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 $i\in I$ ?


1 Answer 1


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.


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