TITLE: Functor from Order Category to Top Category
QUESTION [5 upvotes]: Let Order be the category whose objects are totally ordered sets and whose morphisms are monotone functions, and let Top be the category whose objects are topological spaces and whose morphism are continuous functions.
Then given an ordered sets $A$ with relation $\leq_A$, we can construct a topology $\mathcal{T}_{\leq_A}$ on $A$ with a basis consisting of the following types of sets:
$$(a,b) = \{x|a<_A x <_A b\},$$
$$ (-\infty,b) = \{x|x <_A b\},$$
$$ (a,\infty) = \{x|a<_A x\}.$$
where $a<_A b$ means $a\leq_A b$ but $a\neq b$. I am supposed to show that there is a functor $F:$ Order$ \to$ Top that maps $(A,\leq_A)$ to $(A,\mathcal{T}_{\leq_A})$ and takes a monotone function $f:A\to B$ to the same function.
Now in order for this actually to be a functor, it must hold that the monotone functions $f:A\to B$ are continuous functions in order for them to be morphisms in Top, but I don't see why this must necessarily be true.
For instance, $\Bbb{R}$ is an ordered set. So let us say that $A=B=\Bbb{R}$. Then the topology induced by the order is just the usual topology on $\Bbb{R}$. Then consider the floor function $f:\Bbb{R} \to \Bbb{R}$ given by $f(x) = \lfloor x \rfloor$. This is not a continuous function because $f^{-1}(0.5,1.5) = [1,2)$ which is not an open set in this topology.
So it seems to me that the above is a counterexample to the claim that a monotone function between two ordered sets is necessarily continuous in the topology induced by that order. Is this the case? If this is not a counterexample, I don't see how to prove that the monotone functions are continuous.

REPLY [4 votes]: You're right--monotone functions are not continuous in general.  If you want to get a functor of the sort you're looking for, you'll need to severely restrict the maps you allow in $\mathbf{Order}$.  For instance, it would suffice to consider maps which preserve suprema and infima of arbitrary subsets.