Let $\{A_1,\ldots, A_m\}$ be a family of sets and $I=\{1, \ldots, m\}$. Assume for any $J\subset I$, $B_J=\bigcap_{i\in J}A_j$ satisfies $1\leq |B_J| \leq m-1$ as long as $|J|>1$.

We define a labelling of $J\subseteq I$ as follows. $l(J)=|B_J|$ if $|J|>1$ and $l(J)=m-1$ otherwise. Then we have the labelled poset $(2^I, \subseteq, l)$ (or labelled lattice).

Observe that $l\equiv m-1$ if $l(I)=m-1$. If $l(I)=m-2$, then there are two possibilities:

$l(J)=m-2$ for all $J$ with $|J|=m-1$ or

For a fixed $J_0$ with $|J_0|=m-1$, $l(J_0)=m-1$ and for all other $J$ with $|J|=m-1$, $l(J)=m-2$.

My question is this:

How can I order(partially is fine) this fixed lattice with respect to different labellings so that I will have the labels distributed "nicely"?

As you see I also am not sure of the type of the order. I like to see m-1's close to the top but also many of them. I also like to see large labels more than small labels...

any idea????

**Updates:**

- If $J,K \subset I$ with $|J\cap K|>1$ and $l(J)=l(K)=m-1$, then $l(J\cup K)= m-1$.
- If $K\subset J$, then $l(K)\geq l(J)$.
- If $J_1, \ldots, J_s$ are maximal such that $l(J_i)=m-1$, then $|J_1|+\ldots+|J_s|=m$.