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Is there a bijective function that is discontinuous?

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    $\begingroup$ Take a continuous bijection and exchange the images of two random points. $\endgroup$ – Captain Lama May 11 '16 at 13:58
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    $\begingroup$ It would be more interesting to ask for a everywhere discontinuous bijection $\endgroup$ – Mathematician 42 May 11 '16 at 13:59
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Discontinuous everywhere: $f(x)=x$ if $x$ rational, $x+1$ if $x$ irrational.

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    $\begingroup$ To make this an even more fascinating example: If we use $-x$ and $1-x$ instead of $x$ and $x+1$, then $f$ is discontinuous everywhere and $f\circ f$ is the identity. $\endgroup$ – Hagen von Eitzen May 11 '16 at 14:32
  • $\begingroup$ @HagenvonEitzen Nice variation, since it gives a compositional inverse pair of maps with each everywhere discontinuous. $\endgroup$ – coffeemath May 11 '16 at 19:54
  • $\begingroup$ @HagenvonEitzen Does this work if we make $f(x)=-x$ if $x$ algebraic and $1-x$ if $x$ transcendental? $\endgroup$ – samerivertwice Dec 11 '17 at 11:13
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$$f(x) =\left\lbrace \begin{array}{ll} x & \text{ if } x\not\in\lbrace 0,1\rbrace \\ 1 & \text{ if } x=0 \\ 0 & \text{ if } x=1 \end{array} \right.$$ for example.

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Yes. In fact, one can construct a maximally disconnected function in the sense that it is discontinuous on every open interval $(a, b)$.

This is constructed by having every open interval $(a, b)$ map to $\mathbb{R}$. An example of this is Conway's base 13 function

Another example is the indicator function for rational numbers.

Define a function

$$ f(x) = \begin{cases} x \in \mathbb{Q} & 1 \\x \notin \mathbb{Q} & 0\end{cases} $$

This function is discontinuous everywhere because the rationals are dense in the reals, and so are the irrationals.

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  • $\begingroup$ Is the indicator function a bijection? $\endgroup$ – coffeemath May 11 '16 at 14:07
  • $\begingroup$ No, but $g(x) = (-1)^{f(x)}\cdot x$ is a discontinuous bijection, the identity on the irrationals and $-x$ on the rationals $\endgroup$ – Carl Mummert May 11 '16 at 14:09
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    $\begingroup$ Also if in the first one every open interval maps to $\mathbb{R},$ it won't be a bijection, at least if these maps are each onto. $\endgroup$ – coffeemath May 11 '16 at 14:10
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    $\begingroup$ @CarlMummert The version you have would be continuous at $0.$ But that could probably easily be adjusted for. $\endgroup$ – coffeemath May 11 '16 at 14:12

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