$R$ is a partial order on $A$.
$B_1 \subseteq A$
$B_2 \subseteq A$
$\forall x \in B_1 \exists y \in B_2 (xRy)$
$\forall x \in B_2 \exists y \in B_1 (xRy)$
Prove that if $B_1$ and $B_2$ are disjoint then neither of them has a maximal element.
My attempt:
I will only show that $B_1$ doesn't have maximal element, because proof for $B_2$ is very similar.
Suppose $B_1$ and $B_2$ are disjoint
Suppose $b$ is R-maximal element of $B_1$.
We know that for all $x \in B_1$ there must exist some $y \in B_2$ such that $xRy$. Since $b \in B_1$, there is some $y \in B_2$ such that $bRy$. We know that there also must exist some $p \in B_1$, such that $yRp$.
Since $R$ is transitive, we have $bRp$. We know that $b$ is maximal element, hence $b = p$. We also know that $R$ is antisymmetric, hence $y = b$. But that would imply that $y \in B_1 \cap B_2 $, which is a contradiction, since we know that $B_1 \cap B_2 = \emptyset$. Hence $b$ is not a maximal element.
Proof of $B_2$ not having maximal element is nearly identical. $\Box$
Is it correct?