The Monotonicity Theorem

Let ${\cal M}$ be an o-minimal expansion of a dense linear order $(M,\lt)$.
Let $f:I \longrightarrow M$ be definable, with $I = (a,b)$ an interval in $M$.

Definition
We call $f$ strictly monotone if $f$ is either constant, or strictly increasing, or strictly decreasing.

Our first goal is to prove the

Monotonicity Theorem

There are $n \in {\mathbb N}$ and $a = a_0 \lt a_1 \lt \cdots \lt a_n = b$ such that the restriction of $f$ to the interval $(a_{i-1},a_i)$ is strictly monotone and continuous, for $i=1, \dots, n$.

See Lemma 4 in Pillay and Steinhorn for original statement and van den Dries for a proof. The proof I will give here is due to Peterzil and Starchenko.

The first step in proving the monotonicity theorem is to reduce it to a “local” statement. We say that $f$ is strictly monotone at $x \in I$ if there exist $a’ \in (a,x)$ and $b’ \in (x,b)$ such that the restriction of $f$ to the interval $(a’,b’)$ is strictly monotone.

Exercise
Assume that $f$ is strictly monotone at every $x \in I$. Prove that $f$ is strictly monotone.

Monotonicity Lemma

There is an open subinterval $J \subseteq I$ such that the restriction of $f$ to $J$ is strictly monotone.

We will prove this lemma later; for now, let us see how to derive the Monotonicity Theorem from it. We need one other

Observation
Assume that $f$ is strictly monotone. Then there exists an open subinterval $J \subseteq I$ on which $f$ is continuous.

Proof. We may assume that $f$ is not constant. So by o-minimality, the image $f(I)$ contains an open interval $J’$, and $f^{-1}(J’)$ is an interval $J$ on which $f$ is either an order-preserving or an order-reversing bijection. $\Box$

Proof of the Monotonicity Theorem from the Monotonicity Lemma: Let $A$ be the set of all $x \in I$ such that $f$ is continuous and strictly monotone at $x$. This $A$ is definable and, by the exercise, the restriction of $f$ to any open interval contained in $A$ is continuous and strictly monotone. By the Monotonicity Lemma and the observation, the complement of $A$ in $I$ contains no open interval. So by o-minimality, the complement of $A$ in $I$ is finite. $\Box$

Exercises

1. The limits $\lim_{x \to a^+} f(x)$, $\lim_{x \to b^-} f(x)$ and, for $c \in (a,b)$, the limits $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ exist in $M \cup \{-\infty,+\infty\}$.
2. If $c,d \in M$ and $g:[c,d] \longrightarrow M$ is definable and continuous, then $g$ attains a maximum and a minimum in $[c,d]$.
3. Let $f_1, \dots, f_k:I \into M$ be definable. Show that there are $n \in {\mathbb N}$ and $a = a_0 \lt a_1 \lt \cdots \lt a_n = b$ such that the restriction of $f_j$ to the interval $(a_{i-1},a_i)$ is strictly monotone and continuous, for $i=1, \dots, n$ and $j=1, \dots, k$.

2 thoughts on “The Monotonicity Theorem”

1. Brinda Venkataramani says:

Here is an attempt at the first exercise:

Since $f$ is definable there exists a formula $\phi$ defining the graph of $f$. We note that the set of $x\in I$ for which there is a neighborhood about $x$ where $f$ is constant is definable (with parameters) via the formula $$\exists y\phi(x,y)\wedge\exists s\exists t(s<x<t\wedge\forall z(s<z<t\to f(z)=k)),$$ where $k$ is some constant in $M$. The set of $x$ for which there is a neighborhood about $x$ where $f$ is strictly increasing is also definable is also definable via the formula \exists y\phi(x,y)\wedge\exists s\exists t(s<x<t\wedge\forall u\forall v(s<u<v<t\to f(u)$, and$X_<$. We will show that each is open and closed. To show openness, take$x\in X_=$. Then there is a neighborhood$U$about$x$on which$f$is constant. But then for any$y\in U$there is a neighborhood around$y$for which$f$is constant, namely$U$itself, and so$X_=$is open. To show closedness let$x$be a limit point of$X_=$. By the fact that$f$is strictly monotone at every$y\in I$, we have that there exists an interval$I$about$x$on which$f$is strictly monotone. Since$x$is a limit point of$X_=$, this interval$I$contains a point$y\in X_=$. But then, it must be that$f$is constant on$I$, since$I$has constant monotonicity and contains at least part of a region, where$f$is constant. Showing openness and closedness for$X_$is identical to the above. Since every interval is definably connected, it follows that each of$X_=$,$X_$must be empty or all of$I$. The result follows. For the exercises at the end, I think you can claim that$\lim\limits_{x\to a^+}$is$\text{inf}(f((a_0,a_1)))$, where$(a_0,a_1) $is the first interval on which$f$is strictly monotone and continuous (obtained via monotonicity theorem), assuming$f$is increasing on that interval. Taking an arbitrary neighborhood around$\text{inf}(f((a_0,a_1)))$we get that the intersection is some sub-interval of$f((a_0,a_1))$. Since$f$is continuous the pre-image is open and contains a neighborhood which has$a$as a limit point… I’m not too sure how to formalize this, but it seems to work. 1. Patrick says: In your formula defining local constancy, if you replace the parameter$k$by$f(x)$, then you do not need the parameter$k$. In your definition of$X_\gt$, you mean$f(u) \lt f(v)$? To show closedness in$I$, you also assume that the limit point$x$belongs to$I$(which you seem to without saying it). Otherwise, your argument for the first exercise is correct. For the next exercise, wouldn’t the left endpoint of this first interval have$a_0 = a$as left endpoint? Here is how I would say this: by the Monotonicity Theorem, there exists$a_1 > a$such that the restriction of$f$to the interval$(a,a_1)$is strictly monotone. Now describe$\lim_{t \to a^+} f(t)$depending on the three possible cases of monotonicity of this restriction of$f\$.

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