Consider the partial order $(A^*,\leq)$, where A is a set, and $A^*=\{P$ $|$ $P$ is a prtition on A$\}$
Let the order relation be:
$P_{1} \leq P_{2} \leftrightarrow \forall p_{1} \in P_{1} \exists p_{2} \in P_{2}: p_{1} \subseteq p_{2}$
I already proved that $\leq$ is a partial order, (transitivity, antisymmetry and reflection).
Consider the following set $\{P_{1},P_{2}\}$, then prove that exist $P_{3}$ so it is infimum (same for supremum) of the proposed set.
Notice that every $p \in P$ is a subset of A and $\forall P$
$\bigcup_{p \in P} p =A$
I already found an upper and lower bound. I was thinking that maybe there has to be something like a supremum axiom for this cases, I let you my advances:
Upper bound = $\{a,b,c,...\}$
Lower bound = $\{\{a\},\{b\},\{c\},...\}$
I know that if we have that bounds we can eventually find better ones, but if we can not then we already have these ones, in that sense it has to exist infimum and supremum, but I don't know how to write it formally.