One of the natural topologies on any poset $(X, \le)$ is the Alexandrov topology in which open sets are precisely upper sets: $$\tau=\{G\subseteq X \mid x\in G\land x\le y \implies y\in G\}.$$
Today I was told that being Alexandrov-continuous is the same as being monotonic. That is given two posets $X$ and $Y,$ a function $f: X\to Y$ is monotonic if and only if continuous (after equipping both domain and co-domain with Alexandrov topology).
Is this true? If so,
- can you sketch an argument to show this equivalence?
- does this lift to an equivalence of categories?
For $x$ in a poset $P$, let $${\uparrow} x = \{ p \in P : x \leq p\}$$ and for $A \subseteq P$, $${\uparrow}A = \{{\uparrow}a : a \in A\} = \{p \in P: \exists a \in A (a \leq p)\}.$$ Then $A$ is an upper set iff $A = {\uparrow}A$.
So your question is
To prove that this is so, take $f : X \to Y$, where $X$ and $Y$ are posets and $f$ is order-preserving.
Let $V \subseteq Y$ be such that ${\uparrow}V = V$, and let $U = f^{-1}(V) \subseteq X$.
We want to prove that ${\uparrow}U=U$; so let $a \in U$ and $b \in X$ such that $a \leq b$.
It follows that $f(a) \in V$ and $f(a) \leq f(b)$, because $f$ is order-preserving.
So $f(b) \in {\uparrow}V = V$, whence $b \in U = f^{-1}(V)$.
We conclude that ${\uparrow}U = U$.
Suppose now that $f$ is Alexandrov-continuous, that is, ${\uparrow}(f^{-1}(V)) = f^{-1}(V)$, whenever ${\uparrow}V=V$.
We now want to prove that $f$ is order-preserving; so let $a,b \in X$ with $a \leq b$, and try to prove that $f(a) \leq f(b)$.
Take $V = {\uparrow}f(a)$; clearly, ${\uparrow}V=V$, and so if $U = f^{-1}(V)$, we have ${\uparrow}U=U$, because $f$ is continuous.
Since $f(a) \in V$, we have $a \in U$, whence $b \in {\uparrow}U = U$, and therefore, $f(b) \in V$.
We conclude that $f(a) \leq f(b)$.