I have this question (seemingly simple, though there is something I don't understand). Consider the set $S=\{1,2,...,n\}$ and construct the set $P$ of one element or two element subsets of S. (I understand this as meaning $\{1\},\{1,2\}$ belong to S, etc.). Partially order P by set inclusion: $a\leq b\Leftrightarrow a\subseteq b$
My question has three parts: a) Find the height of P, $h(P)$.
b) Find the width of P, $w(P)$.
c) How many chains of P have $h(P)$ elements?
For a), I know that h(P) is the maximum length of a chain in P. A chain C is a set {a,b,...,c} such that any two elements of C are comparable. So, we have to find the maximum length of a set $\{a,b,...,c\}\subseteq P(S)$=the powerset of S, such that $a\leq b\leq ...\leq c$ (I write $\leq$ for set inclusion). So, a maximal chain would be $C=\{\{1\},\{1,2\}\}$ and so $h(P)=2$. Is this correct?
For b), I know that $w(P)$ is the maximum length of an antichain in P, that is of a set where any two elements are not comparable. So we need construct an antichain that is maximal. I was thinking of {1},{2},....,{n}. This is an antichain and maximal, correct?
For c) I don't quite understand what I am supposed to prove. Any help welcome... Thanks in advance.