\phi,b,d,q)\) is bounded on \(C\) and \(|q|\ge\kappa^{-1}\) there. The function

\[
\widetilde\psi(u,b,d,q)=\psi(u,bq,d/q)
\]

is analytic on a neighbourhood of \(\overline{\Gamma(C)}\), and \(\psi\circ\Phi=\widetilde\psi\circ\Gamma\) on \(C\).

**Proof.** Since \(z\) is bounded, so is \(b\le\kappa|z|\). Next, \(|q|\in[\kappa^{-1},\kappa]\) and \(d=(c/z)q=wq\) is bounded. On the closure of \(\Gamma(C)\) we still have \(|q|\ge\kappa^{-1}\), so the map \(\Lambda(u,b,d,q)=(u,bq,d/q)\) is analytic near that closure. It sends \(\Gamma(C)\) onto \(\Phi(C)\), because \(bq=z\) and \(d/q=c/z\). By continuity it sends \(\overline{\Gamma(C)}\) into \(K\subset U\). The open set \(\Lambda^{-1}(U)\) therefore contains \(\overline{\Gamma(C)}\), and \(\widetilde\psi=\psi\circ\Lambda\) is analytic there. \(\square\)

No sign of \(z\), \(c\) or of the two summands has been assumed. The choice of \(b\) is made in the proof of two-coordinate preparation below.

## Analytic units and the lower-dimensional induction

The preparation theorem is proved for a class of functions closed under the operations it needs. A **restricted analytic function** is a function on a closed cube \(Q\subset\mathbb R^m\) that extends analytically to a neighbourhood of \(Q\). An **\(\mathcal L\)-function** on a set \(E\subset\mathbb R^m\) is a function given on \(E\) by a finite expression built from the coordinate functions and real constants with sums, products, rational powers \(t\mapsto t^\lambda\) and restricted analytic functions, each operation applied only where it is defined. A non-integer power requires \(t>0\), a negative integer power requires \(t\ne0\), and a restricted analytic function requires its argument to lie in its cube.

**\(\mathcal L\)-cells** are defined inductively, one ambient dimension at a time. The only cell in \(\mathbb R^0\) is the point. A cell in \(\mathbb R^n\) is either a **graph** \(\{(x,\zeta(x)):x\in B\}\) or a **band** \(\{(x,z):x\in B,\ \zeta_1(x)<z<\zeta_2(x)\}\). Here \(B\subset\mathbb R^{n-1}\) is a cell, the **base**; \(\zeta\), \(\zeta_1\) and \(\zeta_2\) are analytic \(\mathcal L\)-functions on \(B\) with \(\zeta_1<\zeta_2\); and \(\zeta_1=-\infty\) or \(\zeta_2=+\infty\) is allowed. Cells are connected. A graph is homeomorphic to its base. A band is homeomorphic to \(B\times\mathbb R\) by \((x,\tau)\mapsto(x,z)\), with \(z=\zeta_1+(\zeta_2-\zeta_1)/(1+e^{-\tau})\), \(z=\zeta_1+e^\tau\), \(z=\zeta_2-e^{-\tau}\) or \(z=\tau\), according to which endpoints are finite. Induction on \(n\) proves connectedness.

Let \(C\) be a band with base \(B\). A **translation** for \(C\) is an analytic \(\mathcal L\)-function \(\theta\) on \(B\) whose graph does not meet \(C\). Then \(y=|z-\theta(x)|\) is positive on \(C\). An **analytic \(\mathcal L\)-unit in \(y\)** is a function

\[
U(x,z)=\Psi\bigl(\beta(x),\,u(x)y^{1/s},\,v(x)y^{-1/s}\bigr),
\]

where \(s\ge1\) is an integer, \(\beta\) (vector valued), \(u\) and \(v\) are analytic \(\mathcal L\)-functions on \(B\), the argument map is bounded on \(C\), and \(\Psi\) is analytic and nowhere zero on a neighbourhood of the compact closure of its image. A function \(f\) on \(C\) is **reduced** if

\[
f(x,z)=a(x)\,y^r\,U(x,z),\qquad r\in\mathbb Q,
\]

with an analytic \(\mathcal L\)-function \(a\) on \(B\), possibly zero, and a unit \(U\) in the same \(y\). On a graph cell, a function is reduced if it is an analytic \(\mathcal L\)-function of the base. A function is **reducible** on a cell if it is reduced on each cell of a finite refinement. A unit is bounded and bounded away from zero, by continuity of \(\Psi\) on the compact closure, and it has constant sign on the connected cell. We use the analytic form of units, not only these bounds.

We prove the preparation theorem by induction on \(n\). The **induction hypothesis \((\mathrm H_{<n})\)** is the following statement in every dimension \(m<n\): given finitely many cells of \(\mathbb R^m\) and finitely many \(\mathcal L\)-functions, each defined on one of these cells, there is a finite cylindrical decomposition of \(\mathbb R^m\) into cells, compatible with the given cells, with the following property. On each new cell \(D\), the given functions whose domain contains \(D\) are analytic and of constant sign, and any two of them are equal on \(D\) or strictly ordered there. The theorem for arbitrary subsets of \(\mathbb R^n\) is not assumed.

A function on a graph cell \(\{z=\zeta(x)\}\) is reduced in the lower dimension. Substituting \(z=\zeta(x)\) into an \(\mathcal L\)-expression gives an \(\mathcal L\)-function of \(x\), and \((\mathrm H_{<n})\) makes it analytic on the cells of a refinement of the base. This is how graph cells are handled in every argument below, so those arguments only treat bands.

## Finite endpoint refinements

**Endpoint refinement.** Let finitely many cells of \(\mathbb R^n\) be given, together with finitely many \(\mathcal L\)-functions on their bases. Assume \((\mathrm H_{<n})\). There is a finite decomposition of \(\mathbb R^n\) into cells that is compatible with the given cells and, over each base cell on which a given function is defined, with the graph of that function.

**Proof.** Apply \((\mathrm H_{<n})\) in \(\mathbb R^{n-1}\) to the bases of the given cells and to all given functions together with the finite endpoints of the given cells. On a resulting base cell \(D\), the relevant functions are analytic and any two are equal or strictly ordered. List the distinct ones as \(\zeta_1<\cdots<\zeta_p\). The graphs of the \(\zeta_l\), the bands between consecutive ones, and the two bands below \(\zeta_1\) and above \(\zeta_p\) (all of \(D\times\mathbb R\) if \(p=0\)) partition \(D\times\mathbb R\) into cells. A given cell is a union of these, because its base is a union of base cells and its endpoints occur among the \(\zeta_l\). \(\square\)

A decomposition of this shape is called **cylindrical**: the bases of its cells form a cylindrical decomposition of \(\mathbb R^{n-1}\), and over each base cell its cells are the graphs and bands cut out by finitely many ordered analytic functions. Every point of a base cell lies under each cell above it, so projecting a cylindrical decomposition onto the first \(m\) coordinates gives a cylindrical decomposition of \(\mathbb R^m\). In particular finitely many finite cell decompositions have a common cylindrical refinement. This uses no cell theorem for arbitrary sets. Three conventions follow from the lemma and are used without comment below.

- A function of the base alone is reduced on a band after cutting at \(z=0\): on the bands \(z>0\) and \(z<0\) use \(\theta=0\), \(r=0\) and \(U=1\). The graph \(z=0\) is a graph cell.
- If \(f=ay^rU\) is reduced, refine the base so that \(a\) has constant sign. Since \(y>0\) and \(U\) has constant sign, \(f\) has constant sign on each new cell.
- A reduction restricts to a subcell with the same base data. Its translation still misses the subcell, and its argument image only shrinks.

## Collecting bounded monomials into two coordinates

In this and the following sections \(C\) is a band with base \(B\), \(\theta\) is a translation for \(C\), \(y=|z-\theta(x)|>0\), and \((\mathrm H_{<n})\) is assumed. Every refinement of the base is made with \((\mathrm H_{<n})\), and every refinement of a cylinder is an endpoint refinement.

**Two-coordinate lemma.** Let \(M_i=b_i(x)y^{q_i}\), \(1\le i\le N\), with \(q_i\in\mathbb Q\) and \(\mathcal L\)-functions \(b_i\), be bounded on \(C\). After refining the base, there are an integer \(s\ge1\) and analytic \(\mathcal L\)-functions \(u,v\ge0\) on the base such that \(Z=uy^{1/s}\) and \(W=vy^{-1/s}\) are bounded on \(C\), and each \(M_i\) is a monomial in \(Z\) or in \(W\) times a bounded analytic base function, or a bounded base function if \(q_i=0\).

**Proof.** Choose \(s\) with all \(k_i=sq_i\) integers, and refine the base so that every \(b_i\) is analytic and either identically zero or of constant sign, and the finitely many functions introduced below are ordered. For the indices with \(k_i>0\) and \(b_i\ne0\), put \(u_i=|b_i|^{1/k_i}\) and let \(u\) be the largest of them on each base cell; put \(u=0\) if there are no such indices. If \(u>0\), then

\[
M_i=\operatorname{sign}(b_i)\Bigl(\frac{u_i}{u}\Bigr)^{k_i}Z^{k_i},\qquad 0\le\frac{u_i}u\le1,
\]

and \(Z\) equals \(|M_i|^{1/k_i}\) for the index where \(u_i=u\), so \(Z\) is bounded. Indices with \(k_i<0\) are treated in the same way with \(v_i=|b_i|^{1/|k_i|}\), their maximum \(v\) and \(W=vy^{-1/s}\). For \(k_i=0\), \(M_i=b_i\) is a base function, bounded because every base point has a nonempty fibre in \(C\). Roots, ratios and maxima of \(\mathcal L\)-functions are \(\mathcal L\)-functions, and the refinement makes them analytic. \(\square\)

The recovery is polynomial. Let \(\Gamma=(\eta,Z,W)\) collect the new bounded arguments, with \(\eta\) the bounded base functions, and let \(P\) be the polynomial map with \(P\circ\Gamma=(M_1,\ldots,M_N)\) on \(C\). Then

\[
P\bigl(\overline{\Gamma(C)}\bigr)\subset\overline{P(\Gamma(C))},
\]

because \(P\) is continuous. Hence an analytic function \(\Psi\) given near the compact closure of the old arguments composes with \(P\) to an analytic function near \(\overline{\Gamma(C)}\), and if \(\Psi\) has no zeros near the old closure, \(\Psi\circ P\) has none near the new one. This lets any finite set of bounded monomial arguments be replaced by the two coordinates \(Z\), \(W\) and bounded base functions.

## Closure and substitution for analytic units

**Units form a group.** If \(U_1,U_2\) are units in the same \(y\), then \(U_1U_2\) and \(1/U_1\) are units in \(y\), and so is \(|U_1|^\lambda\) for every \(\lambda\in\mathbb Q\).

**Proof.** Collect the arguments of both units, which are bounded monomials and bounded base functions, by the two-coordinate lemma. With the polynomial recovery map \(P\), both outer functions become analytic and zero-free near the compact closure of the new argument image. So do their product and the reciprocal of the first. For a power, note that the image \(\Gamma(C)\) is connected, since \(C\) is, so its closure is connected and \(\Psi_1\circ P\) has a constant sign \(\epsilon\) there. Then \(\epsilon\,\Psi_1\circ P\) is positive on a neighbourhood of that closure, where its \(\lambda\)-th power is analytic and positive. \(\square\)

**Change of distance.** Let \(h>0\) be another distance on \(C\), \(h=|z-\vartheta(x)|\), and suppose \(h=A(x)\,y^e\,V(x,z)\) with an analytic \(\mathcal L\)-function \(A>0\), \(e\in\mathbb Q\) and a positive unit \(V\) in \(y\). Then every unit in \(h\) is a unit in \(y\). Consequently a reduction \(f=a\,h^r\,U\) in \(h\) becomes the reduction

\[
f=\bigl(aA^r\bigr)\,y^{er}\,\bigl(V^rU\bigr)
\]

in \(y\).

**Proof.** An argument of a unit in \(h\) is a bounded monomial \(b(x)h^q=b(x)A(x)^qy^{eq}\,V^q\) or a base function. Since \(V\) is bounded above and below by positive constants, the monomial \(bA^qy^{eq}\) is bounded. Collect these monomials and the arguments of \(V\) by the two-coordinate lemma. Recover \(V\) through its outer function, then \(V^q\) by an analytic power, then the old arguments by multiplication. The resulting analytic map is defined near the compact closure of the new argument image, coincides with the old argument map on \(C\), and so sends that closure into the closure of the old argument image. The old outer function composes with it to a zero-free analytic function there. The last formula follows, and \(V^rU\) is a unit by the group property. \(\square\)

## A common translation for finitely many reductions

**Common translation lemma.** Let \(f_1,\ldots,f_k\) be reduced on \(C\) with translations \(\theta_1,\ldots,\theta_k\). After an endpoint refinement, on each new band some \(\theta_j\) serves as a common translation. More precisely, each distance \(|z-\theta_i|\) equals \(A_iy^{e_i}V_i\) with \(y=|z-\theta_j|\), \(A_i>0\), \(e_i\in\{0,1\}\) and \(V_i\) a positive unit, so every \(f_i\) is reduced with translation \(\theta_j\).

**Proof.** Since

\[
(z-\theta_i)^2-(z-\theta_l)^2=(\theta_l-\theta_i)\bigl(2z-\theta_i-\theta_l\bigr),
\]

an endpoint refinement at the midpoints \((\theta_i+\theta_l)/2\), together with sign conditions on \(\theta_l-\theta_i\), fixes on each new band which of two distances is smaller. On a new band choose \(j\) with \(y=|z-\theta_j|\) smallest. Fix \(i\), put \(d=z-\theta_j\) and \(\delta=\theta_j-\theta_i\), so that \(z-\theta_i=d+\delta\) and \(|d|\le|d+\delta|\). If \(\delta=0\), take \(A_i=1\), \(e_i=1\), \(V_i=1\). Otherwise refine at \(z=\theta_j\pm|\delta|\) and at \(z=\theta_j\), so that \(d\) has a fixed sign and one of the following cases holds on the band.

If \(|d|\le|\delta|\), then \(\tau=d/\delta=(\operatorname{sign}(d)/\delta)\,y\) is a bounded monomial in \(y\). The inequality \(|\tau|\le|1+\tau|\) means \(\tau\ge-\tfrac12\), so \(\tau\in[-\tfrac12,1]\) and

\[
|z-\theta_i|=|\delta|\,(1+\tau),\qquad 1+\tau\in[\tfrac12,2].
\]

If \(|d|\ge|\delta|\), then \(\tau=\delta/d=\operatorname{sign}(d)\,\delta\,y^{-1}\) lies in \([-1,1]\), and \(|1+\tau|\ge1\) forces \(\tau\ge0\). Hence

\[
|z-\theta_i|=y\,(1+\tau),\qquad 1+\tau\in[1,2].
\]

In both cases \(1+\tau\) is a positive unit, analytic and positive near the compact closure of the possible values of \(\tau\). The change of distance lemma converts the reduction of \(f_i\). \(\square\)

Products of reducible functions are therefore reducible: multiply coefficients and units and add exponents. The same holds for quotients on a domain where the denominator does not vanish, after the base refinement that makes the denominator's coefficient zero-free.

## Dominant and comparable pieces

**Dominance lemma.** Let \(f,g\) be reducible on \(C\) and \(\epsilon>0\). After a refinement, each new cell satisfies one of

\[
|f|\le\epsilon|g|,\qquad |g|\le\epsilon|f|,\qquad c_1|g|\le|f|\le c_2|g|,
\]

with constants \(0<c_1\le c_2\) independent of the point.

**Proof.** Give \(f\) and \(g\) a common translation and separate the cells where a coefficient vanishes; there one of the first two inequalities is trivial. Elsewhere write \(f=ay^rU\) and \(g=by^tV\) with \(a,b\ne0\). The units give constants \(0<m\le m'\) with \(m\le|U/V|\le m'\), so \(mH\le|f/g|\le m'H\) for \(H=|a/b|\,y^{r-t}\). Let \(\eta=\min(\epsilon,\tfrac12)\). Cut at the levels \(H=\eta/m'\) and \(H=1/(\eta m)\), which satisfy \(\eta/m'<1/(\eta m)\). If \(r=t\), these levels are conditions on base functions and are reached by a base refinement. If \(r\ne t\), the level \(H=L\) means \(y=(L|b/a|)^{1/(r-t)}\), a positive analytic base function, and the endpoints \(\theta\pm(L|b/a|)^{1/(r-t)}\) give an endpoint refinement. Below the lower level \(|f|\le\eta|g|\le\epsilon|g|\). Above the upper level \(|g|\le\eta|f|\). Between them \(m\eta/m'\le|f/g|\le m'/(\eta m)\). \(\square\)

## Zero translation or comparison with the coordinate

**Translation dichotomy.** Let \(f\) be reduced on \(C\) with translation \(\theta\). After a refinement, on each new band either \(f\) is reduced with translation \(0\), or \(\theta\ne0\), \(z\) and \(\theta\) have the same sign, and \(\tfrac12|\theta|\le|z|\le2|\theta|\). Each distance is replaced by \(A\,t^e\,V\) with \(t=|z|\), \(A>0\), \(e\in\{0,1\}\) and a positive unit \(V\), so nonnegative exponents remain nonnegative.

**Proof.** Cut at \(z=0\); the graph is lower-dimensional. Fix the sign of \(\theta\) by a base refinement; where \(\theta=0\) there is nothing to do. Refine at the endpoints \(z=\pm\tfrac12|\theta|\) and \(z=\pm2|\theta|\), and put \(t=|z|\). If \(|\theta|\le\tfrac12t\), then

\[
|z-\theta|=t\,(1-\theta/z),\qquad \theta/z=\bigl(\operatorname{sign}(z)\,\theta\bigr)\,t^{-1}\in[-\tfrac12,\tfrac12].
\]

If \(t\le\tfrac12|\theta|\), then \(|z-\theta|=|\theta|\,(1-z/\theta)\), where \(z/\theta=(\operatorname{sign}(z)/\theta)\,t\) also lies in \([-\tfrac12,\tfrac12]\). If \(\tfrac12|\theta|<t<2|\theta|\) and the signs of \(z\) and \(\theta\) are opposite, then \(|z-\theta|=t\,(1+|\theta|\,t^{-1})\) with \(|\theta|t^{-1}\in(\tfrac12,2)\). In these three cases the factor in parentheses is a positive unit in \(t\), and the change of distance lemma gives a reduction with translation \(0\). In the remaining case the signs agree and the old translation is kept. \(\square\)

## Pullback through a power coordinate

Let \(p\in\mathbb Z\setminus\{0\}\), let \(c\) be an analytic \(\mathcal L\)-function on the base, and put

\[
H(x,z)=\bigl(x,\ c(x)|z|^{1/p}\bigr),
\]

on a cell \(C\) where it is defined (so \(z\ne0\) if \(p<0\)). Write \(v\) for the new last coordinate.

**Pullback lemma.** Suppose \(f\) is reducible on cells covering \(H(C)\). Then \(f\circ H\) is reducible on \(C\). If \(p>0\) and the reductions of \(f\) have nonnegative exponents, so do those of \(f\circ H\).

**Proof.** Refine the base so that \(c\) has constant sign, and cut at \(z=0\). Where \(c=0\), \(H\) maps into the graph \(v=0\), on which \(f\) restricts to a base function, so \(f\circ H\) is a base function. Consider a band with \(c\ne0\), on which \(z\) has a fixed sign \(\sigma\). There the map \(z\mapsto c|z|^{1/p}\) is an analytic strictly monotone bijection of each fibre interval onto an interval, with inverse \(z=\sigma(v/c)^p\), where \(v/c>0\). It therefore maps bands to bands and graphs to graphs, with endpoints transformed by this formula. Conversely the preimage of a cell in the image is a cell. So the cells on which \(f\) is reduced pull back to a refinement of \(C\).

On such a piece \(f=a\,|v-\theta|^r\,U\). With \(t=|z|\) and \(\xi=\theta/c\), the pulled-back distance is

\[
h=\bigl|c\,t^{1/p}-\theta\bigr|=|c|\,\bigl|t^{1/p}-\xi\bigr|.
\]

We reduce \(h\) with exponents \(0\), \(1/m\) or \(1\) and positive units.

Let \(p=m>0\) and \(T=t^{1/m}\). If \(\xi=0\), then \(h=|c|\,t^{1/m}\). If \(\xi<0\), put \(\alpha=|\xi|\) and cut at \(t=\alpha^m\): for \(T\ge\alpha\), \(h=|c|\,t^{1/m}(1+\alpha T^{-1})\), and for \(T\le\alpha\), \(h=|c|\,\alpha\,(1+\alpha^{-1}T)\). The ratios lie in \([0,1]\), so the factors are positive units with translation \(0\). If \(\xi>0\), cut at \(T=\xi/2\) and \(T=2\xi\). For \(T\le\xi/2\), \(h=|c|\,\xi\,(1-T/\xi)\), and for \(T\ge2\xi\), \(h=|c|\,t^{1/m}(1-\xi T^{-1})\), with ratios in \([0,\tfrac12]\). For \(\xi/2\le T\le2\xi\), factor

\[
T^m-\xi^m=(T-\xi)\,\xi^{m-1}P_m(T/\xi),\qquad P_m(\rho)=1+\rho+\cdots+\rho^{m-1}>0\ \ (\rho>0).
\]

Since \(h>0\) on the piece, \(z\ne\sigma\xi^m\) there; refine at \(z=\sigma\xi^m\), take the translation \(\sigma\xi^m\), and put \(y=|z-\sigma\xi^m|=|t-\xi^m|\), \(\kappa=\operatorname{sign}(t-\xi^m)\) and \(\omega=\xi^{-m}y\). Then \(T/\xi=(1+\kappa\omega)^{1/m}\) and

\[
h=|c|\,\xi^{1-m}\,y\,\Bigl[P_m\bigl((1+\kappa\omega)^{1/m}\bigr)\Bigr]^{-1}.
\]

For \(\kappa=1\), \(0\le\omega\le2^m-1\), and for \(\kappa=-1\), \(0\le\omega\le1-2^{-m}\). In both cases \(1+\kappa\omega\ge2^{-m}\) on the compact range of \(\omega\), so the bracket is a positive analytic function of the bounded monomial \(\omega\), and the last factor is a positive unit.

Let \(p=-m<0\). If \(\xi=0\), then \(h=|c|\,t^{-1/m}\). Otherwise \(h=|c\,\xi|\,t^{-1/m}\,\bigl|t^{1/m}-\xi^{-1}\bigr|\). The last factor is a distance of the kind just reduced, and \(t^{-1/m}\) is reduced with translation \(0\). A common translation and multiplication reduce \(h\).

In every case the change of distance lemma turns the reduction of \(f\) into a reduction of \(f\circ H\). For \(p>0\), \(h\) carries the exponent \(0\), \(1/m\) or \(1\), so a nonnegative \(r\) stays nonnegative. \(\square\)

## Bounded arguments of a general analytic composition

**Bounded argument lemma.** Let \(g_1,\ldots,g_l\) be reduced on \(C\) with a common translation, let \(G=(g_1,\ldots,g_l)\) be bounded on \(C\), and let \(F\) be analytic near \(\overline{G(C)}\). After a base refinement,

\[
F\circ G=\psi\bigl(\beta(x),\,u(x)y^{1/s},\,v(x)y^{-1/s}\bigr)
\]

with \(u,v\ge0\), a bounded argument map, and \(\psi\) analytic near the compact closure of the argument image.

**Proof.** Write \(g_i=a_iy^{r_i}U_i\). Each monomial \(a_iy^{r_i}=g_i/U_i\) is bounded, since \(|U_i|\) has a positive lower bound on \(C\). Gather these monomials and the arguments of all \(U_i\) with the two-coordinate lemma. Recovering the \(U_i\) through their outer functions and multiplying gives an analytic map near the compact closure of the new argument image. It agrees with \(G\) on \(C\), so it sends that closure into \(\overline{G(C)}\), where \(F\) is analytic. Their composite is the required \(\psi\). \(\square\)

If \(u>0\), put \(Z=uy^{1/s}\) and \(c=uv\). Then \(W=vy^{-1/s}=c/Z\), and \(c=ZW\) is a bounded base function. So the composite has the form \(\psi(\beta,Z,c/Z)\) of the localization lemma, with \(Z\) as last coordinate. The cases \(u=0<v\) and \(u=v=0\) involve only \(W\) or only base functions. These forms are completed in the section on closure under bounded analytic composition.

## Ordinary analytic preparation without a cell-theorem assumption

**Ordinary preparation.** Assume \((\mathrm H_{<n})\). Let \(C\subset\mathbb R^n\) be a cell with base \(B\), let \(\phi=(\phi_1,\ldots,\phi_k)\) be \(\mathcal L\)-functions on \(B\) such that \(\Phi(x,z)=(\phi(x),z)\) is bounded on \(C\), and let \(\psi\) be analytic on a neighbourhood of \(K=\overline{\Phi(C)}\). Then \(\psi\circ\Phi\) is reducible on \(C\), with nonnegative exponents.

A reduction with a nonnegative exponent, or with zero coefficient, is called **nondegenerate**. Two remarks are used throughout. First, if \(\chi\) is analytic near a compact set containing the closure of the image of finitely many base functions, then \(\chi\) composed with them is an \(\mathcal L\)-function piecewise. Cover the compact set by finitely many closed boxes inside the domain of \(\chi\). Refine the base so that on each cell the base functions stay in one box, using sign conditions on their differences from the box bounds. On that cell, \(\chi\) is restricted analytic on the box. Second, by the graph convention, only bands need to be treated.

### Finite localization and signed normalization

For every point \((u_0,z_0)\in K\), the finite Taylor family gives

\[
\psi(u,z)=\sum_{i=0}^dc_i(u)(z-z_0)^iA_i(u,z),\qquad A_i(u_0,z_0)=1,
\]

on a box around the point, which we shrink so that the uniform regularity bound holds on it for every \(m\le d\). Choose a closed box inside it around the point; the interiors of finitely many such boxes cover \(K\). A box is defined by bounds on the \(u\)-coordinates and on \(z\). Refine the base with the sign conditions on the \(\phi_i\) minus the \(u\)-bounds, and refine the cylinders at the constant \(z\)-bounds. Then every cell of \(C\) maps into one of the closed boxes.

Fix a cell and its box, and put \(g_i=c_i\circ\phi\), \(\mathcal L\)-functions on the base by the first remark. Refine the base so that each \(g_i\) has constant sign and the \(|g_i|\) are ordered. If all \(g_i\) vanish on a base cell, the identity gives \(\psi\circ\Phi=0\) there. Otherwise let \(g_m\) have the largest absolute value. Then \(t_i=g_i/g_m\in[-1,1]\), \(t_m=1\), and

\[
\psi\circ\Phi=g_m\cdot Q(t,\phi,z),\qquad Q(t,u,z)=\sum_{i=0}^dt_i(z-z_0)^iA_i(u,z).
\]

The map \(\Theta=(t,\phi,z)\) is bounded. The closure of its image lies in \(K_m\) times the closed box, on which \(Q\) has order at most \(m\) in its last variable at every point. This holds even where \(g_m\) tends to zero, because \(t_m=1\) identically. Multiplying by the base function \(g_m\) preserves nondegenerate reductions. It therefore suffices to prove the following statement.

**(O\(_d\)).** If \(\Phi=(\phi,z)\) is bounded on a band \(C\), and \(F\) is analytic near \(\overline{\Phi(C)}\) with order at most \(d\) in \(z\) at every point of that closure, then \(F\circ\Phi\) has a nondegenerate reduction on a refinement of \(C\).

### Induction on the regularity bound

We prove (O\(_d\)) by induction on \(d\), in the fixed dimension \(n\).

If \(d=0\), then \(F\) has no zero on \(\overline{\Phi(C)}\). Cut \(C\) at \(z=0\). On a band where \(z\) has the sign \(\sigma\), \(F(\phi,z)=F(\phi,\sigma|z|)\) is a unit in \(y=|z|\): its arguments are the base functions \(\phi\) and the bounded monomial \(|z|\), and \(F(u,\sigma T)\) is analytic and zero-free near the closure of their image. So \(F\circ\Phi=1\cdot y^0\cdot U\) with translation \(0\).

Let \(d\ge1\). At every point \((u_1,z_1)\) of \(\overline{\Phi(C)}\), \(F\) has some order \(e\le d\) in \(z-z_1\). The Weierstrass preparation theorem writes it there as an analytic zero-free factor times a monic polynomial in \(z-z_1\) of degree \(e\), whose coefficients are analytic in \(u\) and vanish at \(u_1\). As before, finitely many closed boxes inside such neighbourhoods cover the closure, and \(C\) is refined so that each cell maps into one box. On a band of this refinement

\[
F\circ\Phi=E(\phi,z)\cdot P,\qquad P=Z^e+a_1(x)Z^{e-1}+\cdots+a_e(x),\quad Z=z-z_1,
\]

where \(E\) is zero-free near the closed box and the coefficients \(a_k\) are bounded \(\mathcal L\)-functions of the base. If \(e=0\), \(P=1\). If \(e\ge1\), the substitution

\[
w=Z+\frac{a_1(x)}e=z-\vartheta(x),\qquad \vartheta=z_1-\frac{a_1}e,
\]

removes the term of degree \(e-1\):

\[
P=w^e+\sum_{i=2}^eb_i(x)\,w^{e-i},
\]

with bounded \(\mathcal L\)-functions \(b_i\). A shift of the last coordinate by a base function maps cells to cells. Refine the base so that the \(b_i\) have constant signs and the roots \(|b_i|^{1/i}\) are linearly ordered. Where all \(b_i\) vanish, \(P=w^e\). Off the graph \(w=0\), this is \(\operatorname{sign}(w)^e\,|z-\vartheta|^e\), a reduction with translation \(\vartheta\) and exponent \(e\). Otherwise let

\[
\varrho=\max_{2\le i\le e}|b_i|^{1/i}>0,
\]

attained at an index \(j\), so that \(|b_i|\le\varrho^i\) for all \(i\) and \(|b_j|=\varrho^j\). Cut at \(w=0\) and \(w=\pm2\varrho\); the graphs are lower-dimensional.

### Outside the scale: an explicit polynomial unit

On a band where \(|w|\ge2\varrho\) and \(w\) has the sign \(\sigma\),

\[
P=w^e\Bigl(1+\sum_{i=2}^e\frac{b_i}{w^i}\Bigr),\qquad\Bigl|\sum_{i=2}^e\frac{b_i}{w^i}\Bigr|\le\sum_{i=2}^e2^{-i}<\frac12.
\]

Each quotient \(b_i/w^i=(\sigma^ib_i)\,|w|^{-i}\) is a bounded monomial in \(y=|z-\vartheta|\). By the two-coordinate lemma they are polynomials in two bounded arguments and bounded base functions. The outer function \(1+\sum_i(\text{recovered quotients})\) takes values at least \(\tfrac12\) on the actual argument image. By continuity it stays at least \(\tfrac12\) on the closure, so it is zero-free on a neighbourhood. Hence the parenthesis is a unit, and \(P=\sigma^e\,y^e\,U\) is a reduction with exponent \(e\). The two-coordinate lemma supplies the analytic form of this unit; the inequality alone would only give bounds.

### Inside the scale: the analytic order strictly decreases

On a band where \(0<|w|\le2\varrho\) and \(w\) has the sign \(\sigma\), introduce the coordinate \(\nu=w/\varrho\). This is the power-coordinate map with \(p=1\) and coefficient \(\sigma/\varrho\), so bands correspond to bands. Put \(\beta_i=b_i/\varrho^i\) and

\[
R(\beta,\nu)=\nu^e+\sum_{i=2}^e\beta_i\nu^{e-i},\qquad P=\varrho^eR(\beta(x),\nu).
\]

Here \(|\beta_i|\le1\), \(|\nu|\le2\), and \(\beta_j=\operatorname{sign}(b_j)\) is a constant \(\pm1\) on the band. The map \((\beta,\nu)\) is bounded, and on the closure of its image \(\beta_j\) still equals that constant. At every point of the closure \(R\) has order at most \(e-1\) in \(\nu\). Otherwise \(\partial_\nu^{e-1}R=e!\,\nu\) would vanish, giving \(\nu=0\). Then \(\partial_\nu^{e-i}R(\beta,0)=(e-i)!\,\beta_i\) would vanish for every \(i\ge2\), contradicting \(\beta_j=\pm1\). The bound holds on the whole compact closure, which is what (O\(_{e-1}\)) requires.

By (O\(_{e-1}\)), \(R(\beta,\nu)\) has a nondegenerate reduction on a refinement of the image band. The pullback lemma with \(p=1\) gives a nondegenerate reduction of \(R(\beta,w/\varrho)\), and multiplication by the base factor \(\varrho^e\) one of \(P\).

### Restore the analytic factor and the coordinate

The factor \(E(\phi,z)\) is zero-free near its compact argument closure, so as in the case \(d=0\) it is a unit with translation \(0\) on each band of fixed sign of \(z\). A common translation for \(E\) and the reduction of \(P\) allows multiplication. The common translation lemma replaces distances by base factors times \(y^0\) or \(y^1\) times positive units, so nonnegative exponents stay nonnegative.

The coordinate \(w\) was used only as a name. A reduction in \(w\) with translation \(\eta\) involves the distance \(|w-\eta|=|z-(\vartheta+\eta)|\), which is a reduction in \(z\) with translation \(\vartheta+\eta\). Its graph misses the cell because the graph of \(\eta\) missed the corresponding cell in \(w\).

Every step used finitely many boxes, signs and scales. In the inner case the order bound strictly decreased, and the outer case was settled directly. This proves (O\(_d\)) and hence ordinary preparation. \(\square\)

## Full analytic reduction in a coordinate and its inverse

**Two-coordinate preparation.** Assume \((\mathrm H_{<n})\). Let \(C\subset\mathbb R^n\) be a cell on which \(z\ne0\), with base \(B\). Let \(c\) and \(\phi\) be \(\mathcal L\)-functions on \(B\) such that \(\Phi(x,z)=(\phi(x),z,c(x)/z)\) is bounded on \(C\), and suppose \(\psi\) is analytic on a neighbourhood of \(K=\overline{\Phi(C)}\). Then \(\psi\circ\Phi\) is reducible on \(C\).

The function \(c\) is bounded, since \(c=z\cdot(c/z)\) on \(C\) and each point of the base lies under some point of \(C\). Make the base functions analytic. On the base cells where \(c=0\), \(\psi\circ\Phi=\psi(\phi,z,0)\), and ordinary preparation applies. On the others, \(c\) and \(z\) have fixed signs, and we write \(w=c/z\).

### The finite localization really has compatible cells

Take the open sets of the localization lemma to be open boxes in \((u,z,w)\), whose closures lie inside the domains used there. Membership in such a box is a condition on the base functions \(\phi\), a condition \(\gamma<z<\gamma'\) with constants, and a condition \(\gamma<c/z<\gamma'\). With the signs of \(c\) and \(z\) fixed, the last condition says that \(z\) lies between the base functions \(c/\gamma\) and \(c/\gamma'\). A base refinement and an endpoint refinement therefore make every cell lie in the preimage of one box.

On a cell of form 1, ordinary preparation applies to \(F(\phi,c,z)\), with parameters \((\phi,c)\). On a cell of form 2, \(w=c/z=(\sigma c)|z|^{-1}\) is the power-coordinate map with \(p=-1\), where \(\sigma=\operatorname{sign}z\). Ordinary preparation of \(F(\phi,c,w)\) on the image cells, followed by the pullback lemma, reduces the function. It remains to treat form 3, where

\[
\psi\circ\Phi=X_++X_-,\qquad X_+=z\,\psi_+(\phi,c,z),\quad X_-=\psi_-(\phi,c,w).
\]

### Keep the two nondegenerate reductions before changing coordinates

Ordinary preparation gives a nondegenerate reduction of \(\psi_+(\phi,c,z)\) in \(z\), and one of \(\psi_-(\phi,c,w)\) in \(w\) on the image cells. Multiplying the first by \(z=\operatorname{sign}(z)|z|\) gives a nondegenerate reduction of \(X_+\). Pulling the second back through \(w=c/z\) gives a reduction of \(X_-\) in \(z\), whose exponent may be negative. We keep both the reductions in \(z\) and the original reduction of \(\psi_-\) in \(w\), and refine further where needed; reductions restrict to subcells.

If \(X_+\) or \(X_-\) vanishes on a cell, the other summand is the whole function. Otherwise the dominance lemma with \(\epsilon=\tfrac12\) leaves three cases on each cell: \(|X_+|\le\tfrac12|X_-|\), \(|X_-|\le\tfrac12|X_+|\), or \(X_+\) and \(X_-\) comparable.

### Dominance gives an analytic unit, not just bounds

Suppose \(|X_+|\le\tfrac12|X_-|\). With a common translation, \(\mathcal R=X_+/X_-=A(x)\,y^\lambda V\) is reduced, and \(|\mathcal R|\le\tfrac12\). The monomial \(Ay^\lambda\) is bounded, since \(V\) is bounded away from zero. Collect it with the arguments of \(V\) by the two-coordinate lemma. The analytic recovery map represents \(\mathcal R\) as an analytic function of bounded arguments near the closure of their image. On that closure its values lie in \([-\tfrac12,\tfrac12]\), by continuity. So \(1+\mathcal R\) is represented by an analytic function with values in \([\tfrac12,\tfrac32]\) there, zero-free on a neighbourhood: it is a unit. Then \(\psi\circ\Phi=X_-(1+\mathcal R)\) is reduced. The case \(|X_-|\le\tfrac12|X_+|\) is symmetric.

### The balanced scale completes the analytic reduction

Suppose finally that \(X_+\) and \(X_-\) are comparable. Apply the translation dichotomy to the reduction of \(\psi_+(\phi,c,z)\) in \(z\), with translation \(\theta_+\), and to the reduction of \(\psi_-(\phi,c,w)\) in \(w\), with translation \(\theta_-\). Pull the refinements in \(w\) back through \(w=c/z\), and take a common refinement. The dichotomy keeps the exponents nonnegative. On each resulting cell, choose a positive base function \(b\) comparable to \(|z|\) as follows.

1. If \(\theta_+\ne0\), then \(|z|\) is comparable to \(|\theta_+|\); take \(b=|\theta_+|\).
2. If \(\theta_+=0\) and \(\theta_-\ne0\), then \(|c/z|\) is comparable to \(|\theta_-|\); take \(b=|c|/|\theta_-|\).
3. If \(\theta_+=\theta_-=0\), the reductions read \(\psi_+=a_+|z|^rU_+\) and \(\psi_-=a_-|w|^sU_-\), with \(r,s\ge0\) and \(a_\pm\ne0\) because neither summand vanishes. Comparability of \(X_\pm\) gives

\[
|a_+|\,|z|^{r+1}\ \text{comparable to}\ |a_-|\,|c|^s|z|^{-s},
\]

so \(|z|^{r+s+1}\) is comparable to \(|a_-||c|^s/|a_+|\). Take \(b=\bigl(|a_-||c|^s/|a_+|\bigr)^{1/(r+s+1)}\), using \(r+s+1>0\).

In each case \(b\) is an \(\mathcal L\)-function, analytic and positive on the base cell, and \(|z|\) is comparable to \(b\). The balanced substitution lemma writes \(\psi\circ\Phi=\widetilde\psi(\phi,b,d,q)\) with \(q=z/b\) and \(d=c/b\). Here \(\widetilde\psi\) is analytic near the closure of the image of the bounded map \((\phi,b,d,q)\). Ordinary preparation in the last coordinate \(q\), with parameters \((\phi,b,d)\), reduces \(\widetilde\psi(\phi,b,d,q)\) on the image cells of \((x,z)\mapsto(x,z/b)\). The pullback lemma with \(p=1\) returns to \(z\). All covers, signs and cases are finite. This proves two-coordinate preparation. \(\square\)

## Closure under bounded analytic composition

**Composition theorem.** Assume \((\mathrm H_{<n})\). Let \(g_1,\ldots,g_l\) be reducible on a cell \(C\subset\mathbb R^n\) with \(G=(g_1,\ldots,g_l)\) bounded, and suppose \(F\) is analytic on a neighbourhood of \(\overline{G(C)}\). Then the composite \(F\circ G\) is reducible on \(C\).

**Proof.** After a common refinement and translation, the bounded argument lemma gives

\[
F\circ G=\psi\bigl(\beta,\,uy^{1/s},\,vy^{-1/s}\bigr),\qquad y=|z-\theta|,\quad u,v\ge0.
\]

On a band where \(z-\theta\) has the sign \(\sigma_0\), the map \((x,z)\mapsto(x,y)\) carries cells to cells, and we work in the coordinate \(y\). If \(u>0\), put \(Z=uy^{1/s}\) and \(c=uv\), so that \(F\circ G=\psi(\beta,Z,c/Z)\) with a bounded base function \(c\). Two-coordinate preparation in \(Z\), followed by the pullback lemma with \(p=s\), reduces it. If \(u=0<v\), then \(F\circ G=\psi(\beta,0,W)\) with \(W=vy^{-1/s}\). Ordinary preparation in \(W\), followed by the pullback lemma with \(p=-s\), reduces it. If \(u=v=0\), \(F\circ G\) is a function of the base, an \(\mathcal L\)-function piecewise by the first remark on ordinary preparation. To return to \(z\), note that a translation \(\eta\) for \(y\) corresponds to the translation \(\theta+\sigma_0\eta\) for \(z\), because \(|y-\eta|=|z-(\theta+\sigma_0\eta)|\). \(\square\)

## Close the dimension induction for finite expressions

**Preparation of \(\mathcal L\)-functions.** For every \(n\), the following hold.

1. Finitely many \(\mathcal L\)-functions, each defined on one of finitely many given cells of \(\mathbb R^n\), are reduced on the cells of a finite cylindrical decomposition of \(\mathbb R^n\) compatible with the given cells.
2. The statement \((\mathrm H_{<n+1})\) holds: the decomposition can be chosen so that the functions are moreover analytic and of constant sign, and any two of them are equal or strictly ordered on each cell.

**Proof.** In dimension \(0\) both statements are trivial. Assume them in all dimensions below \(n\), which gives \((\mathrm H_{<n})\). We prove 1 by induction on the length of an expression, for a single function on a single cell. Common refinements, by the endpoint refinement lemma and restriction of reductions to subcells, then give the statement for finitely many.

The coordinate functions of the base and constants are base functions. Off the graph \(z=0\), the last coordinate is \(z=\operatorname{sign}(z)\,|z|\), reduced with translation \(0\) and exponent \(1\). Products are reduced by a common translation.

Let \(f=ay^rU\) be reduced and \(\lambda\in\mathbb Q\), with \(f^\lambda\) defined on the cell. If \(\lambda\) is an integer, \(f^\lambda=a^\lambda y^{r\lambda}U^\lambda\). If \(\lambda\) is not an integer, then \(f>0\) on the cell, so after a base refinement \(a\) and \(U\) have constant signs with positive product, and \(f^\lambda=|a|^\lambda y^{r\lambda}|U|^\lambda\). The group lemma makes \(U^\lambda\) and \(|U|^\lambda\) units.

For a sum \(f+g\), the dominance lemma gives cells on which, after possibly exchanging \(f\) and \(g\), \(|f|\le M|g|\) for a constant \(M\). Where \(g=0\), also \(f=0\). Elsewhere \(\mathcal R=f/g\) is reducible and bounded by \(M\), and the composition theorem applied to \(t\mapsto1+t\) reduces \(1+\mathcal R\). Then \(f+g=g(1+\mathcal R)\) is reducible. Unlike the dominant case above, \(1+\mathcal R\) is not claimed to be a unit; its zeros are allowed.

A restricted analytic function \(F\) on a cube \(Q\), applied to already reducible functions \(g_1,\ldots,g_l\), is defined on the cell only if \(G=(g_1,\ldots,g_l)\) takes values in \(Q\). Then \(G\) is bounded and \(F\) is analytic near \(Q\supset\overline{G(C)}\). The composition theorem applies. This proves 1 in dimension \(n\).

For 2, reduce all the given functions and their pairwise differences on one decomposition. Every reduced function is analytic on its cell, since \(y\), its rational powers, the base data and the outer function are analytic. After a base refinement its coefficient, and hence the function, has constant sign. Constant sign of the differences gives the ordering. \(\square\)

The induction used neither a cell theorem for arbitrary subsets of \(\mathbb R^n\) nor the complement theorem. Both are proved next from the statement just obtained.

## The global definition and its elementary set operations

Let \(\mathbf P=\mathbb R\cup\{\infty\}\) be the real projective line, and regard \(\mathbb R^n\) as a subset of the compact analytic manifold \(\mathbf P^n\). Each factor has the two charts \(x_i\) and \(u_i=1/x_i\), the second around \(\infty\). The products of these give \(2^n\) charts covering \(\mathbf P^n\); a chart is described by the set \(I\) of inverted coordinates.

A subset \(Z\) of an analytic manifold is **semianalytic** if every point of the manifold has a neighbourhood \(V\) such that \(Z\cap V\) is a finite Boolean combination of sets \(\{g=0\}\) and \(\{g>0\}\) with \(g\) analytic on \(V\). Following Lion and Rolin, a set \(Z\subset\mathbb R^n\) is **globally semianalytic** if it is semianalytic as a subset of \(\mathbf P^n\), so the condition is imposed also at the points at infinity. A set \(A\subset\mathbb R^n\) is **globally subanalytic** if \(A=\pi(Z)\) for a globally semianalytic \(Z\subset\mathbb R^{n+p}\) and the projection \(\pi\) onto the first \(n\) coordinates. A map is globally subanalytic if its graph is.

The following constructions use only these definitions.

*Polynomial sets.* A set defined by finitely many polynomial equations and strict inequalities is globally semianalytic. Consider a chart whose inverted coordinates form the set \(I\). There a polynomial \(F(x)\) becomes a rational function of the chart coordinates. Multiply it by \(D(u)=\prod_{i\in I}u_i^{2d_i}\), with even exponents \(2d_i\) at least the degree of \(F\) in \(x_i\). The product is a polynomial in the chart coordinates with the same sign and zeros as \(F\) wherever all \(u_i\ne0\), \(i\in I\). Adding the conditions \(u_i^2>0\) excludes the points at infinity.

*Boolean operations and products.* Local Boolean combinations of local Boolean combinations are again of that kind, so globally semianalytic sets form a Boolean algebra. Pulling back local descriptions along the coordinate projections shows that products of globally semianalytic sets are globally semianalytic. Here \(\mathbb R\subset\mathbf P\) is semianalytic, described near \(\infty\) by \(u^2>0\). Coordinate permutations are analytic automorphisms of \(\mathbf P^n\).

*Projections.* Let \(A=\pi(Z)\) and \(A'=\pi(Z')\) with witnesses \(Z\subset\mathbb R^{n+p}\) and \(Z'\subset\mathbb R^{n+q}\). Then

\[
A\cup A'=\pi\{(x,v,v'):(x,v)\in Z\text{ or }(x,v')\in Z'\},\qquad
A\cap A'=\pi\{(x,v,v'):(x,v)\in Z\text{ and }(x,v')\in Z'\},
\]

and the sets in braces are globally semianalytic by the product and Boolean constructions. Products are treated likewise, and a further projection of \(A\) just forgets more witness coordinates. Thus globally subanalytic sets are closed under finite unions and intersections, products, coordinate permutations and coordinate projections. Complements are not yet available.

*Maps and functions.* Images and preimages of globally subanalytic sets by globally subanalytic maps are obtained by intersecting graphs with cylinders and projecting. Sums, products and compositions of globally subanalytic functions are described through their graphs and a polynomial relation such as \(v=s+t\), followed by projection. A rational power \(t^{a/b}\), \(b>0\), on \(t>0\) has the polynomial graph \(\{y>0,\ y^b=t^a\}\) when \(a\ge0\), and \(\{y>0,\ y^bt^{-a}=1\}\) when \(a<0\); integer powers on their larger domains are polynomial as well.

*Restricted analytic functions.* The graph of a function analytic near a closed cube \(Q\), restricted to \(Q\), is semianalytic near every point of \(\mathbb R^{m+1}\): near \(Q\times\mathbb R\) it is described by its analytic equation and the cube inequalities, and elsewhere it is locally empty. It is bounded, so it is empty near every point at infinity. Hence it is globally semianalytic. If \(K\) is compact and globally subanalytic and \(F\) is analytic on a neighbourhood of \(K\), choose finitely many closed boxes \(Q_i\) covering \(K\) within that neighbourhood. The graph of \(F|_K\) is \((K\times\mathbb R)\cap\bigcup_i\operatorname{graph}(F|_{Q_i})\). Each box graph is globally semianalytic by the preceding argument, so the intersection is globally subanalytic by the already proved product, union and intersection operations.

By induction on the length of an expression, the graph of an \(\mathcal L\)-function over a globally subanalytic set on which the function is defined is globally subanalytic. By induction on the dimension, every \(\mathcal L\)-cell is globally subanalytic. A graph cell is the graph of its endpoint over its base. A band is the projection of the set of \((x,z,s,s')\) with \(x\) in the base, \(s=\zeta_1(x)\), \(s'=\zeta_2(x)\) and \(s<z<s'\), where infinite endpoints are omitted.

## Finite analytic cells for arbitrary globally subanalytic sets

**Cell theorem.** Let \(A_1,\ldots,A_k\subset\mathbb R^n\) be globally subanalytic. Then \(\mathbb R^n\) admits a finite cylindrical decomposition into analytic \(\mathcal L\)-cells in which each \(A_i\) is a union of cells.

**Proof.** First let \(Z\subset\mathbb R^N\) be globally semianalytic. Every point of \(\mathbf P^N\) has a chart neighbourhood on which \(Z\) is a finite Boolean combination of analytic equations and inequalities. Inside it choose a closed box, in the chart coordinates, whose interior contains the point. By compactness, finitely many such boxes \(Q_1,\ldots,Q_L\) have interiors covering \(\mathbf P^N\).

For a chart with inverted coordinates \(I\), write \(\chi(x)\) for the chart coordinates of a finite point: \(\chi_i(x)=x_i\) for \(i\notin I\) and \(\chi_i(x)=1/x_i\) for \(i\in I\). The coordinate hyperplanes cut \(\mathbb R^N\) into finitely many cells, on which every \(\chi_i\) is an \(\mathcal L\)-function wherever it is defined. The trace of a box \(Q_l=\prod_i[\alpha_i,\beta_i]\) on \(\mathbb R^N\) is the set where the \(\chi_i\) are defined and \(\alpha_i\le\chi_i\le\beta_i\). These are sign conditions on finitely many \(\mathcal L\)-functions. By the preparation of \(\mathcal L\)-functions, statement 2, there is a cylindrical decomposition of \(\mathbb R^N\) on whose cells all these functions have constant signs, so each trace is a union of cells. The traces cover \(\mathbb R^N\), so every cell lies in the trace of some box \(Q_l\).

On such a cell \(D\), the analytic functions describing \(Z\) near \(Q_l\) are restricted analytic on \(Q_l\), and their composites with \(\chi\) are \(\mathcal L\)-functions on \(D\). A further cylindrical refinement makes their signs constant. The Boolean formula for \(Z\) then has a constant truth value on each cell. So \(Z\) is a union of cells of a cylindrical decomposition of \(\mathbb R^N\). The analytic functions were used only on their own boxes.

Now let \(A=\pi(Z)\) with \(Z\subset\mathbb R^{n+p}\) globally semianalytic. Take a cylindrical decomposition of \(\mathbb R^{n+p}\) such that \(Z\) is a union of cells. The projection of each cell onto \(\mathbb R^n\) is one of its iterated base cells, and these form a cylindrical decomposition of \(\mathbb R^n\). Since \(A\) is the union of the projections of the cells contained in \(Z\), it is a union of cells. For finitely many sets, a common cylindrical refinement of the finitely many decompositions has the required property. \(\square\)

## Full global preparation and the complement theorem

**Global preparation theorem.** Let \(f:E\to\mathbb R\) be a globally subanalytic function, \(E\subset\mathbb R^n\); continuity is not assumed. There is a finite cylindrical decomposition of \(\mathbb R^n\), compatible with \(E\), on each cell of which in \(E\) the function \(f\) is reduced. In particular, on each such band, \(f=a(x)|z-\theta(x)|^rU\) with an analytic unit \(U\) and \(r\in\mathbb Q\), possibly negative.

**Proof.** Decompose \(\mathbb R^{n+1}\) cylindrically so that the graph \(\Gamma\) of \(f\) is a union of cells. The induced decomposition of \(\mathbb R^n\) is compatible with \(E=\pi(\Gamma)\). Over a base cell \(D\subset E\), the cells of \(\Gamma\) form a part of the stack over \(D\), and every point of \(D\) has exactly one point of \(\Gamma\) above it. A band would contain several points over each base point, and two graph cells would give two points. So \(\Gamma\) contains exactly one graph cell over \(D\), and its endpoint, an analytic \(\mathcal L\)-function, equals \(f\) on \(D\). The preparation of \(\mathcal L\)-functions, applied to these finitely many functions, completes the proof. \(\square\)

**Complement theorem.** The complement of a globally subanalytic subset of \(\mathbb R^n\) is globally subanalytic.

**Proof.** In a cell decomposition compatible with \(A\), the complement of \(A\) is the union of the cells not contained in \(A\). Every cell is globally subanalytic, as shown in the previous section. \(\square\)

This is Gabrielov's complement theorem for the global class. Together with the constructions of the previous section it gives the full **Boolean and projection calculus**. The globally subanalytic subsets of \(\mathbb R^n\) form a Boolean algebra. The class is closed under products, coordinate projections, and images and preimages under globally subanalytic maps. If \(A\subset\mathbb R^{n+m}\) is globally subanalytic, so are

\[
\{x:\exists y\ (x,y)\in A\}\qquad\text{and}\qquad\{x:\forall y\ (x,y)\in A\},
\]

the second being the complement of the projection of the complement. In particular closures, interiors and boundaries are globally subanalytic:

\[
x\in\overline A\iff\forall\epsilon>0\ \exists y\in A:\ \sum_i(x_i-y_i)^2<\epsilon^2,
\]

\(\operatorname{int}A=\mathbb R^n\setminus\overline{\mathbb R^n\setminus A}\), and \(\partial A=\overline A\setminus\operatorname{int}A\).

Cells are connected. If a cell meets a component of a finite union of cells, the cell lies inside that component, so every component is a union of cells. Hence the connected components of a globally subanalytic set are finite in number, and each is globally subanalytic.

## Local subanalytic sets and bounded charts

The global theorems apply to the local subanalytic geometry of analytic manifolds through bounded charts. A set \(A\subset\mathbb R^n\) is **locally subanalytic at** a point \(x_0\) if there are a neighbourhood \(V\) of \(x_0\) and a bounded semianalytic set \(S\subset\mathbb R^{n+p}\) with \(A\cap V=\pi(S)\cap V\). Here \(S\) is semianalytic on an open set containing its compact closure.

**Bounded-chart comparison.**

1. A bounded semianalytic set whose compact closure lies in the open set where it is semianalytic is globally semianalytic.
2. A bounded set which is locally subanalytic at each point of its closure is globally subanalytic.
3. A globally subanalytic set is locally subanalytic at every point of \(\mathbb R^n\).

**Proof.** For 1, cover the closure of \(S\) by finitely many closed boxes \(Q_k\), each inside a neighbourhood on which \(S\) has a finite analytic description. Each \(S\cap Q_k\) is semianalytic near every point of \(\mathbf P^N\). Near points of \(Q_k\) it is described by the local formula together with the box inequalities. Near points of \(\mathbb R^N\setminus Q_k\) and near infinity it is locally empty. So each \(S\cap Q_k\) is globally semianalytic, and so is their finite union \(S\).

For 2, cover the compact closure of the bounded set \(A\) by finitely many closed boxes \(Q_k\), each inside a neighbourhood \(V_k\) with \(A\cap V_k=\pi(S_k)\cap V_k\). Then \(A\cap Q_k=\pi\bigl(S_k\cap(Q_k\times\mathbb R^p)\bigr)\), and the bounded semianalytic set \(S_k\cap(Q_k\times\mathbb R^p)\) is globally semianalytic by 1. Thus \(A\) is a finite union of globally subanalytic sets.

For 3, let \(A=\pi(Z)\) and let \(Q\) be a closed box in \(\mathbb R^n\). The set \(Q\times\mathbf P^p\) is compact. Cover it by finitely many closed boxes in charts that keep the ordinary coordinates on \(\mathbb R^n\) and use projective charts on the witness coordinates, each box inside a neighbourhood with an analytic description of \(Z\). In the chart coordinates \((x,u)\), the points of \(Z\) in such a box with \(u_i\ne0\) for the inverted coordinates form a bounded semianalytic set, semianalytic on a neighbourhood of its closure. Its projection to \(x\) is contained in \(A\), and these projections cover \(A\cap Q\). So \(A\cap Q\) is a finite union of projections of bounded semianalytic sets \(S_1,\ldots,S_L\). It is also the projection of the single bounded semianalytic set \(\bigcup_l S_l\times\{l\}\), which uses one more witness coordinate. Hence \(A\) is locally subanalytic at every interior point of \(Q\). \(\square\)

Consequently the trace of a locally subanalytic set on a closed box inside a witness neighbourhood can be treated with the global cell, preparation and dimension theorems. The comparison is about such bounded traces. It does not make an arbitrary unbounded analytic set or map globally subanalytic.

## Uniform finiteness in families

For a globally subanalytic set \(A\subset\mathbb R^{m+n}\) and \(t\in\mathbb R^m\) put \(A_t=\{x\in\mathbb R^n:(t,x)\in A\}\). Take a cylindrical decomposition of \(\mathbb R^{m+n}\), with the coordinates \(t\) first, such that \(A\) is a union of cells, and let \(N\) be the number of its cells.

For a cell \(C\) and a fixed \(t\), the fibre \(C_t\) is empty or connected. By induction on \(n\): if \(C\) is a graph or band over \(C'\), then \(C_t\) is the graph or band over \(C'_t\) of the restricted endpoint functions, which are continuous, and \(C'_t\) is connected or empty. The nonempty fibres \(C_t\) partition \(\mathbb R^n\) into at most \(N\) connected sets, and \(A_t\) is a union of some of them. Therefore

\[
\#\{\text{connected components of }A_t\}\le N\qquad(t\in\mathbb R^m),
\]

and every component consists of finitely many of these fibres, hence is globally subanalytic.

Families of maps are handled through their graphs. If \((t,x)\mapsto f_t(x)\in\mathbb R^q\), defined for \(x\in A_t\), is globally subanalytic, apply the bound to its graph with the coordinates ordered as \((t,b,x)\). The sets \(f_t^{-1}(b)\) are the fibres of that one set over \((t,b)\), so their numbers of connected components are bounded by one constant independent of \(t\) and \(b\). The family must be globally subanalytic as a whole; a collection of individually globally subanalytic fibres does not suffice.

## Convergent Puiseux expansions

Global preparation gives convergent fractional power series, in one variable and with parameters; these are Valette's results in §1.8. In a reduced form \(a|t-\theta|^r\Psi(\beta,u|t-\theta|^{1/s},v|t-\theta|^{-1/s})\) the signs of \(u\) and \(v\) can be moved into \(\Psi\), so we assume \(u,v\ge0\).

### One variable

**One-variable Puiseux theorem.** Let \(f:(0,\epsilon)\to\mathbb R\) be globally subanalytic. There are \(0<\epsilon'\le\epsilon\), integers \(p\ge1\) and \(m\), and real numbers \(a_i\) such that

\[
f(t)=\sum_{i\ge m}a_i\,t^{i/p}\qquad(0<t<\epsilon'),
\]

with absolute convergence. If \(f\) is bounded, \(a_i=0\) for \(i<0\); if moreover \(f(t)\to0\), also \(a_0=0\). If \(f_1,\ldots,f_k\) are bounded and written with a common \(p\), each function \(\tau\mapsto f_l(\tau^{2p})\) agrees, for \(0<|\tau|<\epsilon'^{1/(2p)}\), with a convergent power series in \(\tau\).

**Proof.** Here the base is a point, so the base functions are constants. Global preparation, with a decomposition of \(\mathbb R\) compatible with \((0,\epsilon)\), gives an interval \((0,\epsilon'')\) on which \(f=a|t-\theta|^r\Psi(\beta,u|t-\theta|^{1/s},v|t-\theta|^{-1/s})\). The point \(\theta\) does not lie in \((0,\epsilon'')\). If \(a=0\), the zero series works.

If \(\theta=0\), the argument \(vt^{-1/s}\) stays bounded as \(t\downarrow0\) only if \(v=0\). The point \((\beta,0,0)\) is a limit of arguments, so \(\Psi\) is analytic near it and \(\Psi(\beta,W,0)=\sum_{j\ge0}b_jW^j\) for small \(|W|\). With \(W=ut^{1/s}\),

\[
f(t)=\sum_{j\ge0}a\,b_j\,u^j\,t^{r+j/s}
\]

for small \(t\). Choose \(p\) divisible by \(s\) and by the denominator of \(r\); every exponent is then an integer divided by \(p\), and only finitely many are negative.

If \(\theta\ne0\), shrink the interval so that \(t<|\theta|/2\). Then \(|t-\theta|=|\theta|(1-\operatorname{sign}(\theta)\,t/|\theta|)\) is analytic and positive on a neighbourhood of \(t=0\), and so are its rational powers. The argument map extends analytically to \(t=0\) with its value at \(0\) in the closure of the argument image, where \(\Psi\) is analytic. So \(f\) extends analytically across \(0\) and is given by its Taylor series near \(0\); take \(p=1\).

A lowest nonzero term \(a_mt^{m/p}\) with \(m<0\) would make \(f\) unbounded, and \(f(t)\to0\) forces \(a_0=0\). Finally let \(p\) be common and \(t=\tau^{2p}\). For \(\tau\ne0\), \((\tau^{2p})^{i/p}=|\tau|^{2i}=\tau^{2i}\), so \(f_l(\tau^{2p})=\sum_{i\ge0}a_i\tau^{2i}\) for both signs of \(\tau\). The exponents produced by the substitution must be even, and an even substitution exponent alone does not suffice. For instance \(t^{1/2}\) becomes \(|\tau|\) under \(t=\tau^2\), but \(\tau^2\) under \(t=\tau^4\). \(\square\)

### A compact analytic Taylor observation

The parameter versions need Taylor expansions with a radius that is uniform over a compact parameter set.

**Uniform Taylor lemma.** Let \(H(b,T)\) be real analytic on a neighbourhood of \(K_0\times\{0\}\), where \(K_0\subset\mathbb R^k\) is compact. There are \(\delta>0\) and an open neighbourhood \(U\) of \(K_0\) such that the functions \(H_j(b)=\frac1{j!}\partial_T^jH(b,0)\) are analytic on \(U\) and

\[
H(b,T)=\sum_{j\ge0}H_j(b)\,T^j\qquad(b\in U,\ |T|<\delta),
\]

uniformly on compact subsets of \(U\) for \(|T|\le\delta/2\).

**Proof.** Around a point \((b_0,0)\), \(H\) extends holomorphically to a complex product neighbourhood \(D\times\{|T|<3r\}\). Shrinking, we may assume the extension is bounded by \(M\) on \(D\times\{|T|\le2r\}\). For real \(b\in D\), Cauchy's estimate in the variable \(T\) gives \(|H_j(b)|\le M(2r)^{-j}\). So the series converges to \(H(b,T)\) for \(|T|<2r\), uniformly for \(|T|\le r\) on such \(b\). The coefficients \(H_j\) are derivatives of the real function \(H\) and do not depend on the chosen extension. Finitely many of the real neighbourhoods \(D\cap\mathbb R^k\) cover \(K_0\); let \(U\) be their union and \(\delta\) the smallest of their radii \(r\). \(\square\)

For example, \((1-\sigma T)^q\), with \(\sigma=\pm1\), \(q\in\mathbb Q\) and the real branch equal to \(1\) at \(T=0\), is analytic for real \(|T|<1\) and can be used in such expansions.

### Continuous functions with parameters

**Puiseux theorem with continuity.** Let \(A\subset\mathbb R^n\) be globally subanalytic, let \(U\) be a relatively open neighbourhood of \(A\times\{0\}\) in \(A\times[0,\infty)\), and let \(f:U\to\mathbb R\) be continuous and globally subanalytic. Then \(A\) is the union of finitely many disjoint analytic manifolds \(C\), each globally subanalytic, and there is an even integer \(p\ge2\) such that, for each \(C\), the function \((x,\tau)\mapsto f(x,\tau^p)\) is real analytic on a neighbourhood of \(C\times\{0\}\) in \(C\times\mathbb R\).

The width of that neighbourhood in \(\tau\) may vary with \(x\).

**Proof.** Order the coordinates as \((x,t)\) and take a cylindrical decomposition of \(\mathbb R^{n+1}\) in which \(U\) and \(A\times\{0\}\) are unions of cells and \(f\) is reduced on the cells inside \(U\). Above a base cell \(C\subset A\), the zero section \(C\times\{0\}\) is one of the cells. The cell directly above it is a band \(0<t<\eta(x)\); it lies in \(U\), because every point \((x,0)\) has points \((x,t)\in U\) with small \(t>0\), and membership in \(U\) is constant on cells. Replace an infinite \(\eta\) by \(1\). On this band

\[
f(x,t)=a(x)\,|t-\theta(x)|^r\,\Psi\bigl(\beta(x),\,u(x)|t-\theta|^{1/s},\,v(x)|t-\theta|^{-1/s}\bigr),
\]

with base data analytic on \(C\). After refining the base, \(a\) vanishes identically or nowhere, and \(\theta\) vanishes identically or has a fixed sign. Since the graph of \(\theta\) misses the band, \(\theta>0\) forces \(\theta\ge\eta\).

If \(a=0\), then \(f=0\) on the band and, by continuity, at \(t=0\).

If \(a\ne0\) and \(\theta=0\), then for fixed \(x\) the boundedness of \(v(x)t^{-1/s}\) as \(t\downarrow0\) forces \(v(x)=0\). As \(t\downarrow0\) the argument tends to \((\beta(x),0,0)\), where \(\Psi\) is not zero, so continuity of \(f\) at \((x,0)\) forces \(r\ge0\). Let \(p\) be a common multiple of \(2s\) and of twice the denominator of \(r\). Then \(p/s\) and \(pr\) are even nonnegative integers, and for \(\tau\ne0\) of either sign

\[
f(x,\tau^p)=a(x)\,\tau^{pr}\,\Psi\bigl(\beta(x),\,u(x)\tau^{p/s},\,0\bigr).
\]

The right side is analytic in \((x,\tau)\) near every \((x_0,0)\), because \((\beta(x_0),0,0)\) lies in the closure of the argument image. At \(\tau=0\) it equals \(\lim_{t\downarrow0}f(x,t)=f(x,0)\).

If \(a\ne0\) and \(\theta\ne0\), put \(T=|\theta|\) and \(\sigma=\operatorname{sign}\theta\). For \(0\le t<T/2\),

\[
|t-\theta|=T(x)\Bigl(1-\sigma\frac t{T(x)}\Bigr),
\]

which is analytic and positive in \((x,t)\) near every \((x_0,0)\), for \(t\) of both signs. Substituting it into the reduced form gives an analytic function near \((x_0,0)\). The value of the argument at \((x_0,0)\) is the limit of actual arguments, so it lies in the domain of \(\Psi\). This function agrees with \(f\) for \(t>0\) and, by continuity, at \(t=0\). Composing with \(t=\tau^p\) keeps it analytic.

Choose one \(p\) for the finitely many cells. On each cell, the local analytic functions found near different points agree where their domains overlap, since they agree for \(t>0\). Their union is the required neighbourhood. \(\square\)

### Parameterized Laurent–Puiseux series without continuity

**Laurent–Puiseux theorem.** Suppose that \(A\subset\mathbb R^n\), \(\zeta:A\to(0,\infty)\) and \(f:E\to\mathbb R\) are globally subanalytic, where \(E=\{(x,t):x\in A,\ 0<t<\zeta(x)\}\). Then \(A\) is the union of finitely many disjoint analytic manifolds \(C\), each globally subanalytic, and there is an integer \(p\ge1\) with the following property. For each \(C\) there are a continuous globally subanalytic function \(\xi:C\to(0,\infty)\) with \(\xi\le\zeta\), an integer \(m_C\), and analytic functions \(a_i\) on \(C\) such that

\[
f(x,t)=\sum_{i\ge m_C}a_i(x)\,t^{i/p}\qquad(x\in C,\ 0<t<\xi(x)).
\]

On compact subsets of \(C\) the series converges uniformly in \(t^{1/p}\) for \(t\) below a smaller positive bound. Neither continuity of \(f\) nor a positive radius uniform over an unbounded cell is assumed.

**Proof.** Prepare \(f\) on a cylindrical decomposition compatible with \(E\) and \(A\times\{0\}\), refining so that \(\zeta\) is analytic on the base cells. As before, over each base cell \(C\subset A\) the cell directly above \(t=0\) is a band \(0<t<\eta(x)\) inside \(E\), with \(\eta\le\zeta\) positive and analytic. On it \(f\) has a reduced form as in the previous proof, and we separate the same cases. If \(a=0\), use the zero series and \(\xi=\eta/2\).

Suppose \(\theta=0\) and \(a\ne0\). Again \(v=0\), but now \(r\) may be negative. Let \(K'\) be the closure of \(\beta(C)\), which is compact since \(\beta\) is bounded. The set \(K'\times\{0\}\times\{0\}\) lies in the closure of the argument image. Indeed, if \(\beta(x_j)\to b\), choose \(t_j<\eta(x_j)\) so small that \(u(x_j)t_j^{1/s}\to0\). So \(\Psi(b,W,0)\) is analytic near \(K'\times\{0\}\). The uniform Taylor lemma gives \(\delta>0\) and analytic \(b_j\) near \(K'\) with \(\Psi(b,W,0)=\sum_jb_j(b)W^j\) for \(|W|<\delta\). Put

\[
\xi(x)=\min\Bigl\{\frac{\eta(x)}2,\ \Bigl(\frac{\delta}{2(1+u(x))}\Bigr)^s\Bigr\}.
\]

For \(0<t<\xi(x)\), \(u(x)t^{1/s}<\delta/2\), and

\[
f(x,t)=\sum_{j\ge0}a(x)\,b_j(\beta(x))\,u(x)^j\,t^{r+j/s}.
\]

The coefficients are analytic on \(C\), and \(\xi\) is positive, continuous and globally subanalytic. Writing the exponents over a common denominator gives the stated indexing, with finitely many negative exponents.

Suppose \(\theta\ne0\), with \(T=|\theta|\) and \(\sigma=\operatorname{sign}\theta\). Put \(U_0=uT^{1/s}\) and \(V_0=vT^{-1/s}\). For fixed \(x\), \((\beta,U_0,V_0)(x)\) is the limit of the argument as \(t\downarrow0\). Hence the map \((\beta,U_0,V_0)\) is bounded, and approximating each value by actual arguments shows that the closure of its image lies in the closure of the argument image. The function

\[
\mathcal H(b,U,V,\tau)=(1-\sigma\tau)^r\,\Psi\bigl(b,\,U(1-\sigma\tau)^{1/s},\,V(1-\sigma\tau)^{-1/s}\bigr)
\]

is analytic near that compact closure times \(\{0\}\). The uniform Taylor lemma gives \(\delta\in(0,\tfrac12)\) and analytic coefficients \(\mathcal H_j\). With \(\xi(x)=\min\{\eta(x)/2,\ \delta T(x)/2\}\) and \(0<t<\xi(x)\), we have \(|t-\theta|=T(1-\sigma t/T)\), and

\[
f(x,t)=\sum_{j\ge0}a(x)\,T(x)^{r-j}\,\mathcal H_j\bigl(\beta(x),U_0(x),V_0(x)\bigr)\,t^j.
\]

This is a power series in \(t\) with analytic coefficients.

Choose one \(p\) for the finitely many cells. The radii \(\xi\) are globally subanalytic by their formulas. On a compact subset of \(C\) the base factors are bounded, and the geometric bounds of the uniform Taylor lemma give uniform convergence after decreasing the radius. \(\square\)

The [curve-selection and Łojasiewicz reading](../curve-selection-and-lojasiewicz.html) uses the cell theorem, the complement theorem and the one-variable Puiseux theorem proved here.

## Dimension, fibrewise closure and the frontier

The results below are those of §2.3 of Valette's notes cited at the head of this reading. In this section **definable** means globally subanalytic, in the projective product convention above. Closures and interiors are taken in the ambient Euclidean space, and the Boolean and projection calculus shows that they are definable. Besides that calculus, the proofs use the cell theorem, the bounded-chart comparison and [definable choice](../curve-selection-and-lojasiewicz.html#definable-choice). They use no resolution, uniformization or regular-locus theorem.

Every cell of a cylindrical decomposition is an analytic manifold. Its dimension is the number of bands used to build it: a graph over a cell has the dimension of that cell, and a band over it has one more. If a cell has dimension \(k\), forgetting its graph coordinates maps it homeomorphically onto an open set \(V\subset\mathbb R^k\). The inverse \(\sigma:V\to\mathbb R^n\) is analytic, because each graph coordinate is an analytic function of the coordinates before it.

For a nonempty definable set \(A\subset\mathbb R^n\) and a finite cylindrical decomposition into analytic cells compatible with \(A\), put

\[
\dim A=\max\{\dim C:\ C\text{ a cell of the decomposition with }C\subset A\},
\tag{D1}
\]

and \(\dim\varnothing=-1\). We also use the frontier and the topological boundary,

\[
\operatorname{fr}A=\overline A\setminus A,\qquad \partial A=\overline A\setminus\operatorname{int}A.
\tag{D2}
\]

They are different sets, and the strict inequality of this section concerns the frontier.

### Why cell dimension is intrinsic

**Covering lemma.** Let \(U\subset\mathbb R^d\) be a nonempty open set, and let \(\varphi_i:V_i\to\mathbb R^d\), \(i=1,2,\ldots\), be continuously differentiable maps on open sets \(V_i\subset\mathbb R^{k_i}\) with \(k_i<d\). Then \(U\) is not contained in \(\bigcup_i\varphi_i(V_i)\).

**Proof.** Each \(V_i\) is a countable union of closed cubes contained in it. Let \(Q\subset V_i\) be such a cube, of side \(\ell\), put \(k=k_i\), and let \(L\) bound the norm of the derivative of \(\varphi_i\) on \(Q\). By the mean value inequality, \(\varphi_i\) is \(L\)-Lipschitz on the convex set \(Q\). Cut \(Q\) into \(N^k\) cubes of side \(\ell/N\). Each has diameter \(\sqrt k\,\ell/N\), so its image lies in a cube of side \(2L\sqrt k\,\ell/N\). The total volume of these cubes is

\[
N^k\Bigl(\frac{2L\sqrt k\,\ell}N\Bigr)^d=(2L\sqrt k\,\ell)^d\,N^{k-d}\longrightarrow0\qquad(N\to\infty).
\tag{D3}
\]

So \(\varphi_i(Q)\) has outer Lebesgue measure zero, and so does the countable union of all these images. The nonempty open set \(U\) has positive measure. \(\square\)

Only volumes of cubes enter here; Sard's theorem is not needed.

For differentiable manifolds \(D\subset C\) in one Euclidean space, \(\dim D\le\dim C\): a tangent vector to \(D\) is the velocity of a curve in \(D\), hence of a curve in \(C\), so \(T_pD\subset T_pC\).

Now take two cell decompositions compatible with \(A\) and, by the cell theorem, a common refinement. Let \(C\) be a cell of the first decomposition, of dimension \(d\), and let \(\pi:C\to V\) forget its graph coordinates. Every refined cell \(D\subset C\) has \(\dim D\le d\). The maps \(\pi\circ\sigma_D\), where \(\sigma_D\) is the analytic inverse belonging to \(D\), are continuously differentiable because \(\pi\) is the restriction of a linear map, and their images cover the open set \(V\subset\mathbb R^d\). By the covering lemma, at least one refined cell in \(C\) has dimension \(d\). Refinement therefore leaves the maximum in (D1) unchanged, and since any two decompositions have a common refinement, the maximum is independent of the decomposition. Taking a decomposition compatible with all the sets involved, we obtain

\[
A\subset B\ \Longrightarrow\ \dim A\le\dim B,\qquad \dim(A_1\cup\cdots\cup A_s)=\max_i\dim A_i.
\tag{D4}
\]

The same argument computes the dimension of a definable set \(M\) that is a differentiable manifold of dimension \(d\) embedded in \(\mathbb R^n\). Its cells have dimension at most \(d\). Near a point \(p\in M\), the orthogonal projection onto the tangent space \(T_pM\) maps a neighbourhood of \(p\) in \(M\) homeomorphically onto an open subset of \(T_pM\cong\mathbb R^d\); this follows from the inverse function theorem applied to a local parametrization. Compose the maps \(\sigma_D\) of the cells meeting that neighbourhood with this linear projection. The covering lemma shows that some cell has dimension \(d\), so \(\dim M=d\) in the sense of (D1). Consequently a permutation of the coordinates, or a definable differentiable diffeomorphism between open sets, carries each cell of \(A\) to a definable manifold of the same dimension, and by (D4) it preserves \(\dim A\).

Since a nonempty open subset of \(\mathbb R^m\) contains an open box, which is a cell of dimension \(m\), (D4) shows that a definable subset of \(\mathbb R^m\) of dimension less than \(m\) has empty interior.

### Images, graphs and fibre dimensions

**Image inequality.** Let \(f:A\to\mathbb R^p\) be a definable map, not necessarily continuous, and let \(E\subset A\) be definable. Then

\[
\dim f(E)\le\dim E.
\tag{D5}
\]

**Proof.** Let \(G\subset\mathbb R^{n+p}\) be the graph of \(f|_E\), and take a cylindrical decomposition of \(\mathbb R^{n+p}\), with the coordinates of \(\mathbb R^n\) first, in which \(G\) is a union of cells. The projection \(E\) of \(G\) is then a union of base cells. Fix a base cell \(D\subset E\). Every cell lying above \(D\) has points over each point of \(D\). A cell built with a band in one of the output coordinates has infinitely many points over each point of \(D\), and two different cells of \(G\) above \(D\) would give two points of \(G\) over one point of \(D\). As \(G\) has exactly one point over each point of \(E\), the part of \(G\) above \(D\) is a single cell built from graphs only, of dimension \(\dim D\). Hence \(\dim G=\dim E\). Now permute the coordinates so that the output coordinates come first, which does not change \(\dim G\), and take a cylindrical decomposition compatible with \(G\). The projection of a cell onto the output coordinates is one of its base cells, of dimension at most that of the cell, and \(f(E)\) is the union of the projections of the cells in \(G\). Therefore \(\dim f(E)\le\dim G=\dim E\). \(\square\)

Applied to a definable bijection and to its inverse, (D5) shows that definable bijections preserve dimension, whether or not they are continuous. For example, the map that forgets the graph coordinates of a cell is a definable bijection onto an open subset of \(\mathbb R^d\), where \(d\) is the dimension of the cell.

For \(A\subset\mathbb R^{m+n}\), \(t\in\mathbb R^m\) and \(B\subset\mathbb R^m\), write

\[
A_t=\{x\in\mathbb R^n:(t,x)\in A\},\qquad A_B=A\cap(B\times\mathbb R^n).
\tag{D6}
\]

Take a cylindrical decomposition compatible with \(A\), with the coordinates \(t\) first, and let \(B\) be a base cell. Over each \(t\in B\), the fibres of the cells above \(B\) are the cells of a cylindrical decomposition of \(\mathbb R^n\). A cell above \(B\) has dimension \(\dim B\) plus the number of its bands in the coordinates \(x\), and that number is the dimension of its fibre. Since \(A_t\) is the union of the fibres of the cells of \(A_B\),

\[
\dim A_B=\dim B+\dim A_t\qquad(t\in B,\ A_t\ne\varnothing).
\tag{D7}
\]

So \(\dim A\) is the largest of the numbers \(\dim B+\dim A_t\) over the base cells \(B\) with \(A_B\ne\varnothing\). If \(\dim A\le m\), then over each base cell of dimension \(m\) every fibre of \(A\) has finitely many points, possibly none. Every definable set of dimension \(0\) is finite.

### Closure commutes with fibres after partitioning the base

For most parameters, taking a fibre commutes with taking a closure. The exceptional parameters form a set of smaller dimension, and a finite partition of the base removes their effect.

**Generic closure lemma.** Let \(A\subset\mathbb R^{m+n}\) be definable. The set \(X\) of parameters \(t\in\mathbb R^m\) with \((\overline A)_t\ne\overline{A_t}\) is definable and has dimension less than \(m\). In particular its complement \(B\) is a dense definable set with

\[
(\overline A)_t=\overline{A_t}\qquad(t\in B).
\tag{D8}
\]

**Proof.** For \(m=0\) the parameter space is a point and there is nothing to prove, so let \(m\ge1\). If \(x_j\in A_t\) converge to \(x\), then \((t,x_j)\to(t,x)\); hence \(\overline{A_t}\subset(\overline A)_t\) for every \(t\). Let \(S\) be the set of triples \((t,a,r)\) with \(r>0\), \((t,a)\in\overline A\), and no point of \(A_t\) within distance \(r\) of \(a\). It is definable by the Boolean and projection calculus, and \(X\) is its projection to \(\mathbb R^m\). Suppose \(\dim X=m\). Then \(X\) contains a cell of dimension \(m\), which is open in \(\mathbb R^m\). Definable choice gives a definable map \(t\mapsto(a(t),r(t))\) on this cell with \((t,a(t),r(t))\in S\). As in the proof of (D5), a cylindrical decomposition compatible with the graph of this map shows that \(a\) and \(r\) are analytic on each cell of a finite decomposition of the open cell. One of these cells, \(C\), has dimension \(m\) and is therefore open. The set

\[
\{(t,x):t\in C,\ |x-a(t)|<r(t)\}
\tag{D9}
\]

is open in \(\mathbb R^{m+n}\), contains the point \((t,a(t))\) of \(\overline A\) for each \(t\in C\), and does not meet \(A\). This is impossible, so \(\dim X<m\). Then \(X\) has empty interior, and \(B\) is dense. \(\square\)

**Partitioned closure lemma.** For a suitable partition \(\mathcal P\) of \(\mathbb R^m\) into finitely many definable sets,

\[
\bigl(\overline{A_B}\bigr)_t=\overline{A_t}\qquad(B\in\mathcal P,\ t\in B).
\tag{D10}
\]

On the left, the closure is taken of the restricted family \(A_B\). At a parameter in \(X\) it can differ from the fibre of \(\overline A\).

**Proof.** We induct on \(m\); for \(m=0\) take \(\mathcal P=\{\mathbb R^0\}\). Decompose \(\mathbb R^m\) into cells compatible with the set \(X\) of the generic closure lemma. For a cell \(C\) outside \(X\) and \(t\in C\),

\[
\overline{A_t}\subset\bigl(\overline{A_C}\bigr)_t\subset(\overline A)_t=\overline{A_t}.
\tag{D11}
\]

A cell \(C\subset X\) has some dimension \(d<m\). Let \(\varpi:C\to V\) forget its graph coordinates; it is a definable homeomorphism onto an open set \(V\subset\mathbb R^d\). Put \(A'=\{(\varpi(t),x):(t,x)\in A_C\}\subset\mathbb R^{d+n}\). The induction hypothesis gives a finite definable partition \(\mathcal P'\) of \(\mathbb R^d\) with (D10) for \(A'\), and the sets \(\varpi^{-1}(P')\), \(P'\in\mathcal P'\), partition \(C\). Let \(B=\varpi^{-1}(P')\) be one of them and \(t\in B\). Because \(t\in C\), a point \((t,x)\) lies in \(\overline{A_B}\) exactly when it lies in the closure of \(A_B\) relative to \(C\times\mathbb R^n\). The homeomorphism \((t,x)\mapsto(\varpi(t),x)\) of \(C\times\mathbb R^n\) onto \(V\times\mathbb R^n\) carries \(A_B\) onto \(A'_{P'}\) and relative closures onto relative closures. Since \(\varpi(t)\in V\), the fibre at \(\varpi(t)\) of the closure of \(A'_{P'}\) relative to \(V\times\mathbb R^n\) is its ambient fibre, which by the choice of \(\mathcal P'\) is \(\overline{A'_{\varpi(t)}}=\overline{A_t}\). So (D10) holds on \(B\). The cells outside \(X\) and the sets \(\varpi^{-1}(P')\) form the required partition. \(\square\)

The property (D10) passes to definable subsets of the pieces. If \(C'\subset B\) and \(t\in C'\), then \(\overline{A_t}\subset(\overline{A_{C'}})_t\subset(\overline{A_B})_t=\overline{A_t}\). So the partition may be refined at will.

### Strict frontier decrease

**Frontier theorem.** Let \(A\) be a nonempty definable set. Then

\[
\dim(\overline A\setminus A)<\dim A,\qquad \dim\overline A=\dim A.
\tag{D12}
\]

For \(A=\varnothing\) both dimensions are \(-1\), and only the equality holds.

**Proof.** Let \(k=\dim A\). If \(k=0\), \(A\) is finite, hence closed, and its frontier is empty. Let \(k\ge1\), and suppose that \(\dim\operatorname{fr}A\ge k\). The definable set \(\operatorname{fr}A\) then contains a cell \(D\) of some dimension \(m\ge k\). After a permutation of the coordinates, which changes no dimension, \(D\) is the graph \(\{(t,g(t)):t\in V_D\}\) of an analytic map \(g:V_D\to\mathbb R^{n-m}\) on an open set \(V_D\subset\mathbb R^m\). Fix \(t_0\in V_D\) and a bounded open box \(W=W_1\times W_2\) around \((t_0,g(t_0))\) with \(\overline{W_1}\subset V_D\). The map \(\Phi(t,x)=(t,x-g(t))\) is analytic near \(\overline W\). Its graph over \(\overline W\) is a bounded semianalytic set whose closure lies where it is semianalytic, so by the bounded-chart comparison \(\Phi|_W\) is definable. It is injective, with inverse \((t,y)\mapsto(t,y+g(t))\), and maps \(W\) homeomorphically onto an open set. Let \(A'=\Phi(A\cap W)\); by (D4) and (D5), \(\dim A'\le k\). The set \(V=\{t\in W_1:g(t)\in W_2\}\) is open and contains \(t_0\), and \(\Phi(D\cap W)=V\times\{0\}\). Every point of \(D\cap W\) lies in \(\overline A\setminus A\), and since \(W\) is open it is a limit of points of \(A\cap W\). Applying \(\Phi\),

\[
V\times\{0\}\subset\overline{A'}\setminus A'.
\tag{D13}
\]

Decompose \(A'\) cylindrically with the coordinates \(t\) first. The base cells of dimension less than \(m\), together with the exceptional set of the generic closure lemma for \(A'\), form a set of dimension less than \(m\), by (D4). It has empty interior, so it does not contain \(V\). Choose \(t\in V\) outside it. Then \(t\) lies in a base cell of dimension \(m\), and if \(A'_t\ne\varnothing\), (D7) gives \(\dim A'_t\le k-m\le0\). So \(A'_t\) is finite, hence closed, and since \(t\) is not exceptional,

\[
0\in(\overline{A'})_t=\overline{A'_t}=A'_t,
\tag{D14}
\]

contrary to (D13). Hence \(\dim\operatorname{fr}A<k\). Finally \(\overline A=A\cup\operatorname{fr}A\), so (D4) gives \(\dim\overline A=k\). \(\square\)

Consequently a definable set \(A\subset\mathbb R^n\) is nowhere dense exactly when \(\dim A<n\). If \(\dim A=n\), then \(A\) contains a cell of dimension \(n\), which is open. If \(\dim A<n\), then \(\dim\overline A<n\) by (D12), so \(\overline A\) has empty interior. The topological boundary is always small compared with the ambient space:

\[
\dim\partial A<n.
\tag{D15}
\]

Indeed an \(n\)-dimensional cell is an open subset of \(\mathbb R^n\), so one lying in \(A\) lies in \(\operatorname{int}A\); hence \(A\setminus\operatorname{int}A\) contains no such cell. The rest of \(\partial A\) lies in \(\operatorname{fr}A\), whose dimension is less than \(\dim A\le n\). In contrast, \(\dim\partial A<\dim A\) can fail: a line in \(\mathbb R^2\) is its own boundary, while its frontier is empty.

### Fibrewise continuous functions become jointly continuous on base pieces

**Joint continuity theorem.** Let \(A\subset\mathbb R^{m+n}\) and \(f:A\to\mathbb R\) be definable, and suppose that every fibre function \(f_t=f(t,\cdot)\) is continuous on \(A_t\). Then \(\mathbb R^m\) can be divided into finitely many definable pieces \(B\), forming a partition \(\mathcal P\), so that each restriction \(f|_{A_B}\) is continuous.

**Proof.** The map

\[
\psi:\mathbb R\to(-1,1),\qquad \psi(s)=\frac s{\sqrt{1+s^2}},\qquad \psi^{-1}(u)=\frac u{\sqrt{1-u^2}},
\tag{D16}
\]

is an analytic homeomorphism with semialgebraic graph. The composite \(h=\psi\circ f\) is definable, takes values in \((-1,1)\), and has continuous fibre functions \(h_t\). Use (D10) for its graph \(\Gamma_h\subset\mathbb R^{m+n+1}\), with \(t\) as the parameters. Let \((t_j,x_j)\in A_B\) converge to \((t,x)\in A_B\), and let \(u\) be a limit point of the bounded sequence \(h(t_j,x_j)\). Then \((t,x,u)\) lies in the closure of \((\Gamma_h)_B\), so

\[
(x,u)\in\bigl(\overline{(\Gamma_h)_B}\bigr)_t=\overline{\Gamma_{h_t}}.
\tag{D17}
\]

As \(h_t\) is continuous on \(A_t\), its graph is closed in \(A_t\times\mathbb R\), and \(x\in A_t\); hence \(u=h_t(x)\). A bounded sequence with a single limit point converges, so \(h\) is continuous on \(A_B\), and so is \(f=\psi^{-1}\circ h\). \(\square\)

Two features of the proof matter. The graph of \(h_t\) is closed only relative to \(A_t\times\mathbb R\), since \(A_t\) need not be closed. The bounded transform must be injective with a continuous inverse: continuity of a non-injective composite does not give continuity of \(f\) (Exercise D3).

### Transfer to local analytic geometry

Let \(A\) be a locally subanalytic subset of an analytic manifold, and let \(x\) be a point of the manifold. In an analytic chart around \(x\), choose a closed box \(K\) with \(x\) in its interior, inside a neighbourhood on which \(A\) has a subanalytic description. By the bounded-chart comparison, \(A\cap K\) is globally subanalytic, and so is \(A\cap B\) for every open ball \(B\subset K\). On the interior of \(K\) the frontier of \(A\) agrees with the frontier of \(A\cap K\), so it is locally subanalytic too.

The **dimension of the germ** of \(A\) at \(x\), written \(\dim_xA\), is the eventual value of \(\dim(A\cap B_\rho)\) for the balls \(B_\rho\) of radius \(\rho\) around \(x\), as \(\rho\to0\). By (D4) these integers do not increase as \(\rho\) decreases, so they are eventually constant, and any other neighbourhood basis of \(x\) gives the same value. An analytic change of chart is definable on closed boxes, so (D5), applied to it and to its inverse, shows that \(\dim_xA\) does not depend on the chart.

Suppose that \(x\in\overline A\), so that the germ of \(A\) at \(x\) is nonempty. Choose \(\rho\) so small that \(\dim(A\cap B_\rho)=\dim_xA\) and \(\dim(\operatorname{fr}A\cap B_\rho)=\dim_x\operatorname{fr}A\). A point of \(\operatorname{fr}A\cap B_\rho\) is a limit of points of \(A\) lying in the open ball \(B_\rho\), so \(\operatorname{fr}A\cap B_\rho\subset\operatorname{fr}(A\cap B_\rho)\). Applying (D12) to the nonempty set \(A\cap B_\rho\) gives

\[
\dim_x(\overline A\setminus A)<\dim_xA.
\tag{D18}
\]

An empty frontier germ has dimension \(-1\). This strict decrease makes the successive adjunction of lower-dimensional frontier pieces terminate in relative triangulation arguments.

In the same way, (D5) applies to an analytic or locally subanalytic map once its graph is restricted to a closed box in a chart, where it becomes globally subanalytic. For a proper map, the preimage of a compact box in the target is compact and is covered by finitely many such boxes, which gives the inequality for the part of the image lying in that target box. The localization is needed because a locally subanalytic graph need not be globally subanalytic.

The analytic regular-locus, uniformization and constant-rank partition theorems do not follow from these dimension counts; they are separate results.

### Exercises with complete solutions

**Exercise D1 (basic: circle and disc).** For the unit circle \(S^1\subset\mathbb R^2\), and then for the open unit disc, determine the frontier, the topological boundary and their dimensions. Which strict inequalities of this section does each example illustrate?

**Solution.** The circle is closed and has empty interior in \(\mathbb R^2\). So \(\operatorname{fr}S^1=\varnothing\), of dimension \(-1\), while \(\partial S^1=S^1\) has dimension \(1=\dim S^1\). The frontier inequality (D12) holds, but the boundary is not smaller than the set. The open disc is two-dimensional and equals its own interior. Removing it from the closed disc, its closure, leaves the circle, so the frontier and the boundary of the disc coincide and are one-dimensional. In both cases the boundary has dimension less than \(2\), as (D15) asserts.

**Exercise D2 (intermediate: one exceptional parameter).** Let \(A=\{(t,x):t>0,\ x=t^2\}\subset\mathbb R^2\). Find all \(t\) with \((\overline A)_t\ne\overline{A_t}\), give a partition of the parameter line satisfying (D10), and explain why (D10) restricts the family before taking closures.

**Solution.** The closure of \(A\) is \(A\cup\{(0,0)\}\). For \(t>0\), \(A_t=\{t^2\}=(\overline A)_t\); for \(t<0\) both fibres are empty. At \(t=0\), \(A_0=\varnothing\) while \((\overline A)_0=\{0\}\), so \(0\) is the only exceptional parameter. The partition \((-\infty,0),\ \{0\},\ (0,\infty)\) satisfies (D10). Over \((-\infty,0)\) and over \(\{0\}\) the set \(A\) has no points, so the restricted families and their closures are empty; over \(\{0\}\) this agrees with \(\overline{A_0}=\varnothing\). On \((0,\infty)\) the closure of \(A_{(0,\infty)}\) adds only the point \((0,0)\), which lies over the parameter \(0\), outside the piece. Taking the closure of the whole family first would put \(0\) into the fibre over the parameter \(0\), although \(A_0\) is empty.

**Exercise D3 (advanced: continuity and the bounded transform).** Let \(f(t,x)=tx/(t^2+x^2)\) for \((t,x)\ne(0,0)\) and \(f(0,0)=0\). Show that every fibre function \(f_t\) is continuous, that \(f\) is not continuous, and find a partition of the \(t\)-line as in the joint continuity theorem. Then show that the transform \(s\mapsto s/(1+s^2)\) cannot replace (D16).

**Solution.** For \(t\ne0\) the denominator is positive for all \(x\), so \(f_t\) is continuous; \(f_0\) is identically zero. Along \(x=t\ne0\) we have \(f=\tfrac12\), while \(f(0,0)=0\), so \(f\) is not continuous at the origin. With the partition \((-\infty,0),\ \{0\},\ (0,\infty)\), on each open half-plane \(f\) is given by the same rational formula with positive denominator, and on the line \(t=0\) it is zero; so it is continuous on each piece. For the last claim, note that \(s\mapsto s/(1+s^2)\) takes the same value at \(s\) and at \(1/s\). Let \(A=\mathbb R\times\{0\}\subset\mathbb R^2\), and let \(g(0,0)=\tfrac13\) and \(g(t,0)=3\) for \(t\ne0\). Each fibre of \(A\) is a single point, so every fibre function of \(g\) is continuous, and \(g/(1+g^2)\) is constantly \(\tfrac3{10}\). The argument with this transform would therefore allow the single piece \(\mathbb R\), although \(g\) is discontinuous at \((0,0)\). The injectivity of (D16) and the continuity of its inverse are what carry continuity from \(h\) back to \(f\).