# Finite fibers of constant size

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:

1. $\Pi_n(S)$ is definably connected,
2. $S$ is locally bounded at every $x \in \Pi_n(S)$,
3. $S$ is closed in $\Pi_n(S) \times M$, and
4. $\Pi_n|_S:S \longrightarrow M^n$ 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.

## 8 thoughts on “Finite fibers of constant size”

1. zeinab says:

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.

1. Patrick Speissegger says:

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.

1. Sam says:

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.

1. Patrick Speissegger says:

Hmm, can I see the proof?

1. Sam says:

Oh! I found the mistake. I was assuming that a non-definable set has a supremum.

2. Patrick Speissegger says:

Can you think of a counterexample in the field of Puiseux series?

2. Sam says:

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.

3. Jamal Kawach says:

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 $\Pi_{n}(S)$.

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