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I'm trying to solve exercise 2.19.7 from Chapter 3 of Van Den Dries's Tame Topology and O-minimal Structures, which is the following:

Suppose $\mathcal S$ is an o-minimal structure on the ordered set $(\Bbb R,<)$ of real numbers. Show that for a definable set $X\subseteq\Bbb R^m$ the following are equivalent:

  1. $X$ is definably connected;
  2. $X$ is connected in the usual topological sense.

Where $X\subseteq \Bbb R^m$ is called definably connected if it is definable and cannot be written as the disjoint union of two definable open subsets. The implication $(2)\implies(1)$ is trivial, but I don't see how to show the other implication, any hints?

Alessandro Codenotti
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  • @AlexKruckman It appears after the cell decomposition theorem, but a proof without using it would be great (of course if using such a big result is unavoidable that would also make a good answer) – Alessandro Codenotti May 27 '21 at 15:56
  • I agree it would be nice if there were a proof not using cell decomposition - but I don't see how to do it. – Alex Kruckman May 27 '21 at 16:18

1 Answers1

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First prove by induction on $n$ that every cell $C\subseteq \mathbb{R}^n$ is (topologically) connected.

When $n = 0$ or $1$, the cells are points or open intervals in $\mathbb{R}$, and hence connected.

For the inductive step, suppose $C\subseteq \mathbb{R}^{n+1}$ is a cell. Let $C' = \pi(C)\subseteq \mathbb{R}^n$. Then $C'$ is a cell in $\mathbb{R}^n$, so it is connected by induction. For each point $a\in C'$, the fiber $C_a = \{b\in \mathbb{R}\mid (a,b)\in C\}$ is a point or an open interval in $\mathbb{R}$, and hence connected.

Assume for contradiction that $C$ is disconnected, i.e., there is a partition $X = U\sqcup V$ such that $U$ and $V$ are both non-empty and open in the subspace topology on $C$. Let $U' = \pi(U)$ and $V' = \pi(V)$. Note that $U'$ and $V'$ are non-empty and open in $C'$, since $\pi\colon C\to C'$ is an open map, and $C' = U'\cup V'$.

For each $a\in C'$, since $C_a$ is connected, either $C_a\subseteq U$ or $C_a\subseteq V$. Thus $U'$ and $V'$ are disjoint subsets of $C'$. This contradicts connectedness of $C'$.

Now let $X$ be a definably connected set. By cell decomposition, we can write $X$ as a finite disjoint union of cells: $$X = C_1\sqcup C_2\sqcup \dots \sqcup C_n.$$ Assume for contradiction that $X$ is disconnected, i.e., there is a partition $X = U\sqcup V$ such that $U$ and $V$ are both non-empty and open in the subspace topology on $X$. Since each cell $C_i$ is connected, either $C_i \subseteq U$ or $C_i\subseteq V$. Then $U$ and $V$ are each a union of some of the cells $C_i$, and hence definable, contradicting the fact that $X$ is definably connected.

Alex Kruckman
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  • Is connectedness and definable connectedness also equivalent in any o-minimal expansion of $(\mathbb{R},<)$? – 656475 Dec 08 '24 at 15:35
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    @EduardoMagalhães Yes, that's what my answer shows. ("an o-minimal structure on the ordered set (R,<)" means the same thing as "an o-minimal expansion of (R,<)".) – Alex Kruckman Dec 08 '24 at 16:25