When a lattice is a lattice of open sets of some topological space?
When a lattice $(L,\leqslant)$ is a lattice of open (or closed) sets of some topological space $(X,\tau)$? Which conditions have to be satisfied? We may assume that $X$ is $T_1$.
Solution 1:
A lattice $L$ is isomorphic to the lattice of open sets in a topological space iff the following two conditions hold:
$L$ is complete: every subset $S\subseteq L$ has a least upper bound.
$L$ has enough prime elements (elements $p\in L\setminus\{1\}$ such that $a\wedge b\leq p$ implies $a\leq p$ or $b\leq p$): that is, if $a,b\in L$ and $a\not\leq b$, then there exist a prime element $p\in L$ such that $b\leq p$ and $a\not\leq p$.
Such a lattice is called a spatial frame. More generally, a frame is a complete lattice which also satisfies the infinite distributive law $a\wedge\bigvee S=\bigvee\{a\wedge s:s\in S\}$ (this infinite distributive law is a consequence of condition (2)). Frames are studied as a generalization of (sober) topological spaces; roughly speaking, a frame is like the lattice of "open sets" in a "topological space" in which an open set may not be determined by the points it contains.
As a sketch of a proof, if $L$ is the lattice of open sets in a topological space $X$, then it satisfies (1) since an arbitrary union of open sets is open. For (2), note that for any $x\in X$, the open set $p_x=X\setminus\overline{\{x\}}$ is prime (the open sets it contains are exactly the open sets that do not contain $x$). If $a,b\subseteq X$ are open with $a\not\subseteq b$, let $x$ be such that $x\in a$ and $x\not\in b$, and then $b\subseteq p_x$ and $a\not\subseteq p_x$.
Conversely, suppose $L$ satisfies (1) and (2). Let $X$ be the set of all prime elements of $L$. We can then put a topology on $X$ by saying that the open sets are the sets of the form $U_a=\{p\in X:a\not\leq p\}$. The map $a\mapsto U_a$ turns arbitrary joins into unions and finite meets into intersections, so (1) implies that this is indeed a topology. Moreover, (2) implies that $a\leq b$ iff $U_a\subseteq U_b$, so that $a\mapsto U_a$ is an order-isomorphism.
If you want to require the topological space to be $T_1$, you need to modify condition (2) to say that $L$ has enough maximal prime elements (that is, prime elements $p$ such that if $q\geq p$ and $q$ is prime then $q=p$). The proof is roughly the same: if $X$ is a $T_1$ space, then $\{x\}=\overline{\{x\}}$ for all $x\in X$ and so the open sets $p_x$ used above are not just prime but maximal prime (since the only set larger than them is $X$ itself). Conversely, if $L$ is complete and has enough maximal primes, you can take $X$ to be the set of maximal primes and topologize it as before, and maximality of the primes will guarantee that $X$ is $T_1$.