Let $I=[0,1]$ be the unit interval, with $\tau_0$ the usual topology.
Inspired by this question, I've been wondering whether it is possible to topologize $I$ so that the continuous functions are precisely the usual continuous functions that are also nondecreasing. That is,
Is there another topology $\tau$ on $I$ such that $f\colon I\to I$ is $\tau$-continuous if and only if it is $\tau_0$-continuous and nondecreasing?
(Here by $\tau$-continuous we mean continuous when both domain and codomain are equipped with $\tau$.)
There are some related topologies in this vein. If we only want to get nondecreasing functions (i.e., $f$ is continuous iff $f$ is nondecreasing), we have the Alexandrov topology given by: $$\tau_a=\{U\in \mathcal P(I)\mid x\in U\wedge y\geq x\implies y\in U\}$$
Similarly, if we let $\tau_-=\{(a,1]\mid 0\leq a<1 \}\cup\{\emptyset\}$, it is routine to verify $f$ is $\tau_-$-continuous if and only if $f$ is nondecreasing and $\tau_0$-continuous from the left. We could similarly get right continuity by defining $\tau_+=\{[0,a)\mid 0<a\leq 1\}\cup\{\emptyset\}$.
However, these kinds of simple constructions don't seem to have an obvious way to modify them to get all nondecreasing continuous functions - trying to combine the open sets in $\tau_+$ and $\tau_-$ gives us back $\tau_0$, so we lose monotonicity.
I'd be very interested in either an example of such a topology $\tau$ or a proof that one cannot exist.
More generally, one can replace $I$ with any totally ordered set $T$ with its order topology, and ask the same question. The topologies $\tau_a$, $\tau_+$, and $\tau_-$ defined above all generalize easily, and as a result, it is easy to see that if $T$ is any well-ordered set, or more generally, any total ordered set in which every element has an immediate successor, the answer is yes, since we can just take $\tau_-$, as every function is already continuous from the right. But it would be interesting to know if there are other examples.