Supremum is in the Closure of a Nonempty Set

Last updated Nov 6, 2022

Let $(X, \leq)$ be a Total Ordering endowed with the Order Topology. Suppose $A \subset X$ is Nonempty and $\sup\limits A \in X$. Then $\sup\limits A \in \text{cl} A$.

There is a Monotone Net Converging to Supremum and Closure of a Set is all its Net Limits. $\square$

Let $U \subset X$ be Open so that $\sup\limits A \in U$. By construction of the Order Topology we know there must be some Open Interval (including Open Rays and $X$) contained in $U$ that contains $\sup\limits A$. Let $(a,b)$ be this Open Interval (where potentially $a = \leftarrow$ or $b = \rightarrow$). Suppose $(a,b) \cap A = \emptyset$. If $a = \leftarrow$, then since $A \subset (\leftarrow, \sup\limits A] \subset (\leftarrow, b)$, we must have that $A = \emptyset$ $\unicode{x21af}$. Otherwise, if $a \neq \leftarrow$ (i.e. $a \in X$), then $$A = A \cap (\leftarrow, \sup\limits A] = A \cap (\leftarrow, b) = (A \cap (\leftarrow, a]) \sqcup (A \cap (a,b)) = A \cap (\leftarrow, a]$$ so $a$ is an Upper Bound for $A$. But then $a < \sup\limits A$, so $\sup\limits A$ is not a Supremum $\unicode{x21af}$. Therefore $(a,b) \cap A \neq \emptyset$ and $\sup\limits A$ is a Limit Point of $A$. Then $\sup\limits A \in \text{cl}A$ since Closure of a Set is all its Limit Points. $\blacksquare$