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These are the properties that such functions should (could?) have:

1) $f(\mathbb R)=\mathbb R$

2) $f$ is everywhere discontinuous

3) $\mathbb Q \subseteq f( \mathbb I)$

4) $f(\mathbb Q) \subset \mathbb I$

That is, such functions should (could?) have all these properties simultaneously: They map the whole $\mathbb R$ onto the whole $\mathbb R$, they are everywhere discontinuous, they map the set of all irrational numbers $\mathbb I$ into some set that contains all rationals, and they map the set of all rationals into some subset of the set of all irrationals.

Remark: This is not a homework, I am just eager into doing research of everywhere discontinuous functions.

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Let $g : \mathbb{I} \to \mathbb{R}$ be onto (such a function exists because $\mathbb{I}$ and $\mathbb{R}$ have the same cardinality), and partition $\mathbb{Q}$ into two disjoint dense sets $Q_1, Q_2$. Then let $$f(x) = \begin{cases} g(x), & x \in \mathbb{I} \\ -\sqrt{2}, & x \in Q_1 \\ \sqrt{2}, & x \in Q_2.\end{cases}$$

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  • $\begingroup$ $Q_1$ and $Q_2$ should be "everywhere close to each other", right? $\endgroup$ – user720692 Nov 3 at 23:10
  • $\begingroup$ @YamMir: Yes, that is accomplished by having them both be dense. For instance, you may take $Q_1$ to be the set of dyadic rationals $n/2^k$, and $Q_2 = \mathbb{Q} \setminus Q_1$. $\endgroup$ – Nate Eldredge Nov 3 at 23:11
  • $\begingroup$ Do you know of some concrete constructed example of such $g$ which is from $\mathbb I$ onto $\mathbb R$? $\endgroup$ – user720692 Nov 3 at 23:15
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    $\begingroup$ @YamMir: Pick your favorite countable set of irrationals, $A = \{a_1, a_2, \dots\}$, and enumerate the rationals as $\{q_1, q_2, \dots\}$. If $n$ is even, let $g(a_n) = a_{n/2}$. If $n$ is odd, let $g(a_n) = q_{(n+1)/2}$. If $x \in \mathbb{I} \setminus A$, let $g(x)=x$. $\endgroup$ – Nate Eldredge Nov 3 at 23:18
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Sure.

$$f(x)=\begin{cases}x+\sqrt 2&\text{if }x=a+b\sqrt 2\text{ with }a\in\Bbb Q, b\in\Bbb Z\\x&\text{otherwise}\end{cases} $$

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  • $\begingroup$ But if $r$ is rational and not of the form $r=a+b\sqrt 2$ then $f(r)=r$ so $f(\mathbb Q) \cap \mathbb I \nsubseteq \mathbb I$, so $0$ is a problem? $\endgroup$ – user720692 Nov 3 at 22:21
  • $\begingroup$ I think I misunderstood something, I will think about your example again. $\endgroup$ – user720692 Nov 3 at 22:28
  • $\begingroup$ But all the irrationals which are not of the form $a+b\sqrt{2}$ are mapped into themselves, how can then $\mathbb Q \subseteq f( \mathbb I)$ be true? What do I miss? $\endgroup$ – user720692 Nov 3 at 22:31

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