# An order ideal in of a poset $P$ is join-irreducible in $J(P)$ if and only if it is generated by a single element?

An order ideal of $$P$$ is a subposet $$Q \subset P$$ with the property that if $$x,y \in P$$, $$x \in Q$$ and $$y \leq x$$ then $$y \in Q$$.

Given a lattice $$L$$, an element $$x \in L$$ is called join-irreducible if it can't be written as the join of two other elements. That is, if $$x = y \vee z$$ then either $$x = y$$ or $$x=z$$.

For a finite poset $$P$$, let $$J(P)$$ denote the set of order ideals of $$P$$. It is a basic result in combinatorics that $$J(P)$$ is a distributive lattice.

I'm trying to show that an order ideal in $$P$$ is join-irreducible in $$J(P)$$ if and only if it is generated by a single element.

I'd appreciate some guidance.. thank you!!

The order ideal generated by a single element $$x\in P$$ is the set $$A_x:=\{y\in P:y\le x\}$$.
If $$A_x=B\cup C$$ for order ideals $$B,C$$, then $$x\in B$$ or $$x\in C$$, but then $$A_x\subseteq B\subseteq A_x$$ in the first case, and similarly $$A_x=C$$ in the second case.
For the converse, we use finiteness of $$P$$ (otherwise it could fail, for example consider $$(-\infty,0)\subseteq\Bbb R$$).
Let $$B$$ be a join irreducible order ideal, and consider the sets $$A_x$$ for all $$x\in B$$, these must all be included in $$B$$, and thus we get $$B=\bigcup_{x\in B} A_x\,.$$ Since $$B$$ is join irreducible, and this is a finite join, we must have $$A_x=B$$ for some $$x\in B$$.