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A converse of sorts to the intermediate value theorem, with an additional property

Definition of Darboux function: Let $S\subset\mathbb{R}$ be given. We say that $f:S\rightarrow\mathbb{R}$ is a Darboux function if it possesses the following property: for each $a,b\in S$, such that $a<b$, and for each $y_0\in(c,d)$ (where $c=\mathrm{min}\{f(a),f(b)\}$ and $d=\mathrm{max}\{f(a),f(b)\}$) there exists an $x_0\in S\cap (a,b)$ such that $f(x_0)=y_0$.

My problem is this: Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ is a Darboux function. Suppose also that $f^{-1}(\{q\})$ is closed for each $q\in\mathbb{Q}$. I have to prove that $f$ is continuous on $\mathbb{R}$.

Could you help me with this problem, please?

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marked as duplicate by Jonas Meyer, Pete L. Clark, t.b., J. M., Pedro Tamaroff Jul 23 '12 at 1:24

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Did you have any hint or attempt at this question? I must say it is quite fancy. =) +1 – Patrick Da Silva Nov 20 '11 at 1:17
Note: Your condition of "f is a Darboux property" can be stated more understandably as "f has the intermediate value property". – Harry Altman Nov 20 '11 at 2:20
up vote 2 down vote accepted

A proof by contraposition can be given in which, assuming the Darboux property and lack of continuity, it is shown that there exists $q$ such that $f^{-1}\{q\}$ is not closed.

For simplicity, I suggest first supposing that $f(0)=0$ and that $f$ is not continuous at $0$. (Straightforward adjustments can be made for the general case.) By definition, this means that there exists $\varepsilon>0$ such that for all $\delta>0$, there exists $x$ in $(-\delta,\delta)$ such that $|f(x)|\geq \varepsilon$. Let $q$ be a rational number such that $0<q<\varepsilon$. By the Darboux property, for all $\delta>0$ there exist $y$ in $(-\delta,\delta)$ such that $f(y)=q$ or $f(y)=-q$. You can use this fact to conclude that one of $f^{-1}\{q\}$ or $f^{-1}\{-q\}$ has $0$ as a limit point, and therefore is not closed.

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