(For some details and references, see Chapter 1 of my notes.)

We fix a structure ${\cal M} = (M,<,\dots)$ such that $(M,<)$ is a dense linear order.
An **interval** is a nonempty interval with endpoints in $M \cup

\{-\infty,+\infty\}$.

We consider ${\cal M}$ with its **interval topology** on $M$ and the corresponding product topologies on the cartesian powers $M^k$.

### Example: Puiseux series

Let $P$ be the set of all **Puiseux series**, that is, series of the form $G(X) = X^{p/d} \cdot F(X^{1/d})$, where $F(X)$ is a formal power series with real coefficients, $d$ is a nonzero natural number and $p$ is an integer. For such series the number $({\rm ord}(F)+p)/d$ is called its **order** and denoted by ${\rm ord}(G)$, and the **leading coefficient** of $G$ is the coefficient ${\rm lc}(G)$ of the monomial $X^{{\rm ord}(G)}$ in the series $G$.

For Puiseux series $G$ and $H$, we define $G+H$ and $G \cdot H$ in the usual way. We set $G \lt 0$ if and only if ${\rm lc}(G) \lt 0$, and $G \lt H$ if and only if $G-H \lt 0$.

**Fact**

*The structure $(P,\lt)$ is a dense linear order, and its expansion ${\cal P} := (P,\lt,+,\cdot)$ is a real closed ordered field.*

For a proof of this fact, see for instance Section 2.6 in the book *Algorithms in Real Algebraic Geometry*, by Saugata Basu, Richard Pollack and Marie-Françoise Coste-Roy.

Here is the gist of this proof: to show that $P$ is a field, let $F$ be a nonzero Puiseux series. Then $F(X) = aX^r(1+\epsilon(X))$, where $a \in \mathbb{R}$ is nonzero, $r \in \mathbb{Q}$ and $\epsilon$ is a Puiseux series of strictly positive order; hence

$\displaystyle \frac1F(X) = \frac1a X^{-r} \frac1{1+\epsilon(X)} = \frac1a X^{-r} \left(\sum_{k=0}^\infty \epsilon(X)^k\right).$

To show that $P$ is real closed, we show that every positive Puiseux series is a square and that every odd-degree polynomial over $P$ has a root. The former is similar to the argument above, using the Taylor series of the real function $x \mapsto \sqrt{1+x}$; the latter is more involved and uses Newton’s method.

One interesting observation about this proof is that the intermediate value theorem, which can be used to prove that every odd-degree polynomial over $\mathbb{R}$ has a real root, is not available over $P$ as the following shows.

**Exercise**

Prove that $P$ does not have the least-upper-bound property and is totally disconnected.

This means that, in the general context of ${\cal M}$, we need to restrict our attention to definable sets.

**Definition**

A definable set $S \subseteq M^n$ is **definably connected** if $S$ is not disconnected by any two nonempty, definable, open subsets.

**Exercises**

- Prove that the image of a definably connected, definable set

under a definable, continuous map is definably connected. - Let $S,T \subseteq M^n$ be definable and definably connected,

and assume that ${\rm cl} S \cap T \ne \emptyset$. Prove that $S \cup

T$ is definably connected.

**Definition**

The structure ${\cal M}$ is **definably complete** if every definable subset of $M$ has a supremum in $M \cup \{-\infty,+\infty\}$.

By the least upper bound principle for $\mathbb{R}$, every expansion of $(\mathbb{R},\lt)$ is definably complete.

**Exercises**

Assume that ${\cal M}$ is definably complete.

- Prove that every interval is definably connected.
**Intermediate Value Theorem**: Let $f,g:I \longrightarrow M$ be definable and continuous, with $I \subseteq M$ an interval, and

assume that $f(x) \ne g(x)$ for $x \in I$. Prove that either

$f(x) \gt g(x)$ for all $x \in I$, or $f(x) \lt g(x)$ for all $x \in

I$.

On Wednesday we asked whether there was an example of a definably complete structure which was not o-minimal. An easy one to construct is $\mathcal M = (\mathbb R, <, \mathbb Z)$ (ie the reals as an ordered structure with an additional predicate for the integers). Every non-empty subset $X$ of $\mathbb Z$ clearly has a supremum: $\sup(X)$ will be the maximum element of $X$, if it exists, and $+\infty$ otherwise. Taking boolean combinations of $\mathbb Z$ and $<$ shouldn't change anything drastically, and so $\mathcal M$ will still be definably complete. However, $\mathbb Z$ is an infinite set which doesn't contain an interval, and hence $\mathcal M$ is not o-minimal.

In fact, I believe it follows from the fact that $\mathbb R$ is complete (as a metric space) that every subset of $\mathbb R$ has a supremum in $\mathbb R\cup\{\pm\infty\}$. If that is the case, then any expansion $\mathcal M$ of $(\mathbb R, <)$ will be definably complete, and so there would be lots of similar examples.

Apparently, you cannot edit comments on wordpress. As wikipedia points out, every subset of the real numbers does have a supremum, and so any expansion of $(\mathbb R, <)$ is definably complete.

That’s right (see the remark after the definition of “definably complete”). I sometimes ask silly things in class…

In the second set of exercises we are asked to show that if $\mathscr{M}$ is definably complete, then every interval is definably connected.

I believe this assertion is false. For a counterexample, consider the integers with the usual ordering. We have

$\mathbb{Z}=(-\infty,+\infty)=(-\infty,0)\cup (-1,+\infty)$.

For the exercise then, we also need the assumption that < is dense.

I am assuming for the entire post that the order is dense. But it is true that an earlier version of this post didn’t make this assumption; I made the change after I noticed this error.