Set partition refinement lattice, maximal chain length Denote $\Pi_n$ the set of all set partitions of [n]. For set partitions $S=\{S_1,S_2,...,S_p\}$ and $T=\{T_1,T_2,...,T_q\}$ of [n], define $S \leq T$ iff for every i=1,2,...,p there exists a j=1,2,...,q such that $S_i \subseteq T_j$.  
a)Show that $(\Pi_n,\leq)$ is a partial order
b)Find the length of a maximal chain in $(\Pi_n,\leq)$
To prove a relation is partial order we have to prove reflexivity, antisymmetry, and transitivity. It seems proving reflexivity and transitivity is easy. I'm just not sure about antisymmetry.
Here is one example. This is called the set partition refinement lattice. It looks like the max chain length is n. I don't know if that is correct.

 A: Well for antisymmetry it's also straight forward:
Assume $S\leq T\leq S$ then for every $i$ there exists a $j$ and a $k$ so that $S_i\subseteq T_j \subseteq S_k$ thus $S_i\subseteq S_k$ and since $S=\{S_1,\ldots,S_p\}$ is a partition $S_i\cap S_k \neq \emptyset$ implies $S_i = S_k$ and hence $S_i = T_j = S_k$ since this is valid for every $i$ and exchanging the roles of $S$ and $T$ this is valid for every $j$ we get that $T=S$ proving anti-symmetry.
As for the maximal chain length in $\Pi_n$ it necessarily is $\geq n$ necessairily since we can always consider the chain 
$$ \{\{1\},\{2\},\ldots, \{n\}\}\leq \{\{1,2\},\{3\},\ldots, \{n\}\}\leq \cdots \leq \{\{1,\ldots, n\}\}$$
(note that $\{\{1\},\{2\},\ldots, \{n\}\}$ is minimal and $\{\{1,\ldots,n\}\}$ is maximal with respect to $\leq$)
Now to prove the maximal chain can use induction:
for $n=1$ the maximal chain has length $1$ (the minimal and maximal partitions coincide)
Now assume that for $n$ we have a maximal chain of length $m \geq n + 1$, of the form $S^1\leq S^2\leq \cdots \leq S^m$.
Note also that if a chain is maximal then only one of the $S^i_j$ is strictly contained in some $S^{i+1}_k$, for all the others equality holds otherwise the chain can be further extended contradicing maximality.
Now without loss of generality let $n\in S^i_1$ for all $i$. Since by hypothesis the chain is maximal, it starts on the minimal elemnt, thus $S^1_1 = \{n\}$ let $j$ be the minimum for which $S^1_1\subsetneq S^j_1$, then if we simply delete $S^i_1$ for all $i< j$ and remove $n$ from $S^i_1$ for all $i\geq j$ we get a sequence in $\Pi_{n-1}$ that is not strict only at step $j$, thus if we delete step $j$ we obtain a chain of length $n-1$ in $\Pi_{n-1}$ contradicting the induction hypothesis. 
