Prove maxA is unique. Let $(S, <)$ be an ordered set and $A \subset S$. Suppose that $A$ contains a largest element, which we denote by $\max A$. Prove that $A$ has exactly one largest element. 
I'm stuck with this question and I can't quite get something written. I thought if used proof by contradiction by saying that A contains no largest element but that doesn't seem to get me anywhere. Any help/suggestions?
 A: The goal here is to show that if $A$ contains a largest element, then it contains exactly one largest element, denoted by $\max(A)$.  This is a uniqueness proof.  Typically, to prove uniqueness, you suppose that there is another element of $A$ that is also has the property of being "largest", then show that it as to be $\max(A)$.  So the proof is as follows:
Proposition: Let $A \subseteq (S,<)$, and suppose that $A$ contains a maximal element, denoted $\max(A)$.  Then $A$ has exactly one maximal element.
Proof:  Suppose that $x \in A$ such that $a < x$ for all $a\in A$.  In particular, this implies $\max(A) < x$.  On the other hand, $\max(A)$ is a largest element of $A$, which implies $a < \max(A)$ for all $a\in A$.  In particular, we have that $x < \max(A)$.  Therefore
$$ \max(A) < x < \max(A). $$
By the anti-symmetry of $<$, this is only possible if $x = \max(A)$, therefore the largest element $\max(A)$ is unique.
A: You can use proof by contradiction by saying suppose there are two greatest elements and then show that they are both greater than each other. 
