Property of an open subset of $\mathbb{R}$ containing $0$ Let $E$ be an open subset of $\mathbb{R}$ with $0\in E$. I wish to show that any $z\in \mathbb{R}$ can be written $z=1/x+1/y$ for some $x,y\in E$.
I know that our assumptions imply $(-\epsilon,\epsilon) \subset E$ for some $\epsilon>0$, but I am not sure where to go from here. Any guidance would be appreciated!
 A: Any real number z can be written as a-b where a and b are positive. Also z=(a+n)-(b+n) for any integer n. If n is large enough then $x= \frac 1 {a+n}$ and $y=- \frac 1 {b+n}$ are both inside the open set since they approach 0 as n approaches $\infty$.  
A: If $|z|>2/\epsilon$ we may define $\delta=2/z\in(-\epsilon,\epsilon)$.  Then setting $x=\delta$ and $y=\delta$ puts $z$ in the desired form.
For small $z$ (where $|z|\leq 2/\epsilon$), we need to write $z$ as a difference, and there are many ways to do this.  For instance, we can declare that $x=\epsilon/4\in(-\epsilon,\epsilon)$.  Since we want
\begin{equation}
z=1/x+1/y,
\end{equation}
we may then set $y=1/(z-1/x)=\epsilon/(\epsilon z-4)$.  Of course we need to check that $y\in E$.  But since $|z|\leq 2/\epsilon$, $|\epsilon z|\leq 2$.  So $|\epsilon z-4|\geq 2 > 1$, meaning that
\begin{equation}
\bigg\vert \frac{\epsilon}{\epsilon z-4}\bigg\vert < \epsilon,
\end{equation}
and indeed $y\in(-\epsilon,\epsilon)$.  Moreover, $z=1/x+1/y$ by construction.
