An open set is contained in an open interval I am wondering if for an arbitrary open set $A$ in $\mathbb{R}$, there exist an open interval such that $A$ is contained in that interval.
My guess is that this is true, and that this interval would be $(\inf(A), \sup(A))$, where the infimum and supremum can go to infinity.
Is this correct?

Clearly $A\subseteq[\inf A,\sup A]$.
On the other hand, $\sup A\notin A$, because otherwise you'd have $(\sup A-\varepsilon,\sup A+\varepsilon)\subseteq A$ for some $\varepsilon>0$, contradicting $\sup A$ being the supremum of $A$.
Similarly $\inf A\notin A$, so we have $A\subseteq(\inf A,\sup A)$.
Can there be a smaller interval containing $A$? It would be $(u,v)$ where either $u>\inf A$ or $v<\sup A$. Either case leads to a contradiction.
