- Show that $Q$ is a down-set in $P$ iff its complement $P \setminus D$ is an up-set in $P$.
- The down-set property is transitive, that is, $D \in \mathcal{O}(E)$ and $E \in \mathcal{O}(F)$ $\Rightarrow$ $D \in \mathcal{O}(F)$
- Show that the union or intersection of any family of down-sets is a down-set.
A subset $D$ of a poset $P$ is a down-set if $x \in D$, $y \leq x$ $\Rightarrow$ $y \in D$. We denote the down-sets by $\mathcal{O}(P)$. Dually, a subset $U$ of $P$ is an up-set if $x \in U$, $y \geq x$ $\Rightarrow$ $y \in U$. We denote the up-sets in $P$ by $\mathcal{U}(P)$.
Somebody help me please.
Thanks.
As an example, I'll give you a proof of the first statement.
Hopefully that will help you getting a grasp on the subject.
To prove that $Q$ is a down-set of $P$ iff $P\setminus Q$ is an up-set, it is enough to prove one implication (the other will follow by duality, or if you prefer, by applying a similar reasoning).
So let us suppose that $Q$ is a down-set, and prove that $P\setminus Q$ is an up-set.
Take $u \in P \setminus Q$ and $v \in P$ such that $u \leq v$. We want to prove that $v \in P \setminus Q$.
So suppose, for a contradiction, that $v \notin P \setminus Q$; thus $v \in Q$, and since $u \leq v$ and $Q$ is a down-set, $u \in Q$, a contradiction because $u \in P \setminus Q$.
Try to understand the reasoning and apply similar ones (they're all very simple) to the proofs of the other statements. If you can't find these proofs, tells where you got stuck.