This is one of the problem I have been solving form Velleman's How to prove book:
Suppose $R$ is a partial order on $A$, $B_1 \subseteq A$, $B_2 \subseteq A$, $\forall x \in B_1 \exists y \in B_2 (xRy)$, and $\forall x \in B_2 \exists y \in B_1(xRy)$.
Prove that if $B_1$ and $B_2$ are disjoint then neither of them has a maximal element.
Now I have tried to prove it by contradiction by assuming that $B_1$ has maximal element. But that is not leading me to anywhere. Any thoughts on how to solve this problem?
We let $\leq$ be our order.
Suppose $B_1$ has a maximal element $x$.
Then by the hypothesis there is an element $y$ in $B_2$ so that $x\leq y$.
by our hypothesis there is an element $z$ in $B_1$ so that $y\leq z$
By transitivity $x\leq z$.
If $x=z$ then $x\leq y$ and $y\leq x$ but we know $x\neq y$ because $B_1$ and $B_2$ are disjoint, this is a contradiction.
Therefore we must have $x<z$ which is also a contradiction because $x$ is maximal. The contradiction comes from assuming $B_1$ has a maximal element, we conclude $B_1$ has no maximal element.
(Proving $B_2$ has no maximal element is analogous).