Let $\ge$ be a total order on a connected set $X$. If $A \subseteq X$ is closed in the order topology, then $\sup{A}$ and $\inf{A}$ belong to $A$.
My proof: Suppose $\text{sup}A \notin A$. Then $\text{sup}A \in X\backslash A$, which is open. By definition, there exists a base element $(a,b)$ such that $$ \text{sup}A \in (a,b) \subseteq X\backslash A.$$ By the linear continuum properties (satisfied as long as $X$ is connected) , $\sup{A}$ exists and there is an element $a^{\star}$ such that $a < a^{\star} <\sup{A}$, then $a^{\star} \in (a,b) \subseteq X\backslash A$, so $a^{\star}$ is an upper bound of $A$ smaller that $\text{sup}A$, reaching a contradiction. The case of $\text{inf}A$ is specular.
I think that the only counterexamples here can be found in the cases $\sup{A}$ does not exist. For example, pick $\mathbb{N} \subset \mathbb{R}$ and $\sup{\mathbb{N}}$ is not in $\mathbb{N}$ as it does not exist here. My question is: if I assume, in addition to the proposition I stated, that $\sup{A}$ exists or I assume for example that $A$ is bounded above (hence the existence of the supremum follows), is the proof correct? Is it correct also in the cases in which I do not assume boundedness?
Let K be a not empty, bounded above, closed subset
of a connected linear order space S.
max K in K. Proof.
Since S is connected, s = sup K in S.
Show for all open (a,b) nhood s, exists x in (a,b) $\cap$ K.
Thus s in K, QED.
An order dual proof will prove the order dual theorem:
not empty, bounded below, closed subsets have a minimum in them.