Given a poset $(P,\leq)$ the *interval topology* $\tau\_i(P)$ 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 $\uparrow x = \{y\in P: y\geq x\}$.


If we look at infinite lattices, the interval topology need not be Hausdorff: Let $\omega$ be endowed with the antichain order, and add a new bottom and top element. The interval topology of this lattice is the cofinite topology (where a set is open if it is empty or its complement is finite).


However if we require that a lattice $L$ can not be "too wide" with respect to its height, can we force the interval topology to be $T\_2$? More formally:


**Question.** If $L$ is a distributive lattice, is $\tau\_i(L)$ Hausdorff?



Hausdorff interval topology on distributive lattices
The countable atomless Boolean algebra is a counterexample. See
E.S. Northam, [The interval topology of a lattice](http://www.ams.org/journals/proc/1953-004-05/S0002-9939-1953-0057534-2/S0002-9939-1953-0057534-2.pdf), 1953 (Propositions 2 and 3).


