Suppose $(P,\preceq)$ is a poset, and let $L=\{a,b\}\subset P$ with $a\preceq b$. Let $S=P\times L$.
Let $A$ be any antichain in $S$.
Let $B$ be the largest possible subset of $P$ such that $B$ does not contain a chain of length of size exceeding $2$.
Show that card $A\leq$card$B$.
I am just stuck at this and don't know even where to start.
Best Answer
One natural way to put an order on $S$ is to say: $(p_1,l_1)\leq (p_2,l_2)$ iff $p_1\preceq p_2$ and $l_1\preceq l_2$. If this is the case, you can consider $A_a=\{x\in P\mid (x,a)\in A\}$ and $A_b=\{x\in P\mid (x,b)\in A\}$. Clearly, $A$ is a disjoint union of $A_a\times\{a\}$ and $A_b\times\{b\}$, so $|A|=|A_a|+|A_b|$.
Note that $A_a\cap A_b=\emptyset$, because if $x\in A_a\cap A_b$ then $(x,a)\prec (x,b)$ in $A$, and $A$ is not an antichain. Thus $|A_a|+|A_b|=|A_a\cup A_b|$.
Both $A_a$ and $A_b$ are antichains: if $x,y\in A_a$ is such that $x\prec y$, then $(x,a)\prec (y,a)$ in $A$, and $A$ is not an antichain. Similarly for $A_b$. So $A_a\cup A_b$ cannot contain a chain of length more than 2, because otherwise some two elements in such a chain would belong to $A_a$ or $A_b$. Since $B$ is some largest subset of $P$ with this property we have $|A_a\cup A_b|\leq |B|$.