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If I have a continuous function $f : \bar{\mathbb{Q}} \to \mathbb{Q}$, where $\bar{\mathbb{Q}}$ denotes the set of algebraic numbers, how could I show that this function has to be constant?

I was trying to prove it in the following way: I assume my function not to be constant. Then, I can always find an $y \in \bar{\mathbb{Q}} \setminus \mathbb{Q}$ from the range of my function $f$ (due to denseness), and then while bringing back the point $y$ via the prei-image of my function $f$, I would get a point in $\bar{\mathbb{Q}}$ (I can't convince myself if this is true completely), which then would bring me to a contradiction. I am not sure if my last argument holds true. Any help or reading suggestion about algebraic numbers, would help me a lot.

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    $\begingroup$ What topology are you using for $\overline\Bbb Q$? For example if you have an embedding of $\overline\Bbb Q$ into $\Bbb C$ and you use the subspace topology, then $f(z) = \Im z$ is a counterexample. $\endgroup$ – Greg Martin Aug 14 at 17:34
  • $\begingroup$ Usual topology @GregMartin $\endgroup$ – b.omega Aug 14 at 17:48
  • $\begingroup$ More generally any polynomial with algebraic coefficients is continuous, and even continuous for all the possible nonarchimedean topologies too. $\endgroup$ – user208649 Aug 14 at 19:17
  • $\begingroup$ @TokenToucan But the range of the function shall be contained in $\mathbb{Q}$, not in $\overline{\mathbb{Q}}$. $\endgroup$ – Daniel Fischer Aug 14 at 19:39
  • $\begingroup$ @DanielFischer ahh didn't spot that! $\endgroup$ – user208649 Aug 14 at 19:46
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The claim is actually false. Proffering the following counterexample $$f(x)=\begin{cases}1,&\ \text{if the real part of $x$ is greater than $\pi$, and}\\ 0,&\ \text{otherwise.}\end{cases}$$

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  • $\begingroup$ First I tried to prove the claim when the domain is $\Bbb{Q}(\sqrt2)$. Then I came up with the same counterexample using $\sqrt3$ instead of $\pi$. After a few more minutes I saw the light. $\endgroup$ – Jyrki Lahtonen Aug 17 at 9:33

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