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I'm working on the following problem from N.L. Carother's Real Analysis:

Let $I=(\mathbb{R}\setminus\mathbb{Q})\cap [0,1]$ with its usual metric. Prove that there is a continuous function $g$ mapping $I$ onto $\mathbb{Q}\cap[0,1]$.

My thoughts:

I feel the preimage of open sets definition of continuity will be the easiest way to prove this. If I could show $V\subset \mathbb{Q}\cap [0,1]$ is open for all open sets $V$, and I could show that $f^{-1}(V)$ is open as well, then that would mean $f$ is continuous. I've considered trying to prove that $(\mathbb{Q}\cap [0,1])^c$ is closed, but that doesn't seem much easier. I know $\mathbb{Q}$ is dense in $\mathbb{R}$, and so maybe I can use that to say that $B_{\epsilon}(x)\setminus\{x\}\cap(\mathbb{Q}\cap[0,1])\neq\emptyset$, which would mean every $x\in\mathbb{Q}\cap[0,1]$ is a limit point of $\mathbb{Q}\cap[0,1]$, but I still don't see how this could be helpful.

Any hints on how to proceed would be appreciated. Thanks.

orangeskid
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Sujaan Kunalan
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    I really doubt that any such function is continuous... – Dustan Levenstein Sep 17 '14 at 14:45
  • @DustanLevenstein A constant function is always continuous. So there are at least some of them. – Arthur Sep 17 '14 at 14:45
  • @Arthur lol I meant I really doubt that all such functions are continuous. Ambiguities arising in the English language. :P – Dustan Levenstein Sep 17 '14 at 14:46
  • @DustanLevenstein: You're right. I meant to say there exists a $f$ which is continuous. Thanks for the catch. – Sujaan Kunalan Sep 17 '14 at 14:49
  • The function $f(x) = 0.5$ if $x$ is algebraic, and $0.3$ otherwise seems discontinuous. The set of algebraic numbers isn't open in the irrationals. Just to have a discontinuous example. – Arthur Sep 17 '14 at 14:50
  • It's sort of weird to say "Let $f$ be blah" and then say "Prove there exists an $f$..." – Nishant Sep 17 '14 at 14:50
  • @SujaanKunalan well as Arthur mentioned, constant functions are always continuous... – Dustan Levenstein Sep 17 '14 at 14:51
  • @DustanLevenstein may be right, since $mathbb{R}\mathbb{Q}$ is uncountably infinite, whereas $mathbb{Q}$ is countably infinite. – Ryan Dougherty Sep 17 '14 at 14:51
  • A more interesting question: is there a non-constant continous $f$? – Winther Sep 17 '14 at 14:56
  • @Winther better yet, one which is not locally constant, as it's easy to concoct a function which "jumps" on the missing rational points in the domain. – Dustan Levenstein Sep 17 '14 at 15:05
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    @DustanLevenstein due to Baireness, it is always locally constant somewhere... – Henno Brandsma Sep 17 '14 at 15:08
  • I don't think it can exist. If it did and we continously extended it to the whole of $[0,1]$ then such an extension must map a countable set onto an uncountable set which is impossible. In other words if such a function exists it must have infinitely many jumps. – Winther Sep 17 '14 at 15:08
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    Why is there a downvote for this question? There is effort shown in this question. – qqo Sep 17 '14 at 15:45
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    @Nameless: I downvoted this because the question was shown as if it is quoted from a book; but was complete nonsense. Now the question has been edited and corrected, I have withdrawn my downvote and replaced it with an upvote. – Asaf Karagila Sep 17 '14 at 16:28

2 Answers2

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Hint: let $a_1=0, a_2=1/2, a_3=2/3, \ldots, a_i = 1-1/i$ and define $f$ to be constant on $(\mathbb R \setminus \mathbb Q) \cap [a_i, a_{i+1}]$.

Dustan Levenstein
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OP's space $\ I\ $ doesn't have endpoints and $\ \mathbb Q\cap[0;1]\ $ does. Thus a perfectly elegant solution is very unlikely. Nevertheless, @DustanLevenstein's solution is as simple as possible and almost elegant. But elegance has more than one dimension. Thus, I hope that my solution still has something attractive about it. It has the main step, and then another which has to overcome the nuisance caused by the endpoints issue; that second step is routine.

MAIN STEP:   every irrational number $\ x\in I\ $ admits a unique chained fraction representation $\ x\ =\ 1/(a+1/(b+ ...)...).\ $ Define $$ \phi(x)\ :\ I\,\rightarrow\,\mathbb Q_{_{>0}} $$ (where $\ Q_{_{>0}}\ :=\ \mathbb Q\cap(0;\infty))\ $ by

$$ \phi(x)\ :=\ \frac ab $$

This $\ \phi\ $ is clearly a continuous surjection.

THE NUISANCE STEP: There exists a homeomorphism $\ \psi: Q_{_{>0}}\rightarrow \mathbb Q\cap[0;1]. $ Thus a required continuous surjection is: $\,\ g\ :=\ \psi\circ\phi\ :\ I\rightarrow\mathbb Q\cap[0;1] .$

That's all.

Wlod AA
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