I'm trying to solve the following problem.
Let $S$ be a set and $<$ a linear order on $S$. Suppose $E \subset S$ is not empty, and $\inf E = \sup E$. Show that $E$ has exactly one member.
Here is my attempt.
Suppose that $x,y \in E$. By the definition of supremum and infimum, we have $\inf E \leq x \leq \sup E$ and $\inf E \leq y \leq \sup E$. As $\inf E = \sup E$, the first inequality implies that $\inf E = x = \sup E$ and the second that $\inf E = y = \sup E$, so $x = y$. Therefore, $E = \{x\} = \{y\}$, so $E$ has exactly one element.
How does this look? Does it suffice to prove only one direction, or do I need to also prove that if $E$ has a single element, then $\inf E = \sup E$?
Best Answer
Your proof sounds good to me, but you do not need to consider two possible elements.
Indeed, it suffices to consider that $x\in E$ and apply the proposed argument as you did.
From there you shall conclude that $x = \inf(E) = \sup(E)$ whenever $x\in E$, and you are done.
Hopefully this helps!