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?


1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.