Exercise 3.2.13

Let $A\ne \varnothing$ be open and closed, and suppose for contradiction that $A\ne R$ and $r\notin A$. Note that every closed set must contain its supremum and infimum, but Exercise 3.2.4b shows that every open set cannot contain its supremum or its infimum; thus $A$ must be unbounded.

$A\cap (-\infty ,r)$ is open and closed since $A\cap (-\infty ,r)$ is an intersection of open sets, and $A\cap (-\infty ,r)=A\cap (-\infty ,r]$ (since $r\notin A$) is an intersection of closed sets. Moreover, since $A$ is unbounded below, $A\cap (-\infty ,r)\ne \varnothing$.

Attempting to take $s=\sup <!--\; FUNCTION\; APPLICATION\; -->A\cap (-\infty ,r)$ gives a contradiction, since $s\in A\cap (-\infty ,r)$ (because closed and bounded above) we can find $𝜖>0$ with $V_{𝜖}(s)\subseteq A\cap (-\infty ,r)$ (because open) which contradictions $s$ being an upper bound of $A\cap (-\infty ,r)$.

Therefore if $A\ne \varnothing$ we must have $A=R$. The converse is simple, suppose $A\ne R$ is open and closed, this happens iff $A^c$ is open and closed, but since $A^c\ne \varnothing$ we have $A^c=R$ implying $A=\varnothing$.