Let $\M$ be an o-minimal expansion of a dense linear order $(M,\lt)$, and let $n \in \NN$ be nonzero.

Inspired by this post, we now make the following

**Definition**

Let $\sigma \in \{0,1\}^n$ and set $\sigma’:= \sigma \rest{\{0,1\}^{n-1}}$. We say that a definable set $C \subseteq M^n$ is a **$\sigma$-cell** whenever the following holds:

- if $n=1$ and $\sigma(1) = 0$, then $C$ is a singleton set;
- if $n=1$ and $\sigma(1) = 1$, then $C$ is a nonempty, open, definable interval;
- if $n\gt 1$ and $\sigma(n) = 0$, then $C’:= \Pi_{n-1}(C)$ is a $\sigma’$-cell and $C$ is the graph of a definable, continuous $f:C’ \into M$;
- if $n\gt 1$ and $\sigma(n) = 1$, then $C’:= \Pi_{n-1}(C)$ is a $\sigma’$-cell and $C$ is the set $$(f,g)_{C’} := \set{(x,y) \in M^{n-1} \times M:\ x \in C’ \text{ and } f(x) \lt y \lt g(x)},$$ where either $f:C’ \into \{-\infty\}$ or $f:C’ \into M$ is continuous and definable, and either $g:C’ \into \{+\infty\}$ or $g:C’ \into M$ is continuous and definable, and where $f(x) \lt g(x)$ for all $x \in C’$.

Note that, if $C \subseteq M^n$ is a cell and $m \le n$, then $\Pi_m(C)$ is a cell and, for each $a \in M^m$, the fiber $C_a$ is a cell.

For the next exercises, let $\sigma \in \{0,1\}^n$ and $C \subseteq M^n$ be a $\sigma$-cell.

We call $C$ **open** if $\sigma(i) = 1$ for all $i$. For $\sigma$, we set $$\sum\sigma:= \sum_{i=1}^n \sigma(i).$$ We also associate to $\sigma$ the unique strictly increasing map $$\iota = \iota_\sigma:\set{1, \dots, \sum\sigma} \into \{1, \dots, n\}$$ such that $\sigma(i) = 1$ if and only if $i = \iota(j)$ for some $j$, and we set $$C^\sigma := \Pi_{\iota_\sigma}(C)$$ and $$\pi_\sigma:= \Pi_{\iota_\sigma} \rest{C}:C \into C^\sigma.$$

**Exercise 4**

- Show that $C$ is definably connected.
- Show that $C^\sigma$ is an open cell and the map $\pi_\sigma$ is a definable homeomorphism (with respect to the subspace topology on $C$).
- Show that $C$ is open iff $C$ is an open set iff $C$ is not sparse (hence the name “open cell”).

The decomposition picture in the plane described in this post generalizes as follows:

**Definition**

Let $\C$ be a finite collection of cells in $M^n$ and $U \subseteq M^n$. We call $\C$ a **cell decomposition of $U$** if $\C$ is a partition of $U$ and, if $n\gt 1$, the collection $$\Pi_{n-1}\C:= \set{\Pi_{n-1}(C):\ C \in \C}$$ is a cell decomposition of $\Pi_{n-1}(U)$. If $\C$ and $\D$ are cell decompositions of $U$, we call $\D$ a **refinement** of $\C$ if $\D$ is compatible with each $C \in \C$.

**Remark**

Let $\C$ be a cell decomposition of $U \subseteq M^{n+m}$ and $x \in M^n$. Then the collection $$\C_x:= \set{C_x:\ C \in \C}$$ is a cell decomposition of $U_x$.

Is part 2 of the exercise meant to be proved by induction on $n$, given a cell $C \subseteq M^{n}$? I have the following proof in mind, but I’m not entirely sure if it works.

Let $C^{\sigma}$ and $\pi_\sigma$ be as above. To show that $\pi_{\sigma}$ is a definable homeomorphism of $C \subseteq M^n$ onto $C^{\sigma}$, we proceed by induction on $n$. In the case $n=1$ there are two possibilities: If $C$ is a $1$-cell then it is open and so we can take $\pi_{\sigma}$ to be the identity on $C$. If $C$ is a $0$-cell then it is definably homeomorphic to the one-point space $M^{0}$ which is an open cell by convention [at least, I think this is the convention…] and so all cells $C$ in $M^1$ are definably homeomorphic via $\pi_{\sigma}$ to the open cell $C^{\sigma}$.

Now suppose the result holds for all lower values of $n$, where $n > 1$. If $\sigma(n) = 0$ then $C$ is the graph of a definable, continuous function $f : C’ \rightarrow M$, where $C’ = \Pi_{n-1}(C)$. In this case, $\Pi_{n-1} \restriction_{C} : C \rightarrow C’$ is a definable homeomorphism given by mapping $(x,y) \in \gr{f}$ to $x \in C’$; in particular, injectivity of $\Pi_{n-1}$ is given by the fact that $C$ is the graph of a function. By the inductive hypothesis, $\pi_\sigma$ is a definable homeomorphism from $C’$ onto the open cell $(C’)^{\sigma}$, and so we can take the composition $\pi_{\sigma} \circ \Pi_{n-1}$ to obtain the corresponding definable homeomorphism, also called $\pi_{\sigma}$, of $C$ onto $C^{\sigma}$.

If $\sigma(n) = 1$, then $C$ is an interval of the form $(f, g)_{C’}$ for functions $f$ and $g$ as above. If $C$ is an open cell then we can take $\pi_{\sigma}$ to be the identity on $C$, so suppose that $C$ is not open. Choose the first index $i \in \{1, \dots, n-1 \}$ such that $\sigma(i) = 0$ and define the projection $\Pi_{\hat{i}}$ to be the projection of $M^n$ onto the $n-1$ coordinates given by $\{1, \dots, \hat{i}, \dots, n-1 \}$, where $\hat{i}$ denotes the omission of the $i^{\mathrm{th}}$ coordinate. Then for $(x_1, \dots, x_{n-1}, y) \in (f,g)_{C’}$, we have $$\Pi_{\hat{i}} (x_1, \dots, x_{n-1}, y) = (x_1, \dots, \hat{x_i}, \dots, x_{n-1}, y)$$ and so $\Pi_{\hat{i}} \restriction_{C} \rightarrow C’$ is a definable homeomorphism [I think a bit of work needs to be done here, unless this claim is obvious and I’m not seeing it]. By the inductive hypothesis, $C’$ is definably homeomorphic to $(C’)^{\sigma}$ via $\pi_{\sigma}$ and so we can take the composition $\pi_{\sigma} \circ \Pi_{\hat{i}}$ to obtain the corresponding definable homeomorphism, also called $\pi_{\sigma}$, of $C$ onto $C^{\sigma}$.

The base step of the induction and the case $\sigma(n) = 0$ in the inductive step are fine. For the case $\sigma(n) = 1$, let $\sigma’$ be the restriction of $\sigma$ to $\{1, \dots, n-1\}$, and work with $\left(C’\right)^{\sigma’}$, an open cell.

Since $C$ is a $\sigma$-cell, there are definable, continuous functions $f,g:C’ \into M$ (or $\pm\infty$…) such that $f(x) \lt g(x)$ for $x \in C’$. Define $f’,g’:\left(C’\right)^{\sigma’} \into M$ by $$f'(y):= f\left(\left(\pi_{\sigma’}\right)^{-1}(y)\right)$$ and $$g'(y):= g\left(\left(\pi_{\sigma’}\right)^{-1}(y)\right).$$ Then $\left(f’,g’\right)_{\left(C’\right)^{\sigma’}}$ is an open cell, and it is equal to $C^\sigma$.