Here is something that is confusing me.

The function $\begin{equation} f(x)=\begin{cases} x, & \text{if $x \in[-1,0]$}\\ x+1, & \text{if $x \in (0,1]$} \end{cases} \end{equation}$

Is clearly discontinuous at $x=0$ yet its inverse

$\begin{equation} f^{-1}(x)=\begin{cases} x, & \text{if $x \in[-1,0]$}\\ x-1, & \text{if $x \in (1,2]$} \end{cases} \end{equation}$

is continuous on the separate counterparts. Now, clearly $f(x)$ doesn't output any values in the range $0 < p \leq 1$ so I don't know if the fact $f^{-1}(x)$ isn't defined in the interval $(0,1]$ is enough to say it's discontinuous or not, because clearly these values aren't of interest for $f(x)$ either. Any clarification would be greatly appreciated!



You must think in terms of intervals.

If f is continuous and injective on an interval, then it has an inverse which is continuous also.

Indeed, if $f$ is continuous and injective on an interval, then it is monotonic and its image is an interval. So the inverse is monotonic on an interval and its image is an interval. Hence it is continuous.

In your case, $f$ is continuous and injective on $[-1,0]$ and on $(0,1]$.

So $f^{-1}$ is continuous on $f([-1,0])=[-1,0]$ and on $f((0,1])=(1,2]$.

The fact that the initial intervals $[-1,0]$ and $(0,1]$ had a common point in their closures does not mean that their images under $f$ have the same property. As your example shows.

The discontinuity of $f$ at $0$ is reflected by a gap in the domain of $f^{-1}$.

  • $\begingroup$ Thank you very much for the excellent explanation. $\endgroup$
    – Noble.
    Feb 25 '13 at 0:19
  • $\begingroup$ @Noble. You're most welcome. $\endgroup$
    – Julien
    Feb 25 '13 at 0:21

Maybe you should draw $f$ and connect $(0,0)$ and $(0,1)$ vertically and invert it graphically. So choosing $f^{-1}$ to be zero on $(0,1]$ gives the continuous inverse. Usually one would start with $f^{-1}$ such defined and one derives that the inverse of that is discontinuous due to zero slop on an open interval.

On the other hand: How is continuity defined on $[-1,0]\cup(1,2]=:A$? $g$ is continuous on that set iff for every sequence $x_n\to x$ in $A$ there holds $f(x_n)\to f(x)$. However since any convergent sequence has to be for $n$ large enough either in $[-1,0]$ or in $(1,2]$ you find that $f(x_n)\to f(x)$ really holds. Hence the function is continuous on $A$. But usually one seeks continuity on (path)-connected sets. Then the previous idea would lead to continuity


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