Monotonicity and continuity
In this post, we show that a function between two preordered sets is monotone if and only if it is continuous for some topology. This is a well-known result that we rediscovered during a discussion with my research group.
Preliminaries
We first recall some definitions that will be useful in the sequel.
Preordered sets and upper sets
Given a set $X$, recall that a preorder $\le_X$ is a relation on $X$ that is reflexive and transitive, i.e.,
- for all $x \in X$, we have $x \le_X x$, and
- Given $x, y, z \in X$, if $x \le_X y$ and $y \le_X z$, then $x \le_X z$.
Given a preordered set $(X, \le_X)$, we say that $U \subseteq X$ is an upper set if and only if $$U = \bigcup_{u \in U} \{ x \in X : u \le_X x \}.$$ In other words, $U$ is an upper set if and only if for all $u \in U$ and $x \in X$, if $u \le_X x$, then $x \in U$. A classical example of an upper set in $\mathbb{R}^n$ with the usual preorder is the nonnegative orthant $\mathbb{R}^n_+$. Upper sets have the following useful property: the union and the intersection of upper sets are upper sets.
Alexandrov topology
Given a set $X$, recall that a topology $\mathcal{T}$ on $X$ is a collection of subsets of $X$ satisfying
- $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$,
- Arbitrary unions of elements of $\mathcal{T}$ are in $\mathcal{T}$, and
- Finite intersections of elements of $\mathcal{T}$ are in $\mathcal{T}$.
Recall also that the elements of $\mathcal{T}$ are called open sets. If infinite intersections of elements of $\mathcal{T}$ are also in $\mathcal{T}$, then $\mathcal{T}$ is referred to as an Alexandrov topology. Given a preordered set $(X, \le_X)$, we can define an Alexandrov topology $\mathcal{T}_X$ on $X$ as the collection of all upper sets of $X$. We will refer to $\mathcal{T}_X$ as the Alexandrov topology associated with $(X, \le_X)$.
Main result
Theorem. Let $(X, \le_X)$ and $(Y, \le_Y)$ be two preordered sets. A function $f \colon X \to Y$ is monotone if and only if $f$ is continuous for the Alexandrov topologies associated with $(X, \le_X)$ and $(Y, \le_Y)$.
The proof of this theorem is straightforward. Let $\mathcal{T}_X$ and $\mathcal{T}_Y$ be the Alexandrov topologies associated with $(X, \le_X)$ and $(Y, \le_Y)$, respectively. If $f$ is monotone, then for all $x \in X$ and $x’ \in X$, if $x \le_X x’$, then $f ( x ) \le_Y f ( x’ )$. Therefore, for any open set $U \in \mathcal{T}_Y$, we have $$f^{-1} ( U ) = \{ x \in X : f ( x ) \in U \} \in \mathcal{T}_X.$$ Indeed, given $x \in f^{-1} ( U )$ and $x’ \in X$, if $x \le_X x’$, then $f ( x ) \le_Y f ( x’ )$ and hence, $x’ \in f^{-1} ( U )$. Therefore, $f$ is continuous.
Conversely, if $f$ is continuous, then for all $U \in \mathcal{T}_Y$, we have $f^{-1} ( U ) \in \mathcal{T}_X$. Let $x \in X$ and let $U_x$ be the open set of $Y$ defined by $$U_x = \{ y \in Y : f ( x ) \le_Y y \}.$$ We then have $f^{-1} ( U_x ) = \{ x’ \in X : f ( x ) \le_Y f ( x’ ) \} \in \mathcal{T}_X$. Hence, for all $x’ \in X$ such that $x \le_X x’$, we have $f ( x ) \le_Y f ( x’ )$. Therefore, $f$ is monotone.
What about univariate real-valued functions?
A natural question arises: what does this theorem say about univariate real-valued functions? In other words, what does this mean when $X = Y = \mathbb{R}$ for the usual order in $\mathbb{R}$? The definition of monotonous functions corresponds to the usual definition of increasing functions. Moreover, the Alexandrov topology associated with $(\mathbb{R}, \le)$ is given by $$\mathcal{T}_{\mathbb{R}} = \{ [x, \infty) : x \in \mathbb{R} \}.$$ The theorem states then than increasing functions in $\mathbb{R}$ are continuous for the above topology.