Let $A$ be a partially ordered set in which every chain has an upper bound, and let $a \in A$. Then $A$ has a maximal element $u$ such that $u \ge a$.

Question

Let $A$ be a partially ordered set in which every chain has an upper bound, and let $a \in A$. Then $A$ has a maximal element $u$ such that $u \ge a$.

I'm trying to solve this by way of contradiction. If we contradict the hypothesis then for every $u \ge a$, there will be a $x \ge u$. My idea is that we can form a chain by adding $z$ such that $z \ge x$, then we will get a chain that has no upper bound. However, I am not sure how to formulate such a chain. How can I make this proof rigorous?

Answer 1

Hint: consider $B=\{\,x\in A:x\ge a\,\}$ and try proving $B$ (with the induced order) satisfies the same property (every chain has an upper bound); by Zorn's lemma, $B$ has a maximal element; is this also maximal in $A$?