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Definition. A function $f:\Bbb R\to\Bbb R$ will be called potentially continuous if there is a bijection $\phi:\Bbb R\to\Bbb R$ so that $f\circ \phi$ is continuous.

So one could say a potentially continuous (p.c.) function is "a continuous function with a mixed up domain". I was wondering whether there is an easy way to characterize such functions.


Some thoughts $\DeclareMathOperator{\im}{im}$

  • If the image $\im(f)$ is not connected (i.e. no interval), then $f$ is not p.c. because even mixing the domain cannot make $f$ satisfy the intermediate value theorem.
  • Bijective functions are always p.c. because we can choose $\phi=f^{-1}$. Every injective function with an open connected image is p.c. for a similar reason. However, only having a connected image is not enough, as e.g. there are bijections, but no continuous bijections $f:\Bbb R \to [0,1]$.
  • Initially I thought a function can never be p.c. if it attains every value (or at least uncountably many values) uncountably often, e.g. like Conways base 13 function. But then I discovered this: take a Peano curve like function $c$ (or any other continuous surjection $\Bbb R\to\Bbb R^2$) and only look at the $x$-component $c_x:\Bbb R\to\Bbb R$. This is a continuous function which attains every value uncountably often.
  • The question can also be asked this way. Given a family of pairs $(r_i,\kappa_i),i\in I$ of real numbers $r_i$ and cardinal numbers $\kappa_i\le\mathfrak c$ so that $\{r_i\mid i\in I\}$ is connected. Can we find a continuous function $f:\Bbb R\to\Bbb R$ with $|f^{-1}(r_i)|=\kappa_i$?
  • There is no continuous function which attains each real number exactly once except zero which is attained twice. So, e.g. the function $$f(x)=\begin{cases}x-1&\text{for $x\in\Bbb N$}\\x&\text{otherwise}\end{cases}$$ is not p.c., even though its image is all of $\Bbb R$.
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    $\begingroup$ (+1) Interesting question. My bet is that any function with a connected range is potentially continuous. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 21:43
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    $\begingroup$ @JackD'Aurizio I am pretty sure I found a counter-example. See the last item above. $\endgroup$ – M. Winter Oct 21 '17 at 0:16
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    $\begingroup$ If $f^{-1}(0) = \{a,b\}$, and $f$ injective on $(a,b)$, then $f$ is strictly increasing or decreasing on $(a,b)$, so $f(a) \ne f(b)$, contradiction. $\endgroup$ – Orest Bucicovschi Oct 21 '17 at 0:41
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    $\begingroup$ Is every Darboux function potentially continuous ? $\endgroup$ – adityaguharoy Dec 18 '17 at 15:50
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    $\begingroup$ It would be interesting to find a pair $f,g$ of potentially continuous functions such that $f+g$ are not p.c, or are the p.c. functions a vector subspace of $\mathbb R^{\mathbb R}$? $\endgroup$ – Thomas Andrews Dec 19 '17 at 19:08

protected by Noah Schweber Jan 4 at 15:10

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