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?
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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.
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.