Let $A\subset \mathbb{R}$. Prove that $\sup(A) \in \mathrm{cl}(A)$.

Question

Consider $\mathbb{R}$ with the standard topology and let $A\subset \mathbb{R}$. The closure of a set, denoted $\mathrm{cl}(A)$, is defined as the intersection of all closed sets containing $A$.

Answer 1

As the first case, consider that $\sup(A) \in A$. In this case, since $A\subseteq \mathrm{cl}(A)$ by definition, $\sup(A)\in \mathrm{cl}(A)$.

Now suppose $\sup(A)\not\in A$. Suppose by way of contradiction, that $\sup(A)\not\in\mathrm{cl}(A)$. Then there exists a closed set $C$, containing $A$, for which $\sup(A)\not\in C$. But this means that $\sup(A)\in \mathbb{R}\setminus C$, which is open. Now in the standard topology on $\mathbb{R}$, open sets are defined to be unions of open intervals. Hence there exists an open interval $O\subseteq \mathbb{R}\setminus C$ for which $\sup(A)\in O$. Let us write $O=(a, b)$ to represent this open interval. By definition, $\sup(A)$ is the smallest element of $\mathbb{R}$ which is greater than all elements of $A$. If $a$ is also greater than all elements of $A$, then $\sup(A)$ was not the supremum, a contradiction. If, on the other hand, $a$ is less than or equal to some element of $A$, then $O\cap A \neq \emptyset$, which is impossible since $O\subseteq \mathbb{R}\setminus C \subseteq \mathbb{R}\setminus A$.