# Discontinuous bijection example

Is there an example of a function that is discontinuous but bijective whose inverse is also a bijection and discontinuous?

• Almost anything, really. Being bijective doesn't have anything to do with being continuous. Just try thinking of your favorite discontinuous function and check if it also happens to be a bijection. – Jair Taylor Sep 20 '16 at 17:46
• Three answers have appeared (including mine) but so far I'm the only one to up-vote the question (and the only one whose answer has no up-votes). $\qquad$ – Michael Hardy Sep 20 '16 at 22:33

$$f(x) = \begin{cases} x & \text{if }x\in \mathbb Q \\ x+1 & \text{if }x\in \mathbb R\setminus\mathbb Q \end{cases}$$
Or even $$g(x) = \begin{cases} -x & \text{if }x\in\mathbb Q \\ 2\sqrt2-x & \text{if }x-\sqrt2 \in \mathbb Q \\ x & \text{otherwise} \end{cases}$$ which is its own inverse and nowhere continuous.
How about this: \begin{align} f &: [0,1)\to [0,1) \\[5pt] f(x) & = \begin{cases} x+\frac 1 2 & \text{for } 0\le x\le \frac 1 2, \\[5pt] x - \frac 1 2 & \text{for } \frac 1 2 \le x < 1. \end{cases} \end{align} This is an involution, i.e. it is its own inverse. It has a jump discontinuity at $x=\frac 1 2.$