Let $I$ be an open interval containing $a\in\mathbb{R}$. I know that $I=(\inf(I),\sup(I))$.
Is it possible to find $\delta>0$ such that $(a-\delta,a+\delta)=I?$
This would imply that $\sup(I)-\delta=a$ and $\inf(I)+\delta=a$. Then we get $\delta=[\sup(I)-\inf(I)]/2$. But picking $\delta:=[\sup(I)-\inf(I)]/2$ does necessarily imply that $(a-\delta,a+\delta)=I$, since this would imply that $a=\frac{\inf(I)+\sup(I)}{2}$ which may not necessarily be the case.
I guess this shows that $I\ne(a-\delta,a+\delta)$ generally. Maybe it's possible to embed $I$ in an interval of the form $(a-\delta,a+\delta)$ or the other way around?
If $\ I\subset\mathbb{R}\ $ is an interval, then $\ I = (\inf (I), \sup (I))\ = (a-\delta, a+\delta),\ $ where
$a=\frac{\inf (I) + \sup (I)}{2}\ $ and $\ \delta = \ \frac{\sup (I) - \inf (I)}{2}.$
You can confirm via calculation that $\ a-\delta = \inf(I)\ $ and that $\ a+\delta = \sup(I).$