Product of posets with Hausdorff interval topology

Question

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 $i\in I$.)

Answer 1

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.

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 $NW_{a b} \cap NW_{a' b'} = NW_{\min(a, a'), \max(b, b')}$.

Answer 2

The answer is Yes, because the product topology of Hausdorff spaces is Hausdorff again, and because the product topology of the interval topologies of a family of posets equals the interval topology of the product:

Lemma. The interval topology $\tau_i = \tau_{\textrm{int}}(\prod_{i\in I}P_i))$ on $P=\prod_{i\in I} P_i$ equals the product topology $\tau_p$ of the topological spaces $(P_i, \tau_{\textrm{int}}(P_i))$.

Proof. Take a subbasic element of $U\in\tau_i$ and show that it is a member of $\tau_p$. W.l.o.g. we let $U = P\setminus (\uparrow(x_i)_{i\in I})$ where $x_i\in P_i$. Note that $\uparrow(x_i)_{i\in I}$ is a product of closed sets in the spaces $(P_i, \tau_{\textrm{int}}(P_i))$, therefore it is closed in the product topology, so $U\in \tau_p$. Conversely, let $U = \pi_i^{-1}(U_j)$ be subbasic in $\tau_p$ where $\pi_j: P\to P_j$ is the projection map and $U_j = P_j\setminus \uparrow x^*$ for some $x^*\in P_j$. Then $$U = \bigcup \{P\setminus (\uparrow (z_i)_{i\in I}): (z_i)_{i\in I} \in P \text{ and } z_j =x^*\}.$$ So $U\in\tau_i$. QED.