# What functions can be made continuous by “mixing up their domain”?

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 $$\Bbb R$$.
• (+1) Interesting question. My bet is that any function with a connected range is potentially continuous. – Jack D'Aurizio Oct 20 '17 at 21:43
• @JackD'Aurizio I am pretty sure I found a counter-example. See the last item above. – M. Winter Oct 21 '17 at 0:16
• 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. – Orest Bucicovschi Oct 21 '17 at 0:41
• Is every Darboux function potentially continuous ? – adityaguharoy Dec 18 '17 at 15:50
• 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 $\mathbb R^{\mathbb R}$? – Thomas Andrews Dec 19 '17 at 19:08

## protected by Noah SchweberJan 4 at 15:10

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