# When is this topology compatible with the partial ordering?

This question was first asked here, on math stack exchange, but wasn't able to attract any attention. Now that I am thinking more, it feels like the most suitable place for this question is here.

Suppose I have a topological space whose underlying set $$X$$ has a partial ordering on it. A prior this partial ordering has no relation to the topology. Suppose there is a meet semi-lattice $$I \subseteq X,$$ and an arbitrary set of open sets $$\{A_i\}_{i\in I}$$ satisfying $$i\in A_i$$ for all $$i\in I$$.

What conditions (or relations between topology and the partial ordering) would guarantee the existence of another family of open sets $$\{B_i\}_{i\in I}$$ satisfying $$i\in B_i\subseteq A_i$$ and $$B_i\cap B_j\subseteq B_{i\wedge j}$$ for all $$i\in I$$ ?

One of the standard topologies to consider would be the lower-cone topology, whose basic open sets are the lower cones $$i{\downarrow}=\{j\mid j\leq i\}$$. In this topology, the open sets are exactly the downsets. If $$i\in A_i$$ is open, then indeed $$i{\downarrow}\subseteq A_i$$, and furthermore $$(i{\downarrow})\cap(j{\downarrow})=(i\wedge j){\downarrow}$$, so these sets $$B_i=i{\downarrow}$$ fulfill exactly your desired criterion.
But since you require only inclusion $$B_i\cap B_j\subseteq B_{i\wedge j}$$ rather than equality, the upper cone topology also has your feature. That is, the open sets are the upsets, then we may take $$B_i=i{\uparrow}\subseteq A_i$$, and simply observe that $$(i{\uparrow})\cap (j\uparrow)\subseteq (i\wedge j){\uparrow}$$ because anything above $$i$$ and $$j$$ is also above $$i\wedge j$$. Indeed, for this topology we would have $$(i{\uparrow})\cup(j{\uparrow})\subseteq(i\wedge j){\uparrow}$$.
But of course, there are other topologies that also have your feature. The indiscrete topology, in which the only open sets are $$X$$ and $$\emptyset$$, we would have $$A_i=X$$, and taking $$B_i=X$$ fulfills your property. Also, in the discrete topology, in which every set is open, we can take $$B_i=\{i\}$$ and fulfill your property.
Finally, let me point out that if one wants to insist on the identity $$B_i\cap B_j=B_{i\wedge j}$$, then this implies that the topology must be included in the lower cone topology as in the first example above. The reason is that if $$j\leq i$$, then $$i\wedge j=j$$ and so $$j\in B_j=B_{i\wedge j}= B_i\cap B_j$$ and so in particular $$j\in A_i$$ and thus $$A_i$$ must contain every element of $$i{\downarrow}$$. In particular, in this case every open neighborhood $$A_i$$ of any $$i$$ must contain the lower cone below $$i$$, and consequently, every open set must be downward closed. Thus, with the stronger $$B_i\cap B_j=B_{i\wedge j}$$ requirement, the topology is contained in the lower cone topology.