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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?

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The countable atomless Boolean algebra is a counterexample. See E.S. Northam, The interval topology of a lattice, 1953 (Propositions 2 and 3).

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