Is $\textbf{NC}_n$ a ranked lattice? A set partition $\prod_n$ is a $\textit{noncrossing partition}$ (NCP) if its associated equivalence relation $\cong$ satifies the following condition: $$\text{for all } i<j<k<l \text{ if } i \cong k \text{ and } j \cong l \text{ then } i \cong j \cong k \cong l$$
The set of NCPs of order $n$ is denoted $\textbf{NC}_n$. Ordering by reverse refinement makes  $\textbf{NC}_n$ into a subposet of the partition lattice $\prod_n$
I'm interesting in showing that $\textbf{NC}_n$ is a ranked lattice, and was also pondering the natural question as to if it is a sublattice of $\prod_n$?
I"ve been relentlessly trying to find a counter example, as my gut is telling me there is no way that $\textbf{NC}_n$ can be ranked... However, such a counter example has not been forthcoming. I've been fiddling around with this for too long and would appreciate if someone could help me escape my tunnel vision... Thanks!
 A: In this lattice, a covering relation merges two parts (under the condition that these two parts are neighbors in the appropriate sense), just as in the lattice of all set partitions.

So the number of parts provides the rank function.

EDIT:
To understand the rank function, it is enough to look for a function on the elements of the posets that is increased by one along every edge of the Hasse diagram, i.e. along every cover relation.
So you need first to understand the cover relations.
The partial order relation between two noncrossing partitions is given by inclusion of each block of one in some block of the other. An increasing relation is therefore obtained by merging some parts together. Cover relations turn out to be obtained if and only if exactly two parts are merged, and for this to be possible, the two parts must see each other, namely not be separated by another part.
Once one understands the cover relations, it is easy to check that the function "minus number of parts" does the job, indeed increases by one along cover relations.
