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?
