In a finite poset $P$, a down-set $I$ is meet-irreducible $\iff I=P\backslash\uparrow x$ for some $x\in P$ Let $P$ be a finite ordered set. Show that a down-set $I$ is meet-irreducible in $\mathcal{O}(P)$, that is $I \in \mathcal{M}(\mathcal{O}(P))$, if and only if it is of the form $P\backslash\uparrow x$ for some $x\in P$.
Proving that $\forall \,x\in P$ taking $U= \downarrow x$ gives you a join-irreducible element of $\mathcal{O}(P)$ is quite easy, because if you write $I \cup J = \downarrow x $, for some $I, J \in \mathcal{O}(P)$, you have that w.l.o.g. $x \in I$, so $\downarrow x \in I$, that means $\downarrow x = I$. I tried to do something similar for the meet-irreducible elements, but it seems not to lead anywhere.
 A: It's not clear if this fully addresses your question (as it doesn't answer the question in the title, for instance), but it seems that this should work as a dual to the join-irreducible proof you gave:

Let $I\cap J=P-\!\uparrow\!\!x$  Then wlog $x\notin I$, so $\uparrow\!\!x\cap I = \varnothing$.
In other words, $I\subseteq P-\!\uparrow\!\!x$ and therefore $P-\!\uparrow\!\!x = I$.

A: This answer intends to complete the one of Eric Nathan Stucky, which proves one of the implications, and I'll refer you to that answer for that part of the exercise (upvote it and/or mark it as the right answer if you find it helpful), and I'll prove that if $I$ is meet-irreducible, then $I = P \setminus {\uparrow} c$, for some $c \in P$, or equivalently, that if $I$ doesn't have that shape, then $I$ is not meet-irreducible.
Although you don't mention it in your proof of the join-irreducible elements (you only include one implication, just like Eric) you also need to prove that all join-irreducible down-sets have that shape, which you might do starting with the observation that there exist $c_1, \ldots, c_k \in P$, with $c_i\|c_j$ whenever $i\neq j$, such that $I = {\downarrow} c_1 \cup \cdots \cup {\downarrow} c_k$, for every down-set $I$ in a finite poset $P$ (here, $x\|y$ means that $x\nleq y$ and $y\nleq x$, i.e., $x$ and $y$ are incomparable).
Just the same, if $U$ is an up-set, there exist $d_1, \ldots, d_k \in P$ such that $U = {\uparrow} d_1 \cup \cdots \cup {\uparrow} d_k$.
The complement of a down-set is an up-set;
so if $I$ is a down-set, then
$$I = P \setminus U = P \setminus \bigcup_{i=1}^k {\uparrow}d_i = \bigcap_{i=1}^k P \setminus{\uparrow}d_i,$$
with $d_i\|d_j$ for $i\neq j$.
Now, either $k=1$ and $I$ has the desired shape, or $k>1$ and clearly $I$ is not meet-irreducible (since it is the intersection of different down-sets larger than $I$).
