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Suppose that $f: [0,1]\to \mathbb{Q}$ is a function such that: for any sequence $x_1,x_2,\ldots,\in [0,1]$ such that $f(x_1)=f(x_2)=\ldots$ and $\lim x_n=x\in [0,1]$ then $f(x)=f(x_1)$. Prove that there exists a non-empty open interval $I$ of $[0,1]$ such that $f$ is constant on that interval.

So far, by the property of $f$, I have proved that for any $r\in \mathbb{Q}$ then the preimage $f^{-1}(r)$ is compact in $[0,1]$. But I don't know how to proceed from there. Any suggestion?

Naruto
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  • Do you know something about the continuity of your function? – pawel Mar 11 '21 at 15:27
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    No, there is no continuity condition on the function. Certainly, if the function is continuous, I know that it must be constant since the image is in Q. – Naruto Mar 11 '21 at 15:28

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If you look at this powerful theorem of Sierpinsky, and consider the countable, pairwise disjoint, closed cover of $[0,1]$ given by $\{f^{-1}(q)\mid q\in\mathbb{Q}\}$, you can conclude that $f$ is a constant function.

G. Ottaviano
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