On page 10 of the book 'Topology and Geometry for Physicists' by Nash and Sen, an example is discussed involving a function
\begin{equation} f(x)=\begin{cases} -x + 1, & \text{if $x \leq 0$}\\ -x+1/2, & \text{if $x >0$} \end{cases} \end{equation}
The function has a discontinuity at $x=0$ and to illustrate how this leads to a failure of the usual topological definition of continuity (the inverse image of an open set in $Y$ is an open set in $X$), the authors compute the inverse of the open set $(1-\epsilon, 1+\epsilon)$.
The authors say that $f^{-1}\{(1 - \epsilon, 1+ \epsilon)\} = (-\epsilon, 1]$. I cannot see how this can be correct and am wondering if I am missing something, or if the authors have made a mistake?
I feel the example is not particularly illuminating unless we take $0 < \epsilon < 1/2$ (the authors do not specify this - they simply say $\epsilon > 0$), and it seems to me that under this assumption we get $f^{-1}\{(1 - \epsilon, 1+ \epsilon)\} = (-\epsilon, 0]$. I cannot work out how the authors got $(-\epsilon, 1]$.
Can someone explain this?
The solution should be $(-\epsilon, 0]$. Most likely just a mistake.