This question was first asked [here](https://math.stackexchange.com/questions/4648121/a-set-with-a-topology-and-a-partial-ordering-question), 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](https://en.wikipedia.org/wiki/Semilattice) $I \subseteq X,$ and an arbitrary set of open sets $\{A\_i\}\_{i\in I}$ satisfying $i\in A\_i$ for all $i\in I$.



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> 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 $i\in I$ ?
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When is this topology compatible with the partial ordering?
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.


