1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ I'd recommend looking at some of the questions listed under "Related". $\endgroup$ – Gerry Myerson May 7 at 13:28
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.