Let $X$ be a topological space and let $C$ be the set of maximal covers on $X$, i.e. covers that are also antichains.
There is a preordering on $C$ given by $$ \mathcal{U} \leq \mathcal{V} \iff \forall{U \in \mathcal{U}} . \exists{V \in \mathcal{V} . U \subseteq V} $$
Is this preorder actually a lattice? I.e. is it a partial order with a meet and join?
Any relevant references are really appreciated. Thanks.
This is too long for a comment, so I post it as an answer.
If I understood correctly your definition of $C$, $\leq$ is partial order.
Let $(X,\tau)$ be topological space and $C$ the set of all $\subseteq$-maximal (in $P(\tau$)) $\subseteq$-anti-chain (in $\tau$) covers of $X$. Observe that if $\mathcal U, \mathcal V \in C$ then $$\mathcal U \subseteq \mathcal V \implies \mathcal U = \mathcal V, \quad\text{and} \quad U_1, U_2 \in \mathcal U\land U_1 \subseteq U_2 \implies U_1 = U_2,$$ using maximality and anti-chain respectively.
I assume you checked that your relation is preorder (quite easy). To make it partial order we need antisymmetry, that is $$ \mathcal U \leq \mathcal V \land \mathcal V \leq \mathcal U \implies \mathcal U = \mathcal V.$$ Take $U_1 \in \mathcal U$ and pick $V \in \mathcal V$ such that $V \subseteq U_1$ using $\mathcal V \leq \mathcal U$. Now for $V$ pick $U_2\in \mathcal U$ such that $U_2 \subseteq V$ using $\mathcal U \leq \mathcal V$. We have $U_2 \subseteq U_1$, so by our second observation $U_1 = U_2 = V$, and therefore $U_1 \in \mathcal V$. We have $\mathcal U \subseteq \mathcal V$, so by first observation $\mathcal U = \mathcal V$ which we wanted to prove.