Why the supremum of a closed set is not necessarily equal to the maximum?

Question

Consider a partially ordered set $A\subseteq \mathbb{R}^k$. I know that if $A$ is compact then $\sup A=\max A$. Could you give an intuition of why having $A$ closed is not enough to conclude $\sup A=\max A$? Assume that I allow the supremum to take an infinite value.

Answer 1

Suppose $(X,<)$ is a linear order in the order topology and for some $A\subseteq X$, $A$ is closed and $\alpha=\sup(A)$ exists but it is not an element of $A$. Then, for every open interval $I$ containing $\alpha$ then by definition of supremum, there is an element $a\in A$ such that $a\in I$.

This shows that $\alpha$ is a limit point of $A$, and since $A$ is closed it should contain all its limit points, so $\alpha\in A$, and $\sup A=\max A$. I believe then that when considering the order topology it is the case that if $A$ is closed and $\sup A$ exists, $\sup A=\max A$. It could happen however that a closed set does not have maximum nor supremum, as for instance $A=[0,\infty)$ in the reals.

However, with a change of topology we might have different things. Consider the topology on $\mathbb{R}$ whose basic opens are sets of the form $[a,b)$. In this case, the set $A=(-\infty,0)$ is closed, does not have a maximum and $\sup A=0$ exists.