The following exercise comes from the book: Introduction to Lattices and order (second edition, B. A. Davey, H.A. Priestley).
Problem 1.14, pag. 27.
Let $P$ be a finite oredered set.
(1) Show that $Q=\downarrow\mbox{Max}(Q)$, for all $Q\in\mathcal{O}(P)$;
(2) Establish a one-to-one correspondence between the elemenents of $\mathcal{O}(P)$ and antichain in $P$.
(3) Hence show that, for all $x\in P$,
$$|\mathcal{O}(P)|=|\mathcal{O}(P\setminus\{x\})|+|\mathcal{O}(P\setminus (\downarrow x\ \cup \ \uparrow x))|$$
Some definitions
Definition 1 (antichain). Let $P$ an ordered set. $P$ is an antichain if $x\le y$ in $P$ only if $x=y$.
Definition 2 (maximal element). Let $P$ an ordered set, $Q\subseteq P$. Then $a\in Q$ is a maximal element of $Q$ if $a\le x$ and $x\in Q$ imply $a=x$. The set of all maximal element of $Q$ is $\mbox{Max}(Q)$.
Definition 3 (Down-set). Let $P$ be an ordered set, and $Q\subseteq P$. $Q$ is a down-set (of $P$) if, whenever $x\in Q,y\in P$ and $y\le x$, we have $y\in Q$.
Definition 4 (down Q, down x, up x). Given an arbitrary subset $Q$ of $P$ and $x\in P$, we define
$\\ \bullet \ \downarrow Q=\{y\in P \ | \ \exists x\in Q\ \mbox{s.t.} \ y\le x\} \ \ \ (\mbox{Down} \ Q)$
$\bullet \ \downarrow x=\{y\in P\ | \ y\le x\} \ \ \ (\mbox{Down} \ x)$
$\bullet \ \uparrow x=\{y\in P\ | \ y\ge x\}\ \ \ (\mbox{Up} \ x)$
Definition 5 (set of all down-set of P). The family of all down-set of a set $P$ is denoted by $\mathcal{O}(P)$.
$|A|$ is the cardinality of the set $A$.
My ideas
(1.a) I want to show that $\downarrow\mbox{Max}(Q)\subseteq Q,\, \forall Q\in\mathcal{O}(P)$.
Let $x\in\downarrow\mbox{Max}(Q)$, then $\exists q\in\mbox{Max}(Q)$ such that $x\le q\implies x\in Q$, because $Q$ is a down-set of $P$. Is it ok?
(1.b) Now I'll try to prove that $Q\subseteq\downarrow \mbox{Max}(Q)$.
Suppose, by contraddiction, that there exists $q\in Q\setminus\downarrow\mbox{Max}(Q)$, i.e. $q\in Q\wedge q\notin\downarrow\mbox{Max}(Q)$, so $\forall s\in\mbox{Max}(Q)$, we have that $(q>s)\vee q\ ||\ s$ ($q$ is not comparable with $s$).
If $q>s$, then $s\notin\mbox{Max}(Q)$, (contraddiction).
If $q\ || \ s$, then $q\in\mbox{Max}(Q)\implies q\in\downarrow\mbox{Max}(Q)$ (contraddiction).
Is this part ok?
(2) Let $\mathcal{A}(P)$ be the set of all antichains of $P$. I think that
$\begin{array}{ccccc}\phi&:&\mathcal{O}(P)&\longrightarrow&\mathcal{A}(P)\\ \\ &&Q=\downarrow\mbox{Max}(Q)&\longmapsto&\phi(Q)=\mbox{Max}(Q)\end{array}$
is a good one-to-one correspondence.
(3) I have no idea to solve this. I have to study this answer to understand what to do.
Please, check my ideas and tell me if they are ok or not: the order theory is so insidous to me. Thank you.
(1) is OK, although the proof that $Q \subseteq {\downarrow}\mathrm{Max}Q$ has at least a minor problem: if $q||s$ for all $s \in \mathrm{Max}Q$, then $q \notin \mathrm{Max}Q$, because $q \leq q$.
But you can just say (I guess) that in a finite poset any element is below a maximal element, and proceed from there.
(2) is OK.
For (3), just note that the number of anti-chains in $P$ is the sum of the number of anti-chains in $P\setminus\{x\}$ with the number of anti-chains in $P \setminus ({\downarrow}x \cup {\uparrow}x)$.
Indeed, if $A$ is an anti-chain in $P$ and $x \in P$, then, if $x \in A$ it follows that $A$ is an anti-chain in $P \setminus ({\downarrow}x \cup {\uparrow}x)$; otherwise (if $x \notin A$), then $A$ is an anti-chain in $P \setminus\{x\}$.
Hence, by (2) you get the desired result.