$(DEF):$ Let $X$ be a totally ordered set. A set $A \subset X$ is called an upper set if $a \in A $ and $x > a $ implies $x \in A $.
Suppose $A \subset X$ is an upper set that has the least element $a$, then $A = [a, \infty) $
I can prove one direction: If $y \in [a, \infty)$, then $y \geq a $ and we know $a \in A$, then by definition $x \in A $. So, $[a, \infty) \subset A $.
I am stuck trying to prove that $A \subset [a, \infty) $. May I ask for some help? thanks