Is there a bijective function that is discontinuous?
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Discontinuous everywhere: $f(x)=x$ if $x$ rational, $x+1$ if $x$ irrational.
coffeemath
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7To 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. – Hagen von Eitzen May 11 '16 at 14:32
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@HagenvonEitzen Nice variation, since it gives a compositional inverse pair of maps with each everywhere discontinuous. – coffeemath May 11 '16 at 19:54
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@HagenvonEitzen Does this work if we make $f(x)=-x$ if $x$ algebraic and $1-x$ if $x$ transcendental? – it's a hire car baby 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.
Zanzi
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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.
Siddharth Bhat
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No, but $g(x) = (-1)^{f(x)}\cdot x$ is a discontinuous bijection, the identity on the irrationals and $-x$ on the rationals – Carl Mummert May 11 '16 at 14:09
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1Also 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. – coffeemath May 11 '16 at 14:10
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1@CarlMummert The version you have would be continuous at $0.$ But that could probably easily be adjusted for. – coffeemath May 11 '16 at 14:12