Given a poset $(P,\leq)$ the *interval topology* on $P$ is generated by
$$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$
where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.


Let $\{P\_i : i\in I\}$ be a family of posets such that the interval topology of each $P\_i$ is Hausdorff. Is the interval topology of $\prod\_{i\in I} P\_i$ also Hausdorff? (We say that $(x\_i)\_{i\in I} \leq (y\_i)\_{i\in I}$ in $\prod\_{i\in I} P\_i$ if and only if $x\_i \leq y\_i$ for all $i\in I$.)



Product of posets with Hausdorff interval topology
The statement in the answer by Dominic that the interval topology of the product poset is the product topology of the interval topologies is incorrect. The argument that the product topology contains the interval topology is correct, but the one for the opposite containment is not, as shown in this [MO answer](https://mathoverflow.net/a/281902/2926). 


And in fact the interval topology for $\mathbb{R} \times \mathbb{R}$ is *not* Hausdorff, because as shown in the linked answer, every basic open $U$ in the interval topology contains a "northwest quadrant", i.e., a subset of the form $NW\_{a b} = \{(x, y): x < a\; \text{and}\; y > b\}$, and the intersection of two northwest quadrants is always nonempty because $NW\_{a b} \cap NW\_{a' b'} = NW\_{\min(a, a'), \max(b, b')}$.


