when $C$ is a maximal chain

Question

let $R$ be partial order on $A$.

suppose $C \subseteq A$ is a chain, then $C$ is maximal iff for every $a\in A \setminus C$ there is a $c\in C$ s.t $a$ and $c$ are $(a,c)\notin R \\and \\(c,a)\notin R$

the only thing that I know that $C$ is maximal chain that if you add an elemt to it it's any more :(

any solution or hint would be great and thank in advance

Answer 1

A slightly easier-to-parse version of the statement of the problem is the following:

Let $R$ be a partial order on the set $A$, and suppose that $C\subseteq A$ is a chain in the poset $P =(A,R)$. Show that $C$ is maximal iff for every element $a\in A$ that is not an element of the chain (equivalently, $a\in\overline{C}$), there exists an element $c$ of the chain $C$ such that $a$ and $c$ are incomparable.

Suppose that $C$ is a maximal chain in $P$. Fix $a\in \overline{C}$. Then, for each element $c\in{C}$, $a$ and $c$ are either comparable or incomparable. Suppose that for every element $c\in{C}$, we have that $a$ and $c$ are comparable. Then, the set $C\cup\{a\}$ is a chain, contradicting the maximality of $C$. Thus, for some $c\in C$, we must have that $a$ and $c$ are incomparable. Since $a$ was chosen arbitrarily, we are done.

Now, suppose that for every $a\in \overline{C}$, there exists $c\in C$ such that $a$ and $c$ are incomparable. We have to show that the chain $C$ is maximal. Assume to the contrary. Then, there exists an element $a\in\overline{C}$ such that $a$ and $c$ are comparable for every element $c$ of the chain, contradiction. Thus, $C$ is maximal.

The proof is complete. $\square$