Let $P,Q$ be posets and endow them with the [interval topology](https://mathoverflow.net/questions/215673/properties-of-the-interval-topology-of-the-lattice-of-functions) $\tau\_i(P)$ and $\tau\_i(Q)$ respectively. Is it true that if $f: P\to Q$ is order-preserving, then it is continuous, and vice versa?



Does order-preserving equal continuous?
Both implications are incorrect in general.


Consider the real interval $I=[0,1]$. Then the interval topology on $I$ is just the topology coming from the Euclidean metric on $I$. Let $f: I\to I$ be defined as $f(1) = 1$ and $f(x) = 0$ for $x\in I\setminus \{0\}$. Then $f$ is clearly order-preserving, but not continuous.


For a continous function $f:I\to I$ that is not order-preserving, let $f(x) = x$ for $x\in [0,0.5]$ and $f(x) = 1 - x$ for $x \in I\setminus [0, 0.5]$. 


