# partitions form a lattice

Let $S$ be a set. Let $\operatorname{Part}(S)$ be the set of all partitions   on $S$. Since each partition is a cover of $S$, $\operatorname{Part}(S)$ is partially ordered by covering  refinement relation   , so that $P_{1}\preceq P_{2}$ if for every $a\in P_{1}$, there is a $b\in P_{2}$ such that $a\subseteq b$. We say that a partition $P$ is finer than a partition $Q$ if $P\preceq Q$, and coarser than $Q$ if $Q\preceq P$.

###### Proposition 1.

$\operatorname{Part}(S)$

###### Proof.

For any set $\mathcal{P}$ of partitions $P_{i}$ of $S$, the intersection   $\bigcap\mathcal{P}$ is a partition of $S$. Take the meet of $P_{i}$ to be this intersection. Next, let $\mathcal{Q}$ be the set of those partitions of $S$ such that each $Q\in\mathcal{Q}$ is coarser than each $P_{i}$. This set is non-empty because $S\times S\in\mathcal{Q}$. Take the meet $P^{\prime}$ of all these partitions which is again coarser than all partitions $P_{i}$. Define the join of $P_{i}$ to be $P^{\prime}$ and the proof is complete       . ∎

Remarks.

Title partitions form a lattice  PartitionsFormALattice 2013-03-22 16:45:22 2013-03-22 16:45:22 CWoo (3771) CWoo (3771) 6 CWoo (3771) Derivation msc 06B20 lattice of equivalence relations