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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?

user569852
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    Looks like $(-\epsilon,0]$ to me. I would presume that it is a typo. – copper.hat Jun 14 '18 at 15:14
  • Many thanks for the speedy reply. – user569852 Jun 14 '18 at 15:21
  • I get completely stumped by typos. from time to time. – copper.hat Jun 14 '18 at 15:30
  • I suppose authors silently assumed the epsilon symbol automatically implies 'small enough' or even 'arbitarily small'. In this case 'small enough' obviously (?) is $0< \epsilon < \frac 12$, as you found, and then the result is, of course, $(-\epsilon,0]$. IMHO you just bumped into a coincidence of a lack of precision in assumptions and a typo in a result. – CiaPan Jun 14 '18 at 15:51

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The solution should be $(-\epsilon, 0]$. Most likely just a mistake.

JohnKnoxV
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