Let $X\neq\emptyset$ be a set. A partition is a collection $P\subseteq {\cal P}(X)\setminus \{\emptyset\}$ such that
- $\bigcup P = X$, and
- $p\neq q \in P$ implies $p\cap q = \emptyset$.
For $P, Q\in \text{Part}(X)$ we say $P\leq Q$ if for every $q\in Q$ there is $p\in P$ such that $p\subseteq q$. Let $\text{Part}(X)$ denote the collection of all partitions of $X$. It turns out that $(\text{Part}(X),\leq)$ is a complete lattice.
Given $P,Q\in\text{Part}(X)$, what is the explicit construction of $P\vee Q$ and $P\wedge Q$?
Hints: for the lower bound, we need a partition to be a refinement of both elements. Consider the partition formed by taking all nonempty intersections of blocks in P and Q.
Edit: thanks for ThomasAndrews pointing out my error.
For an upper bound, we need a partition of which P and Q are both refinements. Consider the partition who's blocks are formed iteratively as follows: start by taking a block $p\in P$ and forming $b_0=p\cup_q q$, where the union is taken over all $q\in Q$ which intersect nontrivially with $p$.
Then create $b_1=b_0\cup_{p'}p'$, where again the union is taken over all $p'\in P$ which intersects nontrivially with $b_0$. Then from $b_1$ create $b_2$ by throwing in more blocks of Q which intersect nontrivially, and continue the process until it stabilizes. This gives one block of the larger partition; repeat this process until all elements are in some block, which will give you a final partition.