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!!


1 Answer 1


The join of order ideals is simply their union.

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 $x\in B$.


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