Ordered Sets - Down-sets

Question

Let $A$ and $B$ be decreasing sets of $P$. Prove that $A \prec B$ in $\langle {\mathcal{O} (P); \subseteq} \rangle$ iff $B = A \cup \{b \}$ for some minimal element $b$ of $P \setminus A$.

Let $P$ be an ordered set and $Q \subseteq P$. Then $a \in Q$ is a minimal element of $Q$ if $x \leqslant a$ and $x \in Q$ imply $a = x$.

$Q$ is a decreasing set (or down-set), whenever $x \in Q$, $y \in P$ and $y \leqslant x$, we have $y \in Q$.

$x \prec y$ ($x$ is covered by $y$), if $x < y$ and $x \leqslant z < y$ implies $z=y$.

The family of all down-sets of $P$ is denoted by $\mathcal{O} (P)$

I understand the exercise, but I don't know how to prove it, the first thing I tried was

$B \subseteq A \cup \{b \}$ and $B \supseteq A \cup \{b \}$ $\Longrightarrow$ $B = A \cup \{b \}$,

and then I wanted to prove both inclusions, but I couldn't. Somebody help me please.

The book is Davey B.A., Priestley H.A. - Introduction to Lattices and Order page 27 - exercise 1.12.

Thanks.

Answer 1

$A \prec B$ implies $A < B$ in the poset $\langle \mathcal{O}(P); \subseteq \rangle$, so $A \subsetneq B$, which implies there is at least some $b \in B\setminus A$. So clearly $A \cup \{b\} \subseteq B$.

You will need to show $A \cup \{b\}$ is a down-set, so a member of $\mathcal{O}(P)$. Can you show that $b$ is minimal in $P \setminus A$, as required? Can there be another element in $B\setminus A$ besides $b$?

Answer 2

Suppose $A,B\in\mathcal O(P)$ and $A\prec B$.

Claim. $B\setminus A$ has exactly one element.

Proof of claim. Clearly $B\setminus A\ne\emptyset$, since $A$ is a proper subset of $B$. Assume for a contradiction that $B\setminus A$ has more than one element, i.e., we have $b,c\in B\setminus A$, $b\ne c$. Without loss of generality we can assume that $b\not\le c$. Let $C=A\cup\{x\in P:x\le c\}$. Then $C\in\mathcal O(P)$ and $A\subsetneq C\subsetneq B$, but this contradicts the assumption that $A\prec B$.

So $B=A\cup\{b\}$ where $b\notin A$. Since $B$ is a decreasing set, $b$ must be a minimal element of $P\setminus A$.