Let $X$ be a nonempty simply ordered set in the order topology having the least upper bound property. If $Y \subseteq X$ is closed and bounded from above, then $\sup{A} \in Y$.
Tentative proof:
Suppose $\text{sup}Y \notin Y$. Then $\text{sup}Y \in X\backslash Y$, open. Then there exists a basis element $(\alpha,\beta)$ s.t.
$$ \text{sup}Y \in (\alpha,\beta) \subseteq X\backslash Y. $$
We distinguish two cases: if there exists at least an element $k$ such that $\alpha \prec k \prec \sup{Y}$ and if it does not exist. In the first case, if $\alpha \prec k \prec \sup{Y}$, then $k \in (\alpha, \beta) \subseteq X\backslash Y$, hence $k$ is an upper bound of $Y$ smaller than $\text{sup}Y$: contradiction.
If $k$ does not exist, then $\sup{Y}$ is the immediate succesor of $\alpha$. If $\alpha \in X\backslash Y$, we get a contradiction for the same reason of before. If $\alpha \in Y$, then $\alpha$ is the maximum of $Y$, so $\alpha = \sup{Y}$, contradicting $\text{sup}Y \in (\alpha,\beta) \subseteq X\backslash Y.$
Am I correct? Any mistakes I am missing? Thanks!
Let $X$ be a LOTS (linearly ordered topological space). A general fact:
( BTW, the same holds for $\inf(A)$ as well.)
Proof: Let $U=(x,\sup(A)]$ or $U=(x,y)$ be a basic neighbourhood of $\sup(A)$ (the first only applies in the case that $\sup(A)=\max(X)$). Then $x$ cannot be an upperbound for $A$ (as $\sup(A)$ is by definition the smallest upperbound of $A$ and in either type of neighbourhood $x < \sup(A)$). It follows that there must be an $a \in A$ such that $a >x$. Also $a \le \sup(A)$ so in either case $U \cap A \neq \emptyset$. As $A$ intersects each basic neighbourhood of $\sup(A)$, $\sup(A) \in \overline{A}$.
So if $A$ is also closed, $\sup(A) \in A$ and $\sup(A)=\max(A)$ follows.