Let ${\cal M}$ be an expansion of a dense linear order $(M,<)$. Let $S \subseteq M^{n+m}$. For $x \in M^n$, we denote by $\displaystyle S_x := \{y \in M^m:\ (x,y) \in S\}$ the fiber of $S$ over $x$.
If all fibers $S_x$ are finite, we are interested in sufficient definable and topological conditions to guarantee that the cardinality of $S_x$ is constant as $x$ ranges over $\Pi_n(S)$. This will be used later in the proof of the cell decomposition theorem.
The conditions discussed here are related to the idea of a topological covering space, but they are more specific to the situation we encounter later.
Recall that the projection $\Pi_n|_S:S \longrightarrow M^n$ is a local homeomorphism if, for every $(x,y) \in S$, there are an open box $B \subseteq M^{n+m}$ containing $(x,y)$ and a continuous function $f:\Pi_n(B) \longrightarrow M^m$ such that $S \cap B = {\rm gr}(f)$.
If $\Pi_n|_S$ is a local homeomorphism, then $\Pi_n(S)$ is open.
We say that $S$ is locally bounded at $x \in M^n$ if there exists an open box $B \subseteq M^n$ containing $x$ such that $S \cap (B \times M^m)$ is bounded.
Exercise 1
Assume that ${\cal M}$ is definably complete, and let $S \subseteq M^{n+1}$ be definable such that $S_x$ is finite for all $x \in \Pi_n(S)$. Assume in addition that:
- $\Pi_n(S)$ is definably connected,
- $S$ is locally bounded at every $x \in \Pi_n(S)$,
- $S$ is closed in $\Pi_n(S) \times M$, and
- $\Pi_n|_S:S \longrightarrow M^n$ is a local homeomorphism.
Prove that $\left|S_x\right|$ is constant as $x$ ranges over $\Pi_n(S)$. Also, for any three of the conditions 1–4, find an example satisfying these three conditions where $\left|S_x\right|$ is not constant.
Here is an attempt of a proof but I am really not sure if it’s correct because I couldn’t prove the main step.
For $k\in \mathbb{N}$, define $S(k):= \{ x\in \Pi_n(S) \mid \lvert S_x \rvert =k \}$. If we prove that the $S(k)$’s are open sets, $\Pi_n(S)$ will be the union of disjoint open definable sets, by property 1, it will follow that $\Pi_n(S)$ is one the $S(k)$’s i.e. all the fibers have the same cardinality.
The $S(k)$’s are definable because if $S=\phi(M^{n+1})$ then $x\in S(k)$ if and only if $\exists y_1, \dots, \exists y_k, \bigwedge\limits_{i=1}^k \phi(x, y_i) \wedge \left( \forall z \phi(x,z) \rightarrow \bigvee\limits_{i=1}^k z=y_i \right) \wedge \bigwedge\limits_{1\leq i < j \leq k} \neg y_i=y_j$.
Let $x\in S(k)$ and let $y_1, \dots, y_k$ be the elements of $S_x$. By property 4, for all $x_i:=(x, y_i)$, there exists an open box $B_i \subset M^{n+1}$ containing $x_i$ and a continuous function $f_i: \Pi_n(B_i) \rightarrow M$ such that $S \cap B_i =gr(f_i)$.
Let $U:= \bigcap\limits_{i=1}^{k} \Pi(B_i)$ so that we obtain continuous functions $g_i:U\rightarrow V_i$ where the $V_i$'s (subsets of $M$) are disjoint open intervals such that $U\times V_i \subset B_i$ and $g_i:=\left.f_i\right|_U$ with $S\cap(U\times V_i)=gr(g_i)$.
By property 2, $S\cap (U\times M)$ is bounded and by property 3, $S$ is closed in $ \Pi_n(S)\times M$ so $S\cap (U\times M)$ is closed in $U\times M \subset \Pi_n(S)\times M$. Also, since $\mathcal{M}$ is definably complete, $S\cap (U\times M)$ is compact (this is the part I'm really not sure about, I read somewhere that closed and bounded implied compact in our topology if we have the least upper bound property).
If we repeat this process of all $w\in U$, we obtain that:
$$S\cap (U \times M) \subset \bigcup\limits_{w\in U} \bigcup\limits_{i=1}^{k_w} U_w \times V_{iw}.$$ By compactness $w_1, \dots, w_N \in U$ such that:
$$S\cap (U \times M) \subset \bigcup\limits_{j=1}^{N} \bigcup\limits_{i=1}^{k_{w_j}} U_{w_j} \times V_{iw_j}.$$
By taking the $U_{w_j}$ containing $x$ (so $k_{w_j}=k$) and setting:
$\mathcal{U}:=U_{w_j}$
$\mathcal{V}_i:=V_{iw_j}$ for $i\in \{1, \dots k \}$
we obtain that $S\cap (\mathcal{U} \times M) \subset \bigcup\limits_{i=1}^{k} \mathcal{U} \times \mathcal{V}_i $.
Hence, for $z\in \mathcal{U}$, we have $S_z \subset \bigcup\limits_{i=1}^{k} \mathcal{V}_i $. Also, $\{g_1(z), \dots, g_k(z) \} \subset S_z$ and has cardinality $k$ (the $\mathcal{V}_i$'s are disjoint) so $\vert S_z \vert =k$. Therefore, $\mathcal{U} \subset S(k)$ and $S(k)$ is open as desired.
It is not true that definable, closed and bounded implies compact, unless you are in the reals. There is a notion of “definable compactness” that is equivalent to “definable, closed and bounded” in the o-minimal setting (and maybe even the definably complete setting), but it cannot be defined in terms of finite subcoverings of open coverings.
I find it difficult to work with $S_k$ as you define it; instead, I suggest defining $S_k$ to be the set of all points whose fiber has at least $k$ elements, and showing that each such $S_k$ is both open and closed. The first part of your argument shows that this $S_k$ is open; you need to work a bit more to show that it is closed.
That is strange. Can’t you prove definable closed and bounded implies compact by the usual proof that closed boxes are compact on the reals? I was doing it yesterday and I thought the proof worked.
Hmm, can I see the proof?
Oh! I found the mistake. I was assuming that a non-definable set has a supremum.
Can you think of a counterexample in the field of Puiseux series?
Here is a counterexample in the field of Puiseux series (which I think is maybe messier than it needs to be). Take the closed interval $[0,\frac{1}{X}]$. It has an open covering $\{n,n+2: n=-1,0,1,2,\ldots \} \cup \{(\frac{1}{X^{n+1}}, \frac{2}{X^{n}}: n=1,2,3,\ldots \}$, which cannot have any finite subcovering since for example any finite subcovering cannot contain every natural number.
Here is the (incorrect) proof I posted before the Math & Stats server went down yesterday, for those who are interested. (I make no use of the definable completeness of $\mathcal{M}$, which is the first hint that the proof is incorrect.)
Let $\mathcal{M}$ be a definably complete expansion of a dense linear order $(M, <)$, and let $S \subseteq M^{n+1}$ be definable such that $S_{x}$ is finite for all $x \in \Pi_{n}(S)$. Fix $x_{0} \in \Pi_{n}(S)$; say $|S_{x_{0}}| = k$ for some $k \in \mathbb{N}$. Let $X = \{ x \in \Pi_{n}(S) : |S_{x}| = k \}$. Note that $X$ is non-empty since $x_{0} \in X$ by definition. Since $\Pi_{n}(S)$ is definably connected, to prove the result it suffices to show that $X$ is open and closed in $\Pi_{n}(S)$.
To show that $X$ is open in $\Pi_{n}(S)$ let $x \in X$ and let $y_{1}, \dots, y_{k} \in M$ be the $k$ points such that $(x, y_{i}) \in S$ for all $i$. Since $\Pi_{n} \restriction_{S}$ is a local homeomorphism, for each $(x, y_{i}) \in S$ we can find an open box $B_{i} \subseteq M^{n+1}$ such that $(x, y_{i}) \in B_{i}$ and $S \cap B_{i} = \mathrm{gr}(f_{i})$ for some continuous function $f_{i} : \Pi_{n}(B_{i}) \rightarrow M$. Furthermore, since $S$ is locally bounded at every point in $\Pi_{n}(S)$, we can find an open box $B$ containing $z$ such that $S \cap (B \times M)$ is bounded. Set $$ U = B \cap \left(\bigcap_{i = 1}^{k} \Pi_{n}(B_{i}) \right)$$ and note that $U$ is open. Then we can write the part of $S$ above $U$ as a disjoint union of graphs of bounded functions as follows:
$$S \cap (U \times M) = \bigcup_{i = 1}^{k} \mathrm{gr}(f_{i}).$$ Hence if $z$ is in $U$ then $z$ takes on exactly $k$ distinct values under the functions $f_{1}, \dots, f_{k}$ given by $f_{1}(z), \dots, f_{k}(z) \in M$, and so there are exactly $k$ points in the fiber of $S$ above $z$ for all $z \in U$, i.e. $x$ is an interior point of $X$.
To show that $X$ is closed in $\Pi_{n}(S)$, let $x \in \Pi_{n}(S)$ be a limit point of $X$. Note that if the fiber $\{ x \} \times M$ contains a limit point $(x, y)$ of $S$, then $S$ contains $(x, y)$ by the fact that $S$ is closed in $\Pi_{n}(S) \times M$, and so any such $y \in M$ belongs to the fiber $S_{x}$. [This previous step is a bit sketchy; my intent is to avoid counterexamples of the following kind: Let $S \subseteq \mathbb{R}^{2}$ be the set given by the union of the line $y = 0$ and the half-open segment $\{ (x, y) \in \mathbb{R}^{2} : 0 < x \leq 1, y = 1\}$. Since the point $(0,1)$ is a limit point of $S$, we cannot write the part of $S$ above $U$ (for any open set $U$ containing the point $x = 0$) as the disjoint union of graphs of continuous functions, as in the above process.] Now suppose $|S_{x}| = m$. Applying the above process to $x$, we can find open boxes and bounded continuous functions as above such that $$S \cap (U \times M) = \bigcup_{i = 1}^{m} \mathrm{gr}(f_{i})$$ for some open set $U$ (depending on the open boxes) containing $x$. Since $x$ is a limit point of $X$, the neighbourhood $U$ contains a point $z$ of $X$. By the above argument, $z$ takes on exactly $m$ distinct values $f_{1}(z), \dots, f_{m}(z) \in M$ and so $|S_{z}| = m$. But $z \in X$ if and only if $|S_{z}| = k$, and so it must be that $k = m$. Thus $x \in X$ and so $X$ contains all of its limit points in $\Pi_{n}(S)$.