Is there a function $f : [0,1] \rightarrow [0,1]$ such that for all dense subsets S of $[0,1]$ $f$ is discontinuous for all points in that subset? Could you give an example?

Is there a function $f : [0,1] \rightarrow [0,1]$ such that for all $(a,b)$ $f$ defined in $(a,b)$ is surjective on $[0,1]$? What if the codomain of f is $\mathbb{R}$? Could you give an example?

I am also interested into learning if this is true for $(0,1)$ and if these properties have a name rather than what I call "entirely" discontinuous/surjective.

edit: from what I understand conway's 13 is "purely" surjective but is not "purely" discontinuous. Since for every $(a,b)$ there is a $c \in (a,b)$ s.t. $f(c) = 0 $ hence on the set of all $c$ (which is dense), $f$ is continuous.

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    $\begingroup$ Do you know Conway's 13? $\endgroup$ – Asaf Karagila May 7 at 13:04
  • $\begingroup$ this question is phrased very incorrectly. You have to move quantifiers around $\endgroup$ – mathworker21 May 7 at 13:09
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    $\begingroup$ I'd recommend looking at some of the questions listed under "Related". $\endgroup$ – Gerry Myerson May 7 at 13:28

Conway's base 13 function, see e.g. here for its definition, is discontinuous everywhere and has the property that it assumes every real value on every open interval, however small.

I think it fits all your requirements.

  • $\begingroup$ As I pointed out in the edit, Conway's base 13 is "entirerly" surjective. But is it satisfy "entirely" discontinuous? $\endgroup$ – Giannis Tyrovolas May 7 at 14:17
  • $\begingroup$ @GiannisTyrovolas according to the Wikipedia page it is. $\endgroup$ – Henno Brandsma May 7 at 14:18
  • $\begingroup$ What I mean by "entirely" discontinuous is not discontinuous at every point. It means it's discontinuous in every point for every dense subset of R $\endgroup$ – Giannis Tyrovolas May 9 at 20:22

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